Pandas dataframe: Remove secondary upcoming same value
up vote
9
down vote
favorite
1
I have a dataframe:
col1 col2
a 0
b 1
c 1
d 0
c 1
d 0
On 'col2' I want to keep only the first 1 from the top and replace every 1 below the first one with a 0, such that the output is:
col1 col2
a 0
b 1
c 0
d 0
c 0
d 0
Thank you very much.
python pandas dataframe
edited Dec 6 at 15:46
timgeb
47.7k116288
asked Dec 6 at 15:33
s900n
425616
add a comment |
up vote
9
down vote
favorite
1
I have a dataframe:
col1 col2
a 0
b 1
c 1
d 0
c 1
d 0
On 'col2' I want to keep only the first 1 from the top and replace every 1 below the first one with a 0, such that the output is:
col1 col2
a 0
b 1
c 0
d 0
c 0
d 0
Thank you very much.
python pandas dataframe
edited Dec 6 at 15:46
timgeb
47.7k116288
asked Dec 6 at 15:33
s900n
425616
add a comment |
up vote
9
down vote
favorite
1
up vote
9
down vote
favorite
1
1
I have a dataframe:
col1 col2
a 0
b 1
c 1
d 0
c 1
d 0
On 'col2' I want to keep only the first 1 from the top and replace every 1 below the first one with a 0, such that the output is:
col1 col2
a 0
b 1
c 0
d 0
c 0
d 0
Thank you very much.
python pandas dataframe
edited Dec 6 at 15:46
timgeb
47.7k116288
asked Dec 6 at 15:33
s900n
425616
I have a dataframe:
col1 col2
a 0
b 1
c 1
d 0
c 1
d 0
On 'col2' I want to keep only the first 1 from the top and replace every 1 below the first one with a 0, such that the output is:
col1 col2
a 0
b 1
c 0
d 0
c 0
d 0
Thank you very much.
python pandas dataframe
python pandas dataframe
edited Dec 6 at 15:46
timgeb
47.7k116288
asked Dec 6 at 15:33
s900n
425616
edited Dec 6 at 15:46
timgeb
47.7k116288
asked Dec 6 at 15:33
s900n
425616
edited Dec 6 at 15:46
timgeb
47.7k116288
edited Dec 6 at 15:46
timgeb
47.7k116288
edited Dec 6 at 15:46
timgeb
47.7k116288
47.7k116288
asked Dec 6 at 15:33
s900n
425616
asked Dec 6 at 15:33
s900n
425616
asked Dec 6 at 15:33
s900n
425616
425616
add a comment |
add a comment |
8 Answers
8
active
oldest
votes
up vote
8
down vote
accepted
You can find the index of the first 1 and set others to 0:
mask = df['col2'].eq(1)
df.loc[mask & (df.index != mask.idxmax()), 'col2'] = 0
For better performance, see Efficiently return the index of the first value satisfying condition in array.
answered Dec 6 at 15:37
jpp
88k195099
Can you think of a good solution for the case when the index is arbitrary, likeIndex(['u', 'v', 'w', 'x', 'y', 'z']AND col2 could be something like[2, 0, 0, 1, 3, 1]?
– timgeb
Dec 6 at 16:18
@timgeb, To adapt this solution, I think you can use positional indexing (instead of index labels). Something likedf.loc[mask & (np.arange(df.shape[0]) != np.where(mask)[0][0]), 'col2'] = 0. But I'm sure there are more Pythonic ways.
– jpp
Dec 6 at 16:28
Ah, I thought of using numpy, too. Just a bit differently. See my case 3. ;)
– timgeb
Dec 6 at 16:30
add a comment |
up vote
4
down vote
np.flatnonzero
Because I thought we needed more answers
df.loc[df.index[np.flatnonzero(df.col2)[1:]], 'col2'] -= 1
df
col1 col2
0 a 0
1 b 1
2 c 0
3 d 0
4 c 0
5 d 0
Same thing but a little more sneaky.
df.col2.values[np.flatnonzero(df.col2.values)[1:]] -= 1
df
col1 col2
0 a 0
1 b 1
2 c 0
3 d 0
4 c 0
5 d 0
answered Dec 6 at 15:51
piRSquared
151k22138282
add a comment |
up vote
4
down vote
Case 1: df has only ones and zeros in col2 and integer indexes.
>>> df
col1 col2
0 a 0
1 b 1
2 c 1
3 d 0
4 c 1
5 d 0
You can use:
>>> df.loc[df['col2'].idxmax() + 1:, 'col2'] = 0
>>> df
col1 col2
0 a 0
1 b 1
2 c 0
3 d 0
4 c 0
5 d 0
Case2: df can have all kinds of values in col2 and has integer indexes.
>>> df # demo dataframe
col1 col2
0 a 0
1 b 1
2 c 2
3 d 2
4 c 3
5 d 3
You can use:
>>> df.loc[(df['col2'] == 1).idxmax() + 1:, 'col2'] = 0
>>> df
col1 col2
0 a 0
1 b 1
2 c 0
3 d 0
4 c 0
5 d 0
Case 3: df can have all kinds of values in col2 and has an arbitrary index.
>>> df
col1 col2
u a -1
v b 1
w c 2
x d 2
y c 3
z d 3
You can use:
>>> df['col2'].iloc[(df['col2'].values == 1).argmax() + 1:] = 0
>>> df
col1 col2
u a -1
v b 1
w c 0
x d 0
y c 0
z d 0
answered Dec 6 at 15:38
timgeb
47.7k116288
add a comment |
up vote
3
down vote
Using drop_duplicates with reindex
df.col2=df.col2.drop_duplicates().reindex(df.index,fill_value=0)
df
Out[1078]:
col1 col2
0 a 0
1 b 1
2 c 0
3 d 0
4 c 0
5 d 0
answered Dec 6 at 15:41
W-B
97.6k73162
add a comment |
up vote
3
down vote
You can use numpy for an effficient solution:
a = df.col2.values
b = np.zeros_like(a)
b[a.argmax()] = 1
df.assign(col2=b)
col1 col2
0 a 0
1 b 1
2 c 0
3 d 0
4 c 0
5 d 0
answered Dec 6 at 15:39
user3483203
29.9k82354
add a comment |
up vote
1
down vote
i like this too
data['col2'][np.where(data['col2'] == 1)[0][0]+1:] = 0
answered Dec 6 at 15:42
iamklaus
78648
Chained indexing is not recommended.
– jpp
Dec 6 at 16:41
Thanks for the update..
– iamklaus
Dec 7 at 8:43
add a comment |
up vote
1
down vote
Sooo many options, here's mine... almost the same as timgebs answer (found independently), but still different ;)
Find the index of col2 that has the first occurence of a 1, and change all row values after that index to 0:
df['col2'].iloc[df.col2.idxmax()+1:] = 0
answered Dec 6 at 15:55
Sander van den Oord
551419
Be careful, this sets all values to0after the specified index, not just the ones equal to1. Though that's the same with some other answers too.
– jpp
Dec 6 at 16:42
Totally agree. Your solution is more general.
– Sander van den Oord
Dec 6 at 17:42
add a comment |
up vote
0
down vote
id = list(df["col2"]).index(1)
df.iloc[id+1:]["col2"].replace(1,0,inplace=True)
answered Dec 6 at 15:43
shyamrag cp
385
3
While this code may answer the question, providing additional context regarding how and/or why it solves the problem would improve the answer's long-term value.
– Nic3500
Dec 6 at 16:00
Chained indexing is not recommended.
– jpp
Dec 6 at 16:41
add a comment |
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8 Answers
8
active
oldest
votes
8 Answers
8
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
8
down vote
accepted
You can find the index of the first 1 and set others to 0:
mask = df['col2'].eq(1)
df.loc[mask & (df.index != mask.idxmax()), 'col2'] = 0
For better performance, see Efficiently return the index of the first value satisfying condition in array.
answered Dec 6 at 15:37
jpp
88k195099
Can you think of a good solution for the case when the index is arbitrary, likeIndex(['u', 'v', 'w', 'x', 'y', 'z']AND col2 could be something like[2, 0, 0, 1, 3, 1]?
– timgeb
Dec 6 at 16:18
@timgeb, To adapt this solution, I think you can use positional indexing (instead of index labels). Something likedf.loc[mask & (np.arange(df.shape[0]) != np.where(mask)[0][0]), 'col2'] = 0. But I'm sure there are more Pythonic ways.
– jpp
Dec 6 at 16:28
Ah, I thought of using numpy, too. Just a bit differently. See my case 3. ;)
– timgeb
Dec 6 at 16:30
add a comment |
up vote
8
down vote
accepted
You can find the index of the first 1 and set others to 0:
mask = df['col2'].eq(1)
df.loc[mask & (df.index != mask.idxmax()), 'col2'] = 0
For better performance, see Efficiently return the index of the first value satisfying condition in array.
answered Dec 6 at 15:37
jpp
88k195099
Can you think of a good solution for the case when the index is arbitrary, likeIndex(['u', 'v', 'w', 'x', 'y', 'z']AND col2 could be something like[2, 0, 0, 1, 3, 1]?
– timgeb
Dec 6 at 16:18
@timgeb, To adapt this solution, I think you can use positional indexing (instead of index labels). Something likedf.loc[mask & (np.arange(df.shape[0]) != np.where(mask)[0][0]), 'col2'] = 0. But I'm sure there are more Pythonic ways.
– jpp
Dec 6 at 16:28
Ah, I thought of using numpy, too. Just a bit differently. See my case 3. ;)
– timgeb
Dec 6 at 16:30
add a comment |
up vote
8
down vote
accepted
up vote
8
down vote
accepted
You can find the index of the first 1 and set others to 0:
mask = df['col2'].eq(1)
df.loc[mask & (df.index != mask.idxmax()), 'col2'] = 0
For better performance, see Efficiently return the index of the first value satisfying condition in array.
answered Dec 6 at 15:37
jpp
88k195099
You can find the index of the first 1 and set others to 0:
mask = df['col2'].eq(1)
df.loc[mask & (df.index != mask.idxmax()), 'col2'] = 0
For better performance, see Efficiently return the index of the first value satisfying condition in array.
answered Dec 6 at 15:37
jpp
88k195099
answered Dec 6 at 15:37
jpp
88k195099
answered Dec 6 at 15:37
jpp
88k195099
answered Dec 6 at 15:37
jpp
88k195099
88k195099
Can you think of a good solution for the case when the index is arbitrary, likeIndex(['u', 'v', 'w', 'x', 'y', 'z']AND col2 could be something like[2, 0, 0, 1, 3, 1]?
– timgeb
Dec 6 at 16:18
@timgeb, To adapt this solution, I think you can use positional indexing (instead of index labels). Something likedf.loc[mask & (np.arange(df.shape[0]) != np.where(mask)[0][0]), 'col2'] = 0. But I'm sure there are more Pythonic ways.
– jpp
Dec 6 at 16:28
Ah, I thought of using numpy, too. Just a bit differently. See my case 3. ;)
– timgeb
Dec 6 at 16:30
add a comment |
Can you think of a good solution for the case when the index is arbitrary, likeIndex(['u', 'v', 'w', 'x', 'y', 'z']AND col2 could be something like[2, 0, 0, 1, 3, 1]?
– timgeb
Dec 6 at 16:18
@timgeb, To adapt this solution, I think you can use positional indexing (instead of index labels). Something likedf.loc[mask & (np.arange(df.shape[0]) != np.where(mask)[0][0]), 'col2'] = 0. But I'm sure there are more Pythonic ways.
– jpp
Dec 6 at 16:28
Ah, I thought of using numpy, too. Just a bit differently. See my case 3. ;)
– timgeb
Dec 6 at 16:30
Can you think of a good solution for the case when the index is arbitrary, like
Index(['u', 'v', 'w', 'x', 'y', 'z'] AND col2 could be something like [2, 0, 0, 1, 3, 1]?– timgeb
Dec 6 at 16:18
Can you think of a good solution for the case when the index is arbitrary, like
Index(['u', 'v', 'w', 'x', 'y', 'z'] AND col2 could be something like [2, 0, 0, 1, 3, 1]?– timgeb
Dec 6 at 16:18
@timgeb, To adapt this solution, I think you can use positional indexing (instead of index labels). Something like
df.loc[mask & (np.arange(df.shape[0]) != np.where(mask)[0][0]), 'col2'] = 0. But I'm sure there are more Pythonic ways.– jpp
Dec 6 at 16:28
@timgeb, To adapt this solution, I think you can use positional indexing (instead of index labels). Something like
df.loc[mask & (np.arange(df.shape[0]) != np.where(mask)[0][0]), 'col2'] = 0. But I'm sure there are more Pythonic ways.– jpp
Dec 6 at 16:28
Ah, I thought of using numpy, too. Just a bit differently. See my case 3. ;)
– timgeb
Dec 6 at 16:30
Ah, I thought of using numpy, too. Just a bit differently. See my case 3. ;)
– timgeb
Dec 6 at 16:30
add a comment |
up vote
4
down vote
np.flatnonzero
Because I thought we needed more answers
df.loc[df.index[np.flatnonzero(df.col2)[1:]], 'col2'] -= 1
df
col1 col2
0 a 0
1 b 1
2 c 0
3 d 0
4 c 0
5 d 0
Same thing but a little more sneaky.
df.col2.values[np.flatnonzero(df.col2.values)[1:]] -= 1
df
col1 col2
0 a 0
1 b 1
2 c 0
3 d 0
4 c 0
5 d 0
answered Dec 6 at 15:51
piRSquared
151k22138282
add a comment |
up vote
4
down vote
np.flatnonzero
Because I thought we needed more answers
df.loc[df.index[np.flatnonzero(df.col2)[1:]], 'col2'] -= 1
df
col1 col2
0 a 0
1 b 1
2 c 0
3 d 0
4 c 0
5 d 0
Same thing but a little more sneaky.
df.col2.values[np.flatnonzero(df.col2.values)[1:]] -= 1
df
col1 col2
0 a 0
1 b 1
2 c 0
3 d 0
4 c 0
5 d 0
answered Dec 6 at 15:51
piRSquared
151k22138282
add a comment |
up vote
4
down vote
up vote
4
down vote
np.flatnonzero
Because I thought we needed more answers
df.loc[df.index[np.flatnonzero(df.col2)[1:]], 'col2'] -= 1
df
col1 col2
0 a 0
1 b 1
2 c 0
3 d 0
4 c 0
5 d 0
Same thing but a little more sneaky.
df.col2.values[np.flatnonzero(df.col2.values)[1:]] -= 1
df
col1 col2
0 a 0
1 b 1
2 c 0
3 d 0
4 c 0
5 d 0
answered Dec 6 at 15:51
piRSquared
151k22138282
np.flatnonzero
Because I thought we needed more answers
df.loc[df.index[np.flatnonzero(df.col2)[1:]], 'col2'] -= 1
df
col1 col2
0 a 0
1 b 1
2 c 0
3 d 0
4 c 0
5 d 0
Same thing but a little more sneaky.
df.col2.values[np.flatnonzero(df.col2.values)[1:]] -= 1
df
col1 col2
0 a 0
1 b 1
2 c 0
3 d 0
4 c 0
5 d 0
answered Dec 6 at 15:51
piRSquared
151k22138282
edited Dec 6 at 15:56
answered Dec 6 at 15:51
piRSquared
151k22138282
answered Dec 6 at 15:51
piRSquared
151k22138282
answered Dec 6 at 15:51
piRSquared
151k22138282
151k22138282
add a comment |
add a comment |
up vote
4
down vote
Case 1: df has only ones and zeros in col2 and integer indexes.
>>> df
col1 col2
0 a 0
1 b 1
2 c 1
3 d 0
4 c 1
5 d 0
You can use:
>>> df.loc[df['col2'].idxmax() + 1:, 'col2'] = 0
>>> df
col1 col2
0 a 0
1 b 1
2 c 0
3 d 0
4 c 0
5 d 0
Case2: df can have all kinds of values in col2 and has integer indexes.
>>> df # demo dataframe
col1 col2
0 a 0
1 b 1
2 c 2
3 d 2
4 c 3
5 d 3
You can use:
>>> df.loc[(df['col2'] == 1).idxmax() + 1:, 'col2'] = 0
>>> df
col1 col2
0 a 0
1 b 1
2 c 0
3 d 0
4 c 0
5 d 0
Case 3: df can have all kinds of values in col2 and has an arbitrary index.
>>> df
col1 col2
u a -1
v b 1
w c 2
x d 2
y c 3
z d 3
You can use:
>>> df['col2'].iloc[(df['col2'].values == 1).argmax() + 1:] = 0
>>> df
col1 col2
u a -1
v b 1
w c 0
x d 0
y c 0
z d 0
answered Dec 6 at 15:38
timgeb
47.7k116288
add a comment |
up vote
4
down vote
Case 1: df has only ones and zeros in col2 and integer indexes.
>>> df
col1 col2
0 a 0
1 b 1
2 c 1
3 d 0
4 c 1
5 d 0
You can use:
>>> df.loc[df['col2'].idxmax() + 1:, 'col2'] = 0
>>> df
col1 col2
0 a 0
1 b 1
2 c 0
3 d 0
4 c 0
5 d 0
Case2: df can have all kinds of values in col2 and has integer indexes.
>>> df # demo dataframe
col1 col2
0 a 0
1 b 1
2 c 2
3 d 2
4 c 3
5 d 3
You can use:
>>> df.loc[(df['col2'] == 1).idxmax() + 1:, 'col2'] = 0
>>> df
col1 col2
0 a 0
1 b 1
2 c 0
3 d 0
4 c 0
5 d 0
Case 3: df can have all kinds of values in col2 and has an arbitrary index.
>>> df
col1 col2
u a -1
v b 1
w c 2
x d 2
y c 3
z d 3
You can use:
>>> df['col2'].iloc[(df['col2'].values == 1).argmax() + 1:] = 0
>>> df
col1 col2
u a -1
v b 1
w c 0
x d 0
y c 0
z d 0
answered Dec 6 at 15:38
timgeb
47.7k116288
add a comment |
up vote
4
down vote
up vote
4
down vote
Case 1: df has only ones and zeros in col2 and integer indexes.
>>> df
col1 col2
0 a 0
1 b 1
2 c 1
3 d 0
4 c 1
5 d 0
You can use:
>>> df.loc[df['col2'].idxmax() + 1:, 'col2'] = 0
>>> df
col1 col2
0 a 0
1 b 1
2 c 0
3 d 0
4 c 0
5 d 0
Case2: df can have all kinds of values in col2 and has integer indexes.
>>> df # demo dataframe
col1 col2
0 a 0
1 b 1
2 c 2
3 d 2
4 c 3
5 d 3
You can use:
>>> df.loc[(df['col2'] == 1).idxmax() + 1:, 'col2'] = 0
>>> df
col1 col2
0 a 0
1 b 1
2 c 0
3 d 0
4 c 0
5 d 0
Case 3: df can have all kinds of values in col2 and has an arbitrary index.
>>> df
col1 col2
u a -1
v b 1
w c 2
x d 2
y c 3
z d 3
You can use:
>>> df['col2'].iloc[(df['col2'].values == 1).argmax() + 1:] = 0
>>> df
col1 col2
u a -1
v b 1
w c 0
x d 0
y c 0
z d 0
answered Dec 6 at 15:38
timgeb
47.7k116288
Case 1: df has only ones and zeros in col2 and integer indexes.
>>> df
col1 col2
0 a 0
1 b 1
2 c 1
3 d 0
4 c 1
5 d 0
You can use:
>>> df.loc[df['col2'].idxmax() + 1:, 'col2'] = 0
>>> df
col1 col2
0 a 0
1 b 1
2 c 0
3 d 0
4 c 0
5 d 0
Case2: df can have all kinds of values in col2 and has integer indexes.
>>> df # demo dataframe
col1 col2
0 a 0
1 b 1
2 c 2
3 d 2
4 c 3
5 d 3
You can use:
>>> df.loc[(df['col2'] == 1).idxmax() + 1:, 'col2'] = 0
>>> df
col1 col2
0 a 0
1 b 1
2 c 0
3 d 0
4 c 0
5 d 0
Case 3: df can have all kinds of values in col2 and has an arbitrary index.
>>> df
col1 col2
u a -1
v b 1
w c 2
x d 2
y c 3
z d 3
You can use:
>>> df['col2'].iloc[(df['col2'].values == 1).argmax() + 1:] = 0
>>> df
col1 col2
u a -1
v b 1
w c 0
x d 0
y c 0
z d 0
answered Dec 6 at 15:38
timgeb
47.7k116288
edited Dec 6 at 16:29
answered Dec 6 at 15:38
timgeb
47.7k116288
answered Dec 6 at 15:38
timgeb
47.7k116288
answered Dec 6 at 15:38
timgeb
47.7k116288
47.7k116288
add a comment |
add a comment |
up vote
3
down vote
Using drop_duplicates with reindex
df.col2=df.col2.drop_duplicates().reindex(df.index,fill_value=0)
df
Out[1078]:
col1 col2
0 a 0
1 b 1
2 c 0
3 d 0
4 c 0
5 d 0
answered Dec 6 at 15:41
W-B
97.6k73162
add a comment |
up vote
3
down vote
Using drop_duplicates with reindex
df.col2=df.col2.drop_duplicates().reindex(df.index,fill_value=0)
df
Out[1078]:
col1 col2
0 a 0
1 b 1
2 c 0
3 d 0
4 c 0
5 d 0
answered Dec 6 at 15:41
W-B
97.6k73162
add a comment |
up vote
3
down vote
up vote
3
down vote
Using drop_duplicates with reindex
df.col2=df.col2.drop_duplicates().reindex(df.index,fill_value=0)
df
Out[1078]:
col1 col2
0 a 0
1 b 1
2 c 0
3 d 0
4 c 0
5 d 0
answered Dec 6 at 15:41
W-B
97.6k73162
Using drop_duplicates with reindex
df.col2=df.col2.drop_duplicates().reindex(df.index,fill_value=0)
df
Out[1078]:
col1 col2
0 a 0
1 b 1
2 c 0
3 d 0
4 c 0
5 d 0
answered Dec 6 at 15:41
W-B
97.6k73162
answered Dec 6 at 15:41
W-B
97.6k73162
answered Dec 6 at 15:41
W-B
97.6k73162
answered Dec 6 at 15:41
W-B
97.6k73162
97.6k73162
add a comment |
add a comment |
up vote
3
down vote
You can use numpy for an effficient solution:
a = df.col2.values
b = np.zeros_like(a)
b[a.argmax()] = 1
df.assign(col2=b)
col1 col2
0 a 0
1 b 1
2 c 0
3 d 0
4 c 0
5 d 0
answered Dec 6 at 15:39
user3483203
29.9k82354
add a comment |
up vote
3
down vote
You can use numpy for an effficient solution:
a = df.col2.values
b = np.zeros_like(a)
b[a.argmax()] = 1
df.assign(col2=b)
col1 col2
0 a 0
1 b 1
2 c 0
3 d 0
4 c 0
5 d 0
answered Dec 6 at 15:39
user3483203
29.9k82354
add a comment |
up vote
3
down vote
up vote
3
down vote
You can use numpy for an effficient solution:
a = df.col2.values
b = np.zeros_like(a)
b[a.argmax()] = 1
df.assign(col2=b)
col1 col2
0 a 0
1 b 1
2 c 0
3 d 0
4 c 0
5 d 0
answered Dec 6 at 15:39
user3483203
29.9k82354
You can use numpy for an effficient solution:
a = df.col2.values
b = np.zeros_like(a)
b[a.argmax()] = 1
df.assign(col2=b)
col1 col2
0 a 0
1 b 1
2 c 0
3 d 0
4 c 0
5 d 0
answered Dec 6 at 15:39
user3483203
29.9k82354
edited Dec 6 at 15:53
answered Dec 6 at 15:39
user3483203
29.9k82354
answered Dec 6 at 15:39
user3483203
29.9k82354
answered Dec 6 at 15:39
user3483203
29.9k82354
29.9k82354
add a comment |
add a comment |
up vote
1
down vote
i like this too
data['col2'][np.where(data['col2'] == 1)[0][0]+1:] = 0
answered Dec 6 at 15:42
iamklaus
78648
Chained indexing is not recommended.
– jpp
Dec 6 at 16:41
Thanks for the update..
– iamklaus
Dec 7 at 8:43
add a comment |
up vote
1
down vote
i like this too
data['col2'][np.where(data['col2'] == 1)[0][0]+1:] = 0
answered Dec 6 at 15:42
iamklaus
78648
Chained indexing is not recommended.
– jpp
Dec 6 at 16:41
Thanks for the update..
– iamklaus
Dec 7 at 8:43
add a comment |
up vote
1
down vote
up vote
1
down vote
i like this too
data['col2'][np.where(data['col2'] == 1)[0][0]+1:] = 0
answered Dec 6 at 15:42
iamklaus
78648
i like this too
data['col2'][np.where(data['col2'] == 1)[0][0]+1:] = 0
answered Dec 6 at 15:42
iamklaus
78648
answered Dec 6 at 15:42
iamklaus
78648
answered Dec 6 at 15:42
iamklaus
78648
answered Dec 6 at 15:42
iamklaus
78648
78648
Chained indexing is not recommended.
– jpp
Dec 6 at 16:41
Thanks for the update..
– iamklaus
Dec 7 at 8:43
add a comment |
Chained indexing is not recommended.
– jpp
Dec 6 at 16:41
Thanks for the update..
– iamklaus
Dec 7 at 8:43
Chained indexing is not recommended.
– jpp
Dec 6 at 16:41
Chained indexing is not recommended.
– jpp
Dec 6 at 16:41
Thanks for the update..
– iamklaus
Dec 7 at 8:43
Thanks for the update..
– iamklaus
Dec 7 at 8:43
add a comment |
up vote
1
down vote
Sooo many options, here's mine... almost the same as timgebs answer (found independently), but still different ;)
Find the index of col2 that has the first occurence of a 1, and change all row values after that index to 0:
df['col2'].iloc[df.col2.idxmax()+1:] = 0
answered Dec 6 at 15:55
Sander van den Oord
551419
Be careful, this sets all values to0after the specified index, not just the ones equal to1. Though that's the same with some other answers too.
– jpp
Dec 6 at 16:42
Totally agree. Your solution is more general.
– Sander van den Oord
Dec 6 at 17:42
add a comment |
up vote
1
down vote
Sooo many options, here's mine... almost the same as timgebs answer (found independently), but still different ;)
Find the index of col2 that has the first occurence of a 1, and change all row values after that index to 0:
df['col2'].iloc[df.col2.idxmax()+1:] = 0
answered Dec 6 at 15:55
Sander van den Oord
551419
Be careful, this sets all values to0after the specified index, not just the ones equal to1. Though that's the same with some other answers too.
– jpp
Dec 6 at 16:42
Totally agree. Your solution is more general.
– Sander van den Oord
Dec 6 at 17:42
add a comment |
up vote
1
down vote
up vote
1
down vote
Sooo many options, here's mine... almost the same as timgebs answer (found independently), but still different ;)
Find the index of col2 that has the first occurence of a 1, and change all row values after that index to 0:
df['col2'].iloc[df.col2.idxmax()+1:] = 0
answered Dec 6 at 15:55
Sander van den Oord
551419
Sooo many options, here's mine... almost the same as timgebs answer (found independently), but still different ;)
Find the index of col2 that has the first occurence of a 1, and change all row values after that index to 0:
df['col2'].iloc[df.col2.idxmax()+1:] = 0
answered Dec 6 at 15:55
Sander van den Oord
551419
answered Dec 6 at 15:55
Sander van den Oord
551419
answered Dec 6 at 15:55
Sander van den Oord
551419
answered Dec 6 at 15:55
Sander van den Oord
551419
551419
Be careful, this sets all values to0after the specified index, not just the ones equal to1. Though that's the same with some other answers too.
– jpp
Dec 6 at 16:42
Totally agree. Your solution is more general.
– Sander van den Oord
Dec 6 at 17:42
add a comment |
Be careful, this sets all values to0after the specified index, not just the ones equal to1. Though that's the same with some other answers too.
– jpp
Dec 6 at 16:42
Totally agree. Your solution is more general.
– Sander van den Oord
Dec 6 at 17:42
Be careful, this sets all values to
0 after the specified index, not just the ones equal to 1. Though that's the same with some other answers too.– jpp
Dec 6 at 16:42
Be careful, this sets all values to
0 after the specified index, not just the ones equal to 1. Though that's the same with some other answers too.– jpp
Dec 6 at 16:42
Totally agree. Your solution is more general.
– Sander van den Oord
Dec 6 at 17:42
Totally agree. Your solution is more general.
– Sander van den Oord
Dec 6 at 17:42
add a comment |
up vote
0
down vote
id = list(df["col2"]).index(1)
df.iloc[id+1:]["col2"].replace(1,0,inplace=True)
answered Dec 6 at 15:43
shyamrag cp
385
3
While this code may answer the question, providing additional context regarding how and/or why it solves the problem would improve the answer's long-term value.
– Nic3500
Dec 6 at 16:00
Chained indexing is not recommended.
– jpp
Dec 6 at 16:41
add a comment |
up vote
0
down vote
id = list(df["col2"]).index(1)
df.iloc[id+1:]["col2"].replace(1,0,inplace=True)
answered Dec 6 at 15:43
shyamrag cp
385
3
While this code may answer the question, providing additional context regarding how and/or why it solves the problem would improve the answer's long-term value.
– Nic3500
Dec 6 at 16:00
Chained indexing is not recommended.
– jpp
Dec 6 at 16:41
add a comment |
up vote
0
down vote
up vote
0
down vote
id = list(df["col2"]).index(1)
df.iloc[id+1:]["col2"].replace(1,0,inplace=True)
answered Dec 6 at 15:43
shyamrag cp
385
id = list(df["col2"]).index(1)
df.iloc[id+1:]["col2"].replace(1,0,inplace=True)
answered Dec 6 at 15:43
shyamrag cp
385
answered Dec 6 at 15:43
shyamrag cp
385
answered Dec 6 at 15:43
shyamrag cp
385
answered Dec 6 at 15:43
shyamrag cp
385
385
3
While this code may answer the question, providing additional context regarding how and/or why it solves the problem would improve the answer's long-term value.
– Nic3500
Dec 6 at 16:00
Chained indexing is not recommended.
– jpp
Dec 6 at 16:41
add a comment |
3
While this code may answer the question, providing additional context regarding how and/or why it solves the problem would improve the answer's long-term value.
– Nic3500
Dec 6 at 16:00
Chained indexing is not recommended.
– jpp
Dec 6 at 16:41
3
3
While this code may answer the question, providing additional context regarding how and/or why it solves the problem would improve the answer's long-term value.
– Nic3500
Dec 6 at 16:00
While this code may answer the question, providing additional context regarding how and/or why it solves the problem would improve the answer's long-term value.
– Nic3500
Dec 6 at 16:00
Chained indexing is not recommended.
– jpp
Dec 6 at 16:41
Chained indexing is not recommended.
– jpp
Dec 6 at 16:41
add a comment |
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