Rank groups within a grouped sequence of TRUE/FALSE and NA





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10















I have a little nut to crack.



I have a data.frame like this:



   group criterium
1 A NA
2 A TRUE
3 A TRUE
4 A TRUE
5 A FALSE
6 A FALSE
7 A TRUE
8 A TRUE
9 A FALSE
10 A TRUE
11 A TRUE
12 A TRUE
13 B NA
14 B FALSE
15 B TRUE
16 B TRUE
17 B TRUE
18 B FALSE

structure(list(group = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L), .Label = c("A",
"B"), class = "factor"), criterium = c(NA, TRUE, TRUE, TRUE,
FALSE, FALSE, TRUE, TRUE, FALSE, TRUE, TRUE, TRUE, NA, FALSE,
TRUE, TRUE, TRUE, FALSE)), class = "data.frame", row.names = c(NA,
-18L))



And I want to rank the groups of TRUE in column criterium in ascending order while disregarding the FALSEand NA. The goal is to have a unique group identifier inside each group of group.



So the result should look like:



    group criterium goal
1 A NA NA
2 A TRUE 1
3 A TRUE 1
4 A TRUE 1
5 A FALSE NA
6 A FALSE NA
7 A TRUE 2
8 A TRUE 2
9 A FALSE NA
10 A TRUE 3
11 A TRUE 3
12 A TRUE 3
13 B NA NA
14 B FALSE NA
15 B TRUE 1
16 B TRUE 1
17 B TRUE 1
18 B FALSE NA



I'm sure there is a relatively easy way to do this, I just can't think of one. I experimented with dense_rank() and other window functions of dplyr, but to no avail.



Thanks for the help!










share|improve this question




















  • 1





    you can just about grab what you need with this work of beauty; as.numeric(as.factor(cumsum(is.na(d$criterium^NA)) + d$criterium^NA)) -- just needs to be applied by group

    – user20650
    2 days ago













  • that is a really funny solution. Very good job!

    – Humpelstielzchen
    2 days ago











  • In your example all of group A comes first, then group B. We don't need to handle cases with group=A, criterium=TRUE interspersed with group=B, criterium=TRUE?

    – smci
    2 days ago











  • No, when group A stops so stops the sequence for group A.

    – Humpelstielzchen
    2 days ago













  • But I'm suggesting if you construct an example with group=A, criterium=TRUE followed by group=B, criterium=TRUE (with no FALSE's in-between), would that get a new 'goal' number or not? Some of the answers here will fail because they don't group-by group or consider the discontinuity in group.

    – smci
    2 days ago




















10















I have a little nut to crack.



I have a data.frame like this:



   group criterium
1 A NA
2 A TRUE
3 A TRUE
4 A TRUE
5 A FALSE
6 A FALSE
7 A TRUE
8 A TRUE
9 A FALSE
10 A TRUE
11 A TRUE
12 A TRUE
13 B NA
14 B FALSE
15 B TRUE
16 B TRUE
17 B TRUE
18 B FALSE

structure(list(group = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L), .Label = c("A",
"B"), class = "factor"), criterium = c(NA, TRUE, TRUE, TRUE,
FALSE, FALSE, TRUE, TRUE, FALSE, TRUE, TRUE, TRUE, NA, FALSE,
TRUE, TRUE, TRUE, FALSE)), class = "data.frame", row.names = c(NA,
-18L))



And I want to rank the groups of TRUE in column criterium in ascending order while disregarding the FALSEand NA. The goal is to have a unique group identifier inside each group of group.



So the result should look like:



    group criterium goal
1 A NA NA
2 A TRUE 1
3 A TRUE 1
4 A TRUE 1
5 A FALSE NA
6 A FALSE NA
7 A TRUE 2
8 A TRUE 2
9 A FALSE NA
10 A TRUE 3
11 A TRUE 3
12 A TRUE 3
13 B NA NA
14 B FALSE NA
15 B TRUE 1
16 B TRUE 1
17 B TRUE 1
18 B FALSE NA



I'm sure there is a relatively easy way to do this, I just can't think of one. I experimented with dense_rank() and other window functions of dplyr, but to no avail.



Thanks for the help!










share|improve this question




















  • 1





    you can just about grab what you need with this work of beauty; as.numeric(as.factor(cumsum(is.na(d$criterium^NA)) + d$criterium^NA)) -- just needs to be applied by group

    – user20650
    2 days ago













  • that is a really funny solution. Very good job!

    – Humpelstielzchen
    2 days ago











  • In your example all of group A comes first, then group B. We don't need to handle cases with group=A, criterium=TRUE interspersed with group=B, criterium=TRUE?

    – smci
    2 days ago











  • No, when group A stops so stops the sequence for group A.

    – Humpelstielzchen
    2 days ago













  • But I'm suggesting if you construct an example with group=A, criterium=TRUE followed by group=B, criterium=TRUE (with no FALSE's in-between), would that get a new 'goal' number or not? Some of the answers here will fail because they don't group-by group or consider the discontinuity in group.

    – smci
    2 days ago
















10












10








10








I have a little nut to crack.



I have a data.frame like this:



   group criterium
1 A NA
2 A TRUE
3 A TRUE
4 A TRUE
5 A FALSE
6 A FALSE
7 A TRUE
8 A TRUE
9 A FALSE
10 A TRUE
11 A TRUE
12 A TRUE
13 B NA
14 B FALSE
15 B TRUE
16 B TRUE
17 B TRUE
18 B FALSE

structure(list(group = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L), .Label = c("A",
"B"), class = "factor"), criterium = c(NA, TRUE, TRUE, TRUE,
FALSE, FALSE, TRUE, TRUE, FALSE, TRUE, TRUE, TRUE, NA, FALSE,
TRUE, TRUE, TRUE, FALSE)), class = "data.frame", row.names = c(NA,
-18L))



And I want to rank the groups of TRUE in column criterium in ascending order while disregarding the FALSEand NA. The goal is to have a unique group identifier inside each group of group.



So the result should look like:



    group criterium goal
1 A NA NA
2 A TRUE 1
3 A TRUE 1
4 A TRUE 1
5 A FALSE NA
6 A FALSE NA
7 A TRUE 2
8 A TRUE 2
9 A FALSE NA
10 A TRUE 3
11 A TRUE 3
12 A TRUE 3
13 B NA NA
14 B FALSE NA
15 B TRUE 1
16 B TRUE 1
17 B TRUE 1
18 B FALSE NA



I'm sure there is a relatively easy way to do this, I just can't think of one. I experimented with dense_rank() and other window functions of dplyr, but to no avail.



Thanks for the help!










share|improve this question
















I have a little nut to crack.



I have a data.frame like this:



   group criterium
1 A NA
2 A TRUE
3 A TRUE
4 A TRUE
5 A FALSE
6 A FALSE
7 A TRUE
8 A TRUE
9 A FALSE
10 A TRUE
11 A TRUE
12 A TRUE
13 B NA
14 B FALSE
15 B TRUE
16 B TRUE
17 B TRUE
18 B FALSE

structure(list(group = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L), .Label = c("A",
"B"), class = "factor"), criterium = c(NA, TRUE, TRUE, TRUE,
FALSE, FALSE, TRUE, TRUE, FALSE, TRUE, TRUE, TRUE, NA, FALSE,
TRUE, TRUE, TRUE, FALSE)), class = "data.frame", row.names = c(NA,
-18L))



And I want to rank the groups of TRUE in column criterium in ascending order while disregarding the FALSEand NA. The goal is to have a unique group identifier inside each group of group.



So the result should look like:



    group criterium goal
1 A NA NA
2 A TRUE 1
3 A TRUE 1
4 A TRUE 1
5 A FALSE NA
6 A FALSE NA
7 A TRUE 2
8 A TRUE 2
9 A FALSE NA
10 A TRUE 3
11 A TRUE 3
12 A TRUE 3
13 B NA NA
14 B FALSE NA
15 B TRUE 1
16 B TRUE 1
17 B TRUE 1
18 B FALSE NA



I'm sure there is a relatively easy way to do this, I just can't think of one. I experimented with dense_rank() and other window functions of dplyr, but to no avail.



Thanks for the help!







r dplyr data.table rank






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share|improve this question








edited 2 days ago







Humpelstielzchen

















asked 2 days ago









HumpelstielzchenHumpelstielzchen

1,4251318




1,4251318








  • 1





    you can just about grab what you need with this work of beauty; as.numeric(as.factor(cumsum(is.na(d$criterium^NA)) + d$criterium^NA)) -- just needs to be applied by group

    – user20650
    2 days ago













  • that is a really funny solution. Very good job!

    – Humpelstielzchen
    2 days ago











  • In your example all of group A comes first, then group B. We don't need to handle cases with group=A, criterium=TRUE interspersed with group=B, criterium=TRUE?

    – smci
    2 days ago











  • No, when group A stops so stops the sequence for group A.

    – Humpelstielzchen
    2 days ago













  • But I'm suggesting if you construct an example with group=A, criterium=TRUE followed by group=B, criterium=TRUE (with no FALSE's in-between), would that get a new 'goal' number or not? Some of the answers here will fail because they don't group-by group or consider the discontinuity in group.

    – smci
    2 days ago
















  • 1





    you can just about grab what you need with this work of beauty; as.numeric(as.factor(cumsum(is.na(d$criterium^NA)) + d$criterium^NA)) -- just needs to be applied by group

    – user20650
    2 days ago













  • that is a really funny solution. Very good job!

    – Humpelstielzchen
    2 days ago











  • In your example all of group A comes first, then group B. We don't need to handle cases with group=A, criterium=TRUE interspersed with group=B, criterium=TRUE?

    – smci
    2 days ago











  • No, when group A stops so stops the sequence for group A.

    – Humpelstielzchen
    2 days ago













  • But I'm suggesting if you construct an example with group=A, criterium=TRUE followed by group=B, criterium=TRUE (with no FALSE's in-between), would that get a new 'goal' number or not? Some of the answers here will fail because they don't group-by group or consider the discontinuity in group.

    – smci
    2 days ago










1




1





you can just about grab what you need with this work of beauty; as.numeric(as.factor(cumsum(is.na(d$criterium^NA)) + d$criterium^NA)) -- just needs to be applied by group

– user20650
2 days ago







you can just about grab what you need with this work of beauty; as.numeric(as.factor(cumsum(is.na(d$criterium^NA)) + d$criterium^NA)) -- just needs to be applied by group

– user20650
2 days ago















that is a really funny solution. Very good job!

– Humpelstielzchen
2 days ago





that is a really funny solution. Very good job!

– Humpelstielzchen
2 days ago













In your example all of group A comes first, then group B. We don't need to handle cases with group=A, criterium=TRUE interspersed with group=B, criterium=TRUE?

– smci
2 days ago





In your example all of group A comes first, then group B. We don't need to handle cases with group=A, criterium=TRUE interspersed with group=B, criterium=TRUE?

– smci
2 days ago













No, when group A stops so stops the sequence for group A.

– Humpelstielzchen
2 days ago







No, when group A stops so stops the sequence for group A.

– Humpelstielzchen
2 days ago















But I'm suggesting if you construct an example with group=A, criterium=TRUE followed by group=B, criterium=TRUE (with no FALSE's in-between), would that get a new 'goal' number or not? Some of the answers here will fail because they don't group-by group or consider the discontinuity in group.

– smci
2 days ago







But I'm suggesting if you construct an example with group=A, criterium=TRUE followed by group=B, criterium=TRUE (with no FALSE's in-between), would that get a new 'goal' number or not? Some of the answers here will fail because they don't group-by group or consider the discontinuity in group.

– smci
2 days ago














4 Answers
4






active

oldest

votes


















7














Another data.table approach:



library(data.table)
setDT(dt)
dt[, cr := rleid(criterium)][
(criterium), goal := rleid(cr), by=.(group)]





share|improve this answer



















  • 1





    Tried with rleid but didn't get it to work. (+1)

    – markus
    2 days ago











  • works for me. And seems to be the most elegant answer.

    – Humpelstielzchen
    2 days ago



















6














Maybe I have over-complicated this but one way with dplyr is



library(dplyr)

df %>%
mutate(temp = replace(criterium, is.na(criterium), FALSE),
temp1 = cumsum(!temp)) %>%
group_by(temp1) %>%
mutate(goal = +(row_number() == which.max(temp) & any(temp))) %>%
group_by(group) %>%
mutate(goal = ifelse(temp, cumsum(goal), NA)) %>%
select(-temp, -temp1)

# group criterium goal
# <fct> <lgl> <int>
# 1 A NA NA
# 2 A TRUE 1
# 3 A TRUE 1
# 4 A TRUE 1
# 5 A FALSE NA
# 6 A FALSE NA
# 7 A TRUE 2
# 8 A TRUE 2
# 9 A FALSE NA
#10 A TRUE 3
#11 A TRUE 3
#12 A TRUE 3
#13 B NA NA
#14 B FALSE NA
#15 B TRUE 1
#16 B TRUE 1
#17 B TRUE 1
#18 B FALSE NA


We first replace NAs in criterium column to FALSE and take cumulative sum over the negation of it (temp1). We group_by temp1 and assign 1 to every first TRUE value in the group. Finally grouping by group we take a cumulative sum for TRUE values or return NA for FALSE and NA values.






share|improve this answer































    4














    A pure Base R solution, we can create a custom function via rle, and use it per group, i.e.



    f1 <- function(x) {
    x[is.na(x)] <- FALSE
    rle1 <- rle(x)
    y <- rle1$values
    rle1$values[!y] <- 0
    rle1$values[y] <- cumsum(rle1$values[y])
    return(inverse.rle(rle1))
    }


    do.call(rbind,
    lapply(split(df, df$group), function(i){i$goal <- f1(i$criterium);
    i$goal <- replace(i$goal, is.na(i$criterium)|!i$criterium, NA);
    i}))


    Of course, If you want you can apply it via dplyr, i.e.



    library(dplyr)

    df %>%
    group_by(group) %>%
    mutate(goal = f1(criterium),
    goal = replace(goal, is.na(criterium)|!criterium, NA))


    which gives,




    # A tibble: 18 x 3
    # Groups: group [2]
    group criterium goal
    <fct> <lgl> <dbl>
    1 A NA NA
    2 A TRUE 1
    3 A TRUE 1
    4 A TRUE 1
    5 A FALSE NA
    6 A FALSE NA
    7 A TRUE 2
    8 A TRUE 2
    9 A FALSE NA
    10 A TRUE 3
    11 A TRUE 3
    12 A TRUE 3
    13 B NA NA
    14 B FALSE NA
    15 B TRUE 1
    16 B TRUE 1
    17 B TRUE 1
    18 B FALSE NA






    share|improve this answer

































      4














      A data.table option using rle



      library(data.table)
      DT <- as.data.table(dat)
      DT[, goal := {
      r <- rle(replace(criterium, is.na(criterium), FALSE))
      r$values <- with(r, cumsum(values) * values)
      out <- inverse.rle(r)
      replace(out, out == 0, NA)
      }, by = group]
      DT
      # group criterium goal
      # 1: A NA NA
      # 2: A TRUE 1
      # 3: A TRUE 1
      # 4: A TRUE 1
      # 5: A FALSE NA
      # 6: A FALSE NA
      # 7: A TRUE 2
      # 8: A TRUE 2
      # 9: A FALSE NA
      #10: A TRUE 3
      #11: A TRUE 3
      #12: A TRUE 3
      #13: B NA NA
      #14: B FALSE NA
      #15: B TRUE 1
      #16: B TRUE 1
      #17: B TRUE 1
      #18: B FALSE NA


      step by step



      When we call r <- rle(replace(criterium, is.na(criterium), FALSE)) we get an object of class rle



      r
      #Run Length Encoding
      # lengths: int [1:9] 1 3 2 2 1 3 2 3 1
      # values : logi [1:9] FALSE TRUE FALSE TRUE FALSE TRUE ...


      We manipulate the values compenent in the following way



      r$values <- with(r, cumsum(values) * values)
      r
      #Run Length Encoding
      # lengths: int [1:9] 1 3 2 2 1 3 2 3 1
      # values : int [1:9] 0 1 0 2 0 3 0 4 0


      That is, we replaced TRUEs with the cumulative sum of values and set the FALSEs to 0. Now inverse.rle returns a vector in which values will repeated lenghts times



      out <- inverse.rle(r)
      out
      # [1] 0 1 1 1 0 0 2 2 0 3 3 3 0 0 4 4 4 0


      This is almost what OP wants but we need to replace the 0s with NA



      replace(out, out == 0, NA)


      This is done for each group.



      data



      dat <- structure(list(group = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 
      1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L), .Label = c("A",
      "B"), class = "factor"), criterium = c(NA, TRUE, TRUE, TRUE,
      FALSE, FALSE, TRUE, TRUE, FALSE, TRUE, TRUE, TRUE, NA, FALSE,
      TRUE, TRUE, TRUE, FALSE)), class = "data.frame", row.names = c(NA,
      -18L))





      share|improve this answer


























      • Wow, impressive. Thanks for introducing me to rleand inverse.rle. Gruß nach Leipzig.

        – Humpelstielzchen
        2 days ago






      • 1





        @Humpelstielzchen Gern geschehen. Will try to simplify and explain the logic a bit.

        – markus
        2 days ago













      • Thanks! I was dissecting your answer just like that. Your answer taught me the most. But chinsoon12 is just a Teufelskerl. ^^

        – Humpelstielzchen
        2 days ago












      Your Answer






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      4 Answers
      4






      active

      oldest

      votes








      4 Answers
      4






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      7














      Another data.table approach:



      library(data.table)
      setDT(dt)
      dt[, cr := rleid(criterium)][
      (criterium), goal := rleid(cr), by=.(group)]





      share|improve this answer



















      • 1





        Tried with rleid but didn't get it to work. (+1)

        – markus
        2 days ago











      • works for me. And seems to be the most elegant answer.

        – Humpelstielzchen
        2 days ago
















      7














      Another data.table approach:



      library(data.table)
      setDT(dt)
      dt[, cr := rleid(criterium)][
      (criterium), goal := rleid(cr), by=.(group)]





      share|improve this answer



















      • 1





        Tried with rleid but didn't get it to work. (+1)

        – markus
        2 days ago











      • works for me. And seems to be the most elegant answer.

        – Humpelstielzchen
        2 days ago














      7












      7








      7







      Another data.table approach:



      library(data.table)
      setDT(dt)
      dt[, cr := rleid(criterium)][
      (criterium), goal := rleid(cr), by=.(group)]





      share|improve this answer













      Another data.table approach:



      library(data.table)
      setDT(dt)
      dt[, cr := rleid(criterium)][
      (criterium), goal := rleid(cr), by=.(group)]






      share|improve this answer












      share|improve this answer



      share|improve this answer










      answered 2 days ago









      chinsoon12chinsoon12

      9,93611420




      9,93611420








      • 1





        Tried with rleid but didn't get it to work. (+1)

        – markus
        2 days ago











      • works for me. And seems to be the most elegant answer.

        – Humpelstielzchen
        2 days ago














      • 1





        Tried with rleid but didn't get it to work. (+1)

        – markus
        2 days ago











      • works for me. And seems to be the most elegant answer.

        – Humpelstielzchen
        2 days ago








      1




      1





      Tried with rleid but didn't get it to work. (+1)

      – markus
      2 days ago





      Tried with rleid but didn't get it to work. (+1)

      – markus
      2 days ago













      works for me. And seems to be the most elegant answer.

      – Humpelstielzchen
      2 days ago





      works for me. And seems to be the most elegant answer.

      – Humpelstielzchen
      2 days ago













      6














      Maybe I have over-complicated this but one way with dplyr is



      library(dplyr)

      df %>%
      mutate(temp = replace(criterium, is.na(criterium), FALSE),
      temp1 = cumsum(!temp)) %>%
      group_by(temp1) %>%
      mutate(goal = +(row_number() == which.max(temp) & any(temp))) %>%
      group_by(group) %>%
      mutate(goal = ifelse(temp, cumsum(goal), NA)) %>%
      select(-temp, -temp1)

      # group criterium goal
      # <fct> <lgl> <int>
      # 1 A NA NA
      # 2 A TRUE 1
      # 3 A TRUE 1
      # 4 A TRUE 1
      # 5 A FALSE NA
      # 6 A FALSE NA
      # 7 A TRUE 2
      # 8 A TRUE 2
      # 9 A FALSE NA
      #10 A TRUE 3
      #11 A TRUE 3
      #12 A TRUE 3
      #13 B NA NA
      #14 B FALSE NA
      #15 B TRUE 1
      #16 B TRUE 1
      #17 B TRUE 1
      #18 B FALSE NA


      We first replace NAs in criterium column to FALSE and take cumulative sum over the negation of it (temp1). We group_by temp1 and assign 1 to every first TRUE value in the group. Finally grouping by group we take a cumulative sum for TRUE values or return NA for FALSE and NA values.






      share|improve this answer




























        6














        Maybe I have over-complicated this but one way with dplyr is



        library(dplyr)

        df %>%
        mutate(temp = replace(criterium, is.na(criterium), FALSE),
        temp1 = cumsum(!temp)) %>%
        group_by(temp1) %>%
        mutate(goal = +(row_number() == which.max(temp) & any(temp))) %>%
        group_by(group) %>%
        mutate(goal = ifelse(temp, cumsum(goal), NA)) %>%
        select(-temp, -temp1)

        # group criterium goal
        # <fct> <lgl> <int>
        # 1 A NA NA
        # 2 A TRUE 1
        # 3 A TRUE 1
        # 4 A TRUE 1
        # 5 A FALSE NA
        # 6 A FALSE NA
        # 7 A TRUE 2
        # 8 A TRUE 2
        # 9 A FALSE NA
        #10 A TRUE 3
        #11 A TRUE 3
        #12 A TRUE 3
        #13 B NA NA
        #14 B FALSE NA
        #15 B TRUE 1
        #16 B TRUE 1
        #17 B TRUE 1
        #18 B FALSE NA


        We first replace NAs in criterium column to FALSE and take cumulative sum over the negation of it (temp1). We group_by temp1 and assign 1 to every first TRUE value in the group. Finally grouping by group we take a cumulative sum for TRUE values or return NA for FALSE and NA values.






        share|improve this answer


























          6












          6








          6







          Maybe I have over-complicated this but one way with dplyr is



          library(dplyr)

          df %>%
          mutate(temp = replace(criterium, is.na(criterium), FALSE),
          temp1 = cumsum(!temp)) %>%
          group_by(temp1) %>%
          mutate(goal = +(row_number() == which.max(temp) & any(temp))) %>%
          group_by(group) %>%
          mutate(goal = ifelse(temp, cumsum(goal), NA)) %>%
          select(-temp, -temp1)

          # group criterium goal
          # <fct> <lgl> <int>
          # 1 A NA NA
          # 2 A TRUE 1
          # 3 A TRUE 1
          # 4 A TRUE 1
          # 5 A FALSE NA
          # 6 A FALSE NA
          # 7 A TRUE 2
          # 8 A TRUE 2
          # 9 A FALSE NA
          #10 A TRUE 3
          #11 A TRUE 3
          #12 A TRUE 3
          #13 B NA NA
          #14 B FALSE NA
          #15 B TRUE 1
          #16 B TRUE 1
          #17 B TRUE 1
          #18 B FALSE NA


          We first replace NAs in criterium column to FALSE and take cumulative sum over the negation of it (temp1). We group_by temp1 and assign 1 to every first TRUE value in the group. Finally grouping by group we take a cumulative sum for TRUE values or return NA for FALSE and NA values.






          share|improve this answer













          Maybe I have over-complicated this but one way with dplyr is



          library(dplyr)

          df %>%
          mutate(temp = replace(criterium, is.na(criterium), FALSE),
          temp1 = cumsum(!temp)) %>%
          group_by(temp1) %>%
          mutate(goal = +(row_number() == which.max(temp) & any(temp))) %>%
          group_by(group) %>%
          mutate(goal = ifelse(temp, cumsum(goal), NA)) %>%
          select(-temp, -temp1)

          # group criterium goal
          # <fct> <lgl> <int>
          # 1 A NA NA
          # 2 A TRUE 1
          # 3 A TRUE 1
          # 4 A TRUE 1
          # 5 A FALSE NA
          # 6 A FALSE NA
          # 7 A TRUE 2
          # 8 A TRUE 2
          # 9 A FALSE NA
          #10 A TRUE 3
          #11 A TRUE 3
          #12 A TRUE 3
          #13 B NA NA
          #14 B FALSE NA
          #15 B TRUE 1
          #16 B TRUE 1
          #17 B TRUE 1
          #18 B FALSE NA


          We first replace NAs in criterium column to FALSE and take cumulative sum over the negation of it (temp1). We group_by temp1 and assign 1 to every first TRUE value in the group. Finally grouping by group we take a cumulative sum for TRUE values or return NA for FALSE and NA values.







          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered 2 days ago









          Ronak ShahRonak Shah

          46.7k104269




          46.7k104269























              4














              A pure Base R solution, we can create a custom function via rle, and use it per group, i.e.



              f1 <- function(x) {
              x[is.na(x)] <- FALSE
              rle1 <- rle(x)
              y <- rle1$values
              rle1$values[!y] <- 0
              rle1$values[y] <- cumsum(rle1$values[y])
              return(inverse.rle(rle1))
              }


              do.call(rbind,
              lapply(split(df, df$group), function(i){i$goal <- f1(i$criterium);
              i$goal <- replace(i$goal, is.na(i$criterium)|!i$criterium, NA);
              i}))


              Of course, If you want you can apply it via dplyr, i.e.



              library(dplyr)

              df %>%
              group_by(group) %>%
              mutate(goal = f1(criterium),
              goal = replace(goal, is.na(criterium)|!criterium, NA))


              which gives,




              # A tibble: 18 x 3
              # Groups: group [2]
              group criterium goal
              <fct> <lgl> <dbl>
              1 A NA NA
              2 A TRUE 1
              3 A TRUE 1
              4 A TRUE 1
              5 A FALSE NA
              6 A FALSE NA
              7 A TRUE 2
              8 A TRUE 2
              9 A FALSE NA
              10 A TRUE 3
              11 A TRUE 3
              12 A TRUE 3
              13 B NA NA
              14 B FALSE NA
              15 B TRUE 1
              16 B TRUE 1
              17 B TRUE 1
              18 B FALSE NA






              share|improve this answer






























                4














                A pure Base R solution, we can create a custom function via rle, and use it per group, i.e.



                f1 <- function(x) {
                x[is.na(x)] <- FALSE
                rle1 <- rle(x)
                y <- rle1$values
                rle1$values[!y] <- 0
                rle1$values[y] <- cumsum(rle1$values[y])
                return(inverse.rle(rle1))
                }


                do.call(rbind,
                lapply(split(df, df$group), function(i){i$goal <- f1(i$criterium);
                i$goal <- replace(i$goal, is.na(i$criterium)|!i$criterium, NA);
                i}))


                Of course, If you want you can apply it via dplyr, i.e.



                library(dplyr)

                df %>%
                group_by(group) %>%
                mutate(goal = f1(criterium),
                goal = replace(goal, is.na(criterium)|!criterium, NA))


                which gives,




                # A tibble: 18 x 3
                # Groups: group [2]
                group criterium goal
                <fct> <lgl> <dbl>
                1 A NA NA
                2 A TRUE 1
                3 A TRUE 1
                4 A TRUE 1
                5 A FALSE NA
                6 A FALSE NA
                7 A TRUE 2
                8 A TRUE 2
                9 A FALSE NA
                10 A TRUE 3
                11 A TRUE 3
                12 A TRUE 3
                13 B NA NA
                14 B FALSE NA
                15 B TRUE 1
                16 B TRUE 1
                17 B TRUE 1
                18 B FALSE NA






                share|improve this answer




























                  4












                  4








                  4







                  A pure Base R solution, we can create a custom function via rle, and use it per group, i.e.



                  f1 <- function(x) {
                  x[is.na(x)] <- FALSE
                  rle1 <- rle(x)
                  y <- rle1$values
                  rle1$values[!y] <- 0
                  rle1$values[y] <- cumsum(rle1$values[y])
                  return(inverse.rle(rle1))
                  }


                  do.call(rbind,
                  lapply(split(df, df$group), function(i){i$goal <- f1(i$criterium);
                  i$goal <- replace(i$goal, is.na(i$criterium)|!i$criterium, NA);
                  i}))


                  Of course, If you want you can apply it via dplyr, i.e.



                  library(dplyr)

                  df %>%
                  group_by(group) %>%
                  mutate(goal = f1(criterium),
                  goal = replace(goal, is.na(criterium)|!criterium, NA))


                  which gives,




                  # A tibble: 18 x 3
                  # Groups: group [2]
                  group criterium goal
                  <fct> <lgl> <dbl>
                  1 A NA NA
                  2 A TRUE 1
                  3 A TRUE 1
                  4 A TRUE 1
                  5 A FALSE NA
                  6 A FALSE NA
                  7 A TRUE 2
                  8 A TRUE 2
                  9 A FALSE NA
                  10 A TRUE 3
                  11 A TRUE 3
                  12 A TRUE 3
                  13 B NA NA
                  14 B FALSE NA
                  15 B TRUE 1
                  16 B TRUE 1
                  17 B TRUE 1
                  18 B FALSE NA






                  share|improve this answer















                  A pure Base R solution, we can create a custom function via rle, and use it per group, i.e.



                  f1 <- function(x) {
                  x[is.na(x)] <- FALSE
                  rle1 <- rle(x)
                  y <- rle1$values
                  rle1$values[!y] <- 0
                  rle1$values[y] <- cumsum(rle1$values[y])
                  return(inverse.rle(rle1))
                  }


                  do.call(rbind,
                  lapply(split(df, df$group), function(i){i$goal <- f1(i$criterium);
                  i$goal <- replace(i$goal, is.na(i$criterium)|!i$criterium, NA);
                  i}))


                  Of course, If you want you can apply it via dplyr, i.e.



                  library(dplyr)

                  df %>%
                  group_by(group) %>%
                  mutate(goal = f1(criterium),
                  goal = replace(goal, is.na(criterium)|!criterium, NA))


                  which gives,




                  # A tibble: 18 x 3
                  # Groups: group [2]
                  group criterium goal
                  <fct> <lgl> <dbl>
                  1 A NA NA
                  2 A TRUE 1
                  3 A TRUE 1
                  4 A TRUE 1
                  5 A FALSE NA
                  6 A FALSE NA
                  7 A TRUE 2
                  8 A TRUE 2
                  9 A FALSE NA
                  10 A TRUE 3
                  11 A TRUE 3
                  12 A TRUE 3
                  13 B NA NA
                  14 B FALSE NA
                  15 B TRUE 1
                  16 B TRUE 1
                  17 B TRUE 1
                  18 B FALSE NA







                  share|improve this answer














                  share|improve this answer



                  share|improve this answer








                  edited 2 days ago

























                  answered 2 days ago









                  SotosSotos

                  31.5k51741




                  31.5k51741























                      4














                      A data.table option using rle



                      library(data.table)
                      DT <- as.data.table(dat)
                      DT[, goal := {
                      r <- rle(replace(criterium, is.na(criterium), FALSE))
                      r$values <- with(r, cumsum(values) * values)
                      out <- inverse.rle(r)
                      replace(out, out == 0, NA)
                      }, by = group]
                      DT
                      # group criterium goal
                      # 1: A NA NA
                      # 2: A TRUE 1
                      # 3: A TRUE 1
                      # 4: A TRUE 1
                      # 5: A FALSE NA
                      # 6: A FALSE NA
                      # 7: A TRUE 2
                      # 8: A TRUE 2
                      # 9: A FALSE NA
                      #10: A TRUE 3
                      #11: A TRUE 3
                      #12: A TRUE 3
                      #13: B NA NA
                      #14: B FALSE NA
                      #15: B TRUE 1
                      #16: B TRUE 1
                      #17: B TRUE 1
                      #18: B FALSE NA


                      step by step



                      When we call r <- rle(replace(criterium, is.na(criterium), FALSE)) we get an object of class rle



                      r
                      #Run Length Encoding
                      # lengths: int [1:9] 1 3 2 2 1 3 2 3 1
                      # values : logi [1:9] FALSE TRUE FALSE TRUE FALSE TRUE ...


                      We manipulate the values compenent in the following way



                      r$values <- with(r, cumsum(values) * values)
                      r
                      #Run Length Encoding
                      # lengths: int [1:9] 1 3 2 2 1 3 2 3 1
                      # values : int [1:9] 0 1 0 2 0 3 0 4 0


                      That is, we replaced TRUEs with the cumulative sum of values and set the FALSEs to 0. Now inverse.rle returns a vector in which values will repeated lenghts times



                      out <- inverse.rle(r)
                      out
                      # [1] 0 1 1 1 0 0 2 2 0 3 3 3 0 0 4 4 4 0


                      This is almost what OP wants but we need to replace the 0s with NA



                      replace(out, out == 0, NA)


                      This is done for each group.



                      data



                      dat <- structure(list(group = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 
                      1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L), .Label = c("A",
                      "B"), class = "factor"), criterium = c(NA, TRUE, TRUE, TRUE,
                      FALSE, FALSE, TRUE, TRUE, FALSE, TRUE, TRUE, TRUE, NA, FALSE,
                      TRUE, TRUE, TRUE, FALSE)), class = "data.frame", row.names = c(NA,
                      -18L))





                      share|improve this answer


























                      • Wow, impressive. Thanks for introducing me to rleand inverse.rle. Gruß nach Leipzig.

                        – Humpelstielzchen
                        2 days ago






                      • 1





                        @Humpelstielzchen Gern geschehen. Will try to simplify and explain the logic a bit.

                        – markus
                        2 days ago













                      • Thanks! I was dissecting your answer just like that. Your answer taught me the most. But chinsoon12 is just a Teufelskerl. ^^

                        – Humpelstielzchen
                        2 days ago
















                      4














                      A data.table option using rle



                      library(data.table)
                      DT <- as.data.table(dat)
                      DT[, goal := {
                      r <- rle(replace(criterium, is.na(criterium), FALSE))
                      r$values <- with(r, cumsum(values) * values)
                      out <- inverse.rle(r)
                      replace(out, out == 0, NA)
                      }, by = group]
                      DT
                      # group criterium goal
                      # 1: A NA NA
                      # 2: A TRUE 1
                      # 3: A TRUE 1
                      # 4: A TRUE 1
                      # 5: A FALSE NA
                      # 6: A FALSE NA
                      # 7: A TRUE 2
                      # 8: A TRUE 2
                      # 9: A FALSE NA
                      #10: A TRUE 3
                      #11: A TRUE 3
                      #12: A TRUE 3
                      #13: B NA NA
                      #14: B FALSE NA
                      #15: B TRUE 1
                      #16: B TRUE 1
                      #17: B TRUE 1
                      #18: B FALSE NA


                      step by step



                      When we call r <- rle(replace(criterium, is.na(criterium), FALSE)) we get an object of class rle



                      r
                      #Run Length Encoding
                      # lengths: int [1:9] 1 3 2 2 1 3 2 3 1
                      # values : logi [1:9] FALSE TRUE FALSE TRUE FALSE TRUE ...


                      We manipulate the values compenent in the following way



                      r$values <- with(r, cumsum(values) * values)
                      r
                      #Run Length Encoding
                      # lengths: int [1:9] 1 3 2 2 1 3 2 3 1
                      # values : int [1:9] 0 1 0 2 0 3 0 4 0


                      That is, we replaced TRUEs with the cumulative sum of values and set the FALSEs to 0. Now inverse.rle returns a vector in which values will repeated lenghts times



                      out <- inverse.rle(r)
                      out
                      # [1] 0 1 1 1 0 0 2 2 0 3 3 3 0 0 4 4 4 0


                      This is almost what OP wants but we need to replace the 0s with NA



                      replace(out, out == 0, NA)


                      This is done for each group.



                      data



                      dat <- structure(list(group = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 
                      1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L), .Label = c("A",
                      "B"), class = "factor"), criterium = c(NA, TRUE, TRUE, TRUE,
                      FALSE, FALSE, TRUE, TRUE, FALSE, TRUE, TRUE, TRUE, NA, FALSE,
                      TRUE, TRUE, TRUE, FALSE)), class = "data.frame", row.names = c(NA,
                      -18L))





                      share|improve this answer


























                      • Wow, impressive. Thanks for introducing me to rleand inverse.rle. Gruß nach Leipzig.

                        – Humpelstielzchen
                        2 days ago






                      • 1





                        @Humpelstielzchen Gern geschehen. Will try to simplify and explain the logic a bit.

                        – markus
                        2 days ago













                      • Thanks! I was dissecting your answer just like that. Your answer taught me the most. But chinsoon12 is just a Teufelskerl. ^^

                        – Humpelstielzchen
                        2 days ago














                      4












                      4








                      4







                      A data.table option using rle



                      library(data.table)
                      DT <- as.data.table(dat)
                      DT[, goal := {
                      r <- rle(replace(criterium, is.na(criterium), FALSE))
                      r$values <- with(r, cumsum(values) * values)
                      out <- inverse.rle(r)
                      replace(out, out == 0, NA)
                      }, by = group]
                      DT
                      # group criterium goal
                      # 1: A NA NA
                      # 2: A TRUE 1
                      # 3: A TRUE 1
                      # 4: A TRUE 1
                      # 5: A FALSE NA
                      # 6: A FALSE NA
                      # 7: A TRUE 2
                      # 8: A TRUE 2
                      # 9: A FALSE NA
                      #10: A TRUE 3
                      #11: A TRUE 3
                      #12: A TRUE 3
                      #13: B NA NA
                      #14: B FALSE NA
                      #15: B TRUE 1
                      #16: B TRUE 1
                      #17: B TRUE 1
                      #18: B FALSE NA


                      step by step



                      When we call r <- rle(replace(criterium, is.na(criterium), FALSE)) we get an object of class rle



                      r
                      #Run Length Encoding
                      # lengths: int [1:9] 1 3 2 2 1 3 2 3 1
                      # values : logi [1:9] FALSE TRUE FALSE TRUE FALSE TRUE ...


                      We manipulate the values compenent in the following way



                      r$values <- with(r, cumsum(values) * values)
                      r
                      #Run Length Encoding
                      # lengths: int [1:9] 1 3 2 2 1 3 2 3 1
                      # values : int [1:9] 0 1 0 2 0 3 0 4 0


                      That is, we replaced TRUEs with the cumulative sum of values and set the FALSEs to 0. Now inverse.rle returns a vector in which values will repeated lenghts times



                      out <- inverse.rle(r)
                      out
                      # [1] 0 1 1 1 0 0 2 2 0 3 3 3 0 0 4 4 4 0


                      This is almost what OP wants but we need to replace the 0s with NA



                      replace(out, out == 0, NA)


                      This is done for each group.



                      data



                      dat <- structure(list(group = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 
                      1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L), .Label = c("A",
                      "B"), class = "factor"), criterium = c(NA, TRUE, TRUE, TRUE,
                      FALSE, FALSE, TRUE, TRUE, FALSE, TRUE, TRUE, TRUE, NA, FALSE,
                      TRUE, TRUE, TRUE, FALSE)), class = "data.frame", row.names = c(NA,
                      -18L))





                      share|improve this answer















                      A data.table option using rle



                      library(data.table)
                      DT <- as.data.table(dat)
                      DT[, goal := {
                      r <- rle(replace(criterium, is.na(criterium), FALSE))
                      r$values <- with(r, cumsum(values) * values)
                      out <- inverse.rle(r)
                      replace(out, out == 0, NA)
                      }, by = group]
                      DT
                      # group criterium goal
                      # 1: A NA NA
                      # 2: A TRUE 1
                      # 3: A TRUE 1
                      # 4: A TRUE 1
                      # 5: A FALSE NA
                      # 6: A FALSE NA
                      # 7: A TRUE 2
                      # 8: A TRUE 2
                      # 9: A FALSE NA
                      #10: A TRUE 3
                      #11: A TRUE 3
                      #12: A TRUE 3
                      #13: B NA NA
                      #14: B FALSE NA
                      #15: B TRUE 1
                      #16: B TRUE 1
                      #17: B TRUE 1
                      #18: B FALSE NA


                      step by step



                      When we call r <- rle(replace(criterium, is.na(criterium), FALSE)) we get an object of class rle



                      r
                      #Run Length Encoding
                      # lengths: int [1:9] 1 3 2 2 1 3 2 3 1
                      # values : logi [1:9] FALSE TRUE FALSE TRUE FALSE TRUE ...


                      We manipulate the values compenent in the following way



                      r$values <- with(r, cumsum(values) * values)
                      r
                      #Run Length Encoding
                      # lengths: int [1:9] 1 3 2 2 1 3 2 3 1
                      # values : int [1:9] 0 1 0 2 0 3 0 4 0


                      That is, we replaced TRUEs with the cumulative sum of values and set the FALSEs to 0. Now inverse.rle returns a vector in which values will repeated lenghts times



                      out <- inverse.rle(r)
                      out
                      # [1] 0 1 1 1 0 0 2 2 0 3 3 3 0 0 4 4 4 0


                      This is almost what OP wants but we need to replace the 0s with NA



                      replace(out, out == 0, NA)


                      This is done for each group.



                      data



                      dat <- structure(list(group = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 
                      1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L), .Label = c("A",
                      "B"), class = "factor"), criterium = c(NA, TRUE, TRUE, TRUE,
                      FALSE, FALSE, TRUE, TRUE, FALSE, TRUE, TRUE, TRUE, NA, FALSE,
                      TRUE, TRUE, TRUE, FALSE)), class = "data.frame", row.names = c(NA,
                      -18L))






                      share|improve this answer














                      share|improve this answer



                      share|improve this answer








                      edited 2 days ago

























                      answered 2 days ago









                      markusmarkus

                      15.5k11336




                      15.5k11336













                      • Wow, impressive. Thanks for introducing me to rleand inverse.rle. Gruß nach Leipzig.

                        – Humpelstielzchen
                        2 days ago






                      • 1





                        @Humpelstielzchen Gern geschehen. Will try to simplify and explain the logic a bit.

                        – markus
                        2 days ago













                      • Thanks! I was dissecting your answer just like that. Your answer taught me the most. But chinsoon12 is just a Teufelskerl. ^^

                        – Humpelstielzchen
                        2 days ago



















                      • Wow, impressive. Thanks for introducing me to rleand inverse.rle. Gruß nach Leipzig.

                        – Humpelstielzchen
                        2 days ago






                      • 1





                        @Humpelstielzchen Gern geschehen. Will try to simplify and explain the logic a bit.

                        – markus
                        2 days ago













                      • Thanks! I was dissecting your answer just like that. Your answer taught me the most. But chinsoon12 is just a Teufelskerl. ^^

                        – Humpelstielzchen
                        2 days ago

















                      Wow, impressive. Thanks for introducing me to rleand inverse.rle. Gruß nach Leipzig.

                      – Humpelstielzchen
                      2 days ago





                      Wow, impressive. Thanks for introducing me to rleand inverse.rle. Gruß nach Leipzig.

                      – Humpelstielzchen
                      2 days ago




                      1




                      1





                      @Humpelstielzchen Gern geschehen. Will try to simplify and explain the logic a bit.

                      – markus
                      2 days ago







                      @Humpelstielzchen Gern geschehen. Will try to simplify and explain the logic a bit.

                      – markus
                      2 days ago















                      Thanks! I was dissecting your answer just like that. Your answer taught me the most. But chinsoon12 is just a Teufelskerl. ^^

                      – Humpelstielzchen
                      2 days ago





                      Thanks! I was dissecting your answer just like that. Your answer taught me the most. But chinsoon12 is just a Teufelskerl. ^^

                      – Humpelstielzchen
                      2 days ago


















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