Extracting numeric values from a string containing key:value pairs
$begingroup$
I am writing function numericValues(text: String): List[Int] to extract patterns """([a-z]+)s*:s*(d+)""" and return the list of the numeric values :
numericValues("a123 : 0 abc:123 123:abc xyz:1") // List(123, 1)
I would write numericValues like this:
def numericValues(text: String): List[Int] = {
val regex = """([a-z]+)s*:s*(d+)""".r
regex.findAllIn(text).toList.flatMap {s =>
PartialFunction.condOpt(s) { case regex(_, num) => num.toInt }
}
}
I guess the condOpt invocation is redundant and I wonder how to simplify this implementation. Also, I'd appreciate comments on improvements to style and best practice.
regex scala
asked yesterday
MichaelMichael
1285
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add a comment |
$begingroup$
I am writing function numericValues(text: String): List[Int] to extract patterns """([a-z]+)s*:s*(d+)""" and return the list of the numeric values :
numericValues("a123 : 0 abc:123 123:abc xyz:1") // List(123, 1)
I would write numericValues like this:
def numericValues(text: String): List[Int] = {
val regex = """([a-z]+)s*:s*(d+)""".r
regex.findAllIn(text).toList.flatMap {s =>
PartialFunction.condOpt(s) { case regex(_, num) => num.toInt }
}
}
I guess the condOpt invocation is redundant and I wonder how to simplify this implementation. Also, I'd appreciate comments on improvements to style and best practice.
regex scala
asked yesterday
MichaelMichael
1285
New contributor
Michael is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
I am writing function numericValues(text: String): List[Int] to extract patterns """([a-z]+)s*:s*(d+)""" and return the list of the numeric values :
numericValues("a123 : 0 abc:123 123:abc xyz:1") // List(123, 1)
I would write numericValues like this:
def numericValues(text: String): List[Int] = {
val regex = """([a-z]+)s*:s*(d+)""".r
regex.findAllIn(text).toList.flatMap {s =>
PartialFunction.condOpt(s) { case regex(_, num) => num.toInt }
}
}
I guess the condOpt invocation is redundant and I wonder how to simplify this implementation. Also, I'd appreciate comments on improvements to style and best practice.
regex scala
asked yesterday
MichaelMichael
1285
New contributor
Michael is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I am writing function numericValues(text: String): List[Int] to extract patterns """([a-z]+)s*:s*(d+)""" and return the list of the numeric values :
numericValues("a123 : 0 abc:123 123:abc xyz:1") // List(123, 1)
I would write numericValues like this:
def numericValues(text: String): List[Int] = {
val regex = """([a-z]+)s*:s*(d+)""".r
regex.findAllIn(text).toList.flatMap {s =>
PartialFunction.condOpt(s) { case regex(_, num) => num.toInt }
}
}
I guess the condOpt invocation is redundant and I wonder how to simplify this implementation. Also, I'd appreciate comments on improvements to style and best practice.
regex scala
regex scala
asked yesterday
MichaelMichael
1285
New contributor
Michael is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
MichaelMichael
1285
New contributor
Michael is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited yesterday
Michael
asked yesterday
MichaelMichael
1285
New contributor
Michael is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
MichaelMichael
1285
asked yesterday
MichaelMichael
1285
1285
New contributor
Michael is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Michael is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Michael is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
1 Answer
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$begingroup$
You usually want a zero-width word boundary b on either end of the regex, to avoid matching things like 1a:1a.
There's no need to capture [a-z]+ since you are throwing it away.
You can use lookbehind (?<=…) assertions to require a match to be preceded by whatever, without including the whatever in the match result. This means no need for capturing parenthesis, only the integer is included in the match, and the final map is simply _.toInt. Variable-width lookbehind was introduced in Java 9; older versions have only fixed-width lookbehind.
Finally, removing the variable makes braces unnecessary.
def numericValues(text: String): List[Int] = """(?<=b[a-z]+s*:s*)d+b""".r.findAllIn(text).toList.map(_.toInt)
With only fixed-width lookbehind, you could postprocess the matches to remove non-numerics:
….map(_.replaceAll("[^0-9]", "").toInt)
Or, more idiomatically and less hacky, just capture the digits and extract the capture groups:
"""b[a-z]+s*:s*(d+)b""".r.findAllIn(text).matchData.toList.map(_.group(1).toInt)
answered yesterday
Oh My GoodnessOh My Goodness
1,649314
$endgroup$
$begingroup$
Thanks a lot. What if I have only fixed-width lookbehind ?
$endgroup$
– Michael
yesterday
1
$begingroup$
then you can't use variable-width lookbehind :) See edit.
$endgroup$
– Oh My Goodness
yesterday
add a comment |
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1 Answer
1
active
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1 Answer
1
active
oldest
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oldest
votes
$begingroup$
You usually want a zero-width word boundary b on either end of the regex, to avoid matching things like 1a:1a.
There's no need to capture [a-z]+ since you are throwing it away.
You can use lookbehind (?<=…) assertions to require a match to be preceded by whatever, without including the whatever in the match result. This means no need for capturing parenthesis, only the integer is included in the match, and the final map is simply _.toInt. Variable-width lookbehind was introduced in Java 9; older versions have only fixed-width lookbehind.
Finally, removing the variable makes braces unnecessary.
def numericValues(text: String): List[Int] = """(?<=b[a-z]+s*:s*)d+b""".r.findAllIn(text).toList.map(_.toInt)
With only fixed-width lookbehind, you could postprocess the matches to remove non-numerics:
….map(_.replaceAll("[^0-9]", "").toInt)
Or, more idiomatically and less hacky, just capture the digits and extract the capture groups:
"""b[a-z]+s*:s*(d+)b""".r.findAllIn(text).matchData.toList.map(_.group(1).toInt)
answered yesterday
Oh My GoodnessOh My Goodness
1,649314
$endgroup$
$begingroup$
Thanks a lot. What if I have only fixed-width lookbehind ?
$endgroup$
– Michael
yesterday
1
$begingroup$
then you can't use variable-width lookbehind :) See edit.
$endgroup$
– Oh My Goodness
yesterday
add a comment |
$begingroup$
You usually want a zero-width word boundary b on either end of the regex, to avoid matching things like 1a:1a.
There's no need to capture [a-z]+ since you are throwing it away.
You can use lookbehind (?<=…) assertions to require a match to be preceded by whatever, without including the whatever in the match result. This means no need for capturing parenthesis, only the integer is included in the match, and the final map is simply _.toInt. Variable-width lookbehind was introduced in Java 9; older versions have only fixed-width lookbehind.
Finally, removing the variable makes braces unnecessary.
def numericValues(text: String): List[Int] = """(?<=b[a-z]+s*:s*)d+b""".r.findAllIn(text).toList.map(_.toInt)
With only fixed-width lookbehind, you could postprocess the matches to remove non-numerics:
….map(_.replaceAll("[^0-9]", "").toInt)
Or, more idiomatically and less hacky, just capture the digits and extract the capture groups:
"""b[a-z]+s*:s*(d+)b""".r.findAllIn(text).matchData.toList.map(_.group(1).toInt)
answered yesterday
Oh My GoodnessOh My Goodness
1,649314
$endgroup$
$begingroup$
Thanks a lot. What if I have only fixed-width lookbehind ?
$endgroup$
– Michael
yesterday
1
$begingroup$
then you can't use variable-width lookbehind :) See edit.
$endgroup$
– Oh My Goodness
yesterday
add a comment |
$begingroup$
You usually want a zero-width word boundary b on either end of the regex, to avoid matching things like 1a:1a.
There's no need to capture [a-z]+ since you are throwing it away.
You can use lookbehind (?<=…) assertions to require a match to be preceded by whatever, without including the whatever in the match result. This means no need for capturing parenthesis, only the integer is included in the match, and the final map is simply _.toInt. Variable-width lookbehind was introduced in Java 9; older versions have only fixed-width lookbehind.
Finally, removing the variable makes braces unnecessary.
def numericValues(text: String): List[Int] = """(?<=b[a-z]+s*:s*)d+b""".r.findAllIn(text).toList.map(_.toInt)
With only fixed-width lookbehind, you could postprocess the matches to remove non-numerics:
….map(_.replaceAll("[^0-9]", "").toInt)
Or, more idiomatically and less hacky, just capture the digits and extract the capture groups:
"""b[a-z]+s*:s*(d+)b""".r.findAllIn(text).matchData.toList.map(_.group(1).toInt)
answered yesterday
Oh My GoodnessOh My Goodness
1,649314
$endgroup$
You usually want a zero-width word boundary b on either end of the regex, to avoid matching things like 1a:1a.
There's no need to capture [a-z]+ since you are throwing it away.
You can use lookbehind (?<=…) assertions to require a match to be preceded by whatever, without including the whatever in the match result. This means no need for capturing parenthesis, only the integer is included in the match, and the final map is simply _.toInt. Variable-width lookbehind was introduced in Java 9; older versions have only fixed-width lookbehind.
Finally, removing the variable makes braces unnecessary.
def numericValues(text: String): List[Int] = """(?<=b[a-z]+s*:s*)d+b""".r.findAllIn(text).toList.map(_.toInt)
With only fixed-width lookbehind, you could postprocess the matches to remove non-numerics:
….map(_.replaceAll("[^0-9]", "").toInt)
Or, more idiomatically and less hacky, just capture the digits and extract the capture groups:
"""b[a-z]+s*:s*(d+)b""".r.findAllIn(text).matchData.toList.map(_.group(1).toInt)
answered yesterday
Oh My GoodnessOh My Goodness
1,649314
edited yesterday
answered yesterday
Oh My GoodnessOh My Goodness
1,649314
answered yesterday
Oh My GoodnessOh My Goodness
1,649314
answered yesterday
Oh My GoodnessOh My Goodness
1,649314
1,649314
$begingroup$
Thanks a lot. What if I have only fixed-width lookbehind ?
$endgroup$
– Michael
yesterday
1
$begingroup$
then you can't use variable-width lookbehind :) See edit.
$endgroup$
– Oh My Goodness
yesterday
add a comment |
$begingroup$
Thanks a lot. What if I have only fixed-width lookbehind ?
$endgroup$
– Michael
yesterday
1
$begingroup$
then you can't use variable-width lookbehind :) See edit.
$endgroup$
– Oh My Goodness
yesterday
$begingroup$
Thanks a lot. What if I have only fixed-width lookbehind ?
$endgroup$
– Michael
yesterday
$begingroup$
Thanks a lot. What if I have only fixed-width lookbehind ?
$endgroup$
– Michael
yesterday
1
1
$begingroup$
then you can't use variable-width lookbehind :) See edit.
$endgroup$
– Oh My Goodness
yesterday
$begingroup$
then you can't use variable-width lookbehind :) See edit.
$endgroup$
– Oh My Goodness
yesterday
add a comment |
Michael is a new contributor. Be nice, and check out our Code of Conduct.
Michael is a new contributor. Be nice, and check out our Code of Conduct.
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