regexp strangeness

[Dale Amon]

This finds nothing:

import re
import string
card = "abcdef"
DEC029 = re.compile("[^&0-9A-Z/ $*,.\-:#@'=\"[<(+\^!);\\\]%_>?]")
errs = DEC029.findall(card.strip("\n\r"))
print errs

This works correctly:

import re
import string
card = "abcdef"
DEC029 = re.compile("[^&0-9A-Z/ $*,.\-:#@'=\"[<(+\^!)\\;\]%_>?]")
errs = DEC029.findall(card.strip("\n\r"))
print errs

They differ only in the positioning of the quoted backslash.

Just in case it is of interest to anyone.




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[Peter Otten]

This finds nothing:

import re
import string
card = "abcdef"
DEC029 = re.compile("[^&0-9A-Z/ $*,.\-:#@'=\"[<(+\^!);\\\]%_>?]")
errs = DEC029.findall(card.strip("\n\r"))
print errs

This works correctly:

import re
import string
card = "abcdef"
DEC029 = re.compile("[^&0-9A-Z/ $*,.\-:#@'=\"[<(+\^!)\\;\]%_>?]")
errs = DEC029.findall(card.strip("\n\r"))
print errs

They differ only in the positioning of the quoted backslash.

Just in case it is of interest to anyone.

You have to escape twice; once for Python and once for the regular
expression. Or use raw strings, denoted by an r"..." prefix:

re.findall("[^&0-9A-Z/ $*,.\-:#@'=\"[<(+\^!);\\\]%_>?]", "abc") []
re.findall("[^&0-9A-Z/ $*,.\-:#@'=\"[<(+\^!);\\\\\\]%_>?]", "abc") ['a', 'b', 'c']
re.findall(r"[^&0-9A-Z/ $*,.\-:#@'=\"[<(+\^!);\\\]%_>?]", "abc")

['a', 'b', 'c']

Peter

 

[MRAB]

This finds nothing:

import re
import string
card = "abcdef"
DEC029 = re.compile("[^&0-9A-Z/ $*,.\-:#@'=\"[<(+\^!);\\\]%_>?]")

The regular expression you're actually providing is:

>>> print "[^&0-9A-Z/ $*,.\-:#@'=\"[<(+\^!);\\\]%_>?]"

[^&0-9A-Z/ $*,.\-:#@'="[<(+\^!);\\]%_>?]
^^^

The backslash is escaped (the "\\") and the set ends at the first "]".

errs = DEC029.findall(card.strip("\n\r"))
print errs

This works correctly:

import re
import string
card = "abcdef"
DEC029 = re.compile("[^&0-9A-Z/ $*,.\-:#@'=\"[<(+\^!)\\;\]%_>?]")

The regular expression you're actually providing is:

>>> print "[^&0-9A-Z/ $*,.\-:#@'=\"[<(+\^!)\\;\]%_>?]"

[^&0-9A-Z/ $*,.\-:#@'="[<(+\^!)\;\]%_>?]
^^ ^

The first "]" is escaped (the "\]") and the set ends at the second "]".

errs = DEC029.findall(card.strip("\n\r"))
print errs

They differ only in the positioning of the quoted backslash.

Just in case it is of interest to anyone.

You have to escape twice; once for Python and once for the regular
expression. Or use raw strings, denoted by an r"..." prefix:

re.findall("[^&0-9A-Z/ $*,.\-:#@'=\"[<(+\^!);\\\]%_>?]", "abc") []
re.findall("[^&0-9A-Z/ $*,.\-:#@'=\"[<(+\^!);\\\\\\]%_>?]", "abc") ['a', 'b', 'c']
re.findall(r"[^&0-9A-Z/ $*,.\-:#@'=\"[<(+\^!);\\\]%_>?]", "abc")

['a', 'b', 'c']

 

[Steven D'Aprano]

This finds nothing: ....
DEC029 = re.compile("[^&0-9A-Z/ $*,.\-:#@'=\"[<(+\^!);\\\]%_>?]")

This works correctly: ....
DEC029 = re.compile("[^&0-9A-Z/ $*,.\-:#@'=\"[<(+\^!)\\;\]%_>?]")

They differ only in the positioning of the quoted backslash.

So you're telling us that two different regexs do different things? Gosh.
Thanks for the heads up!

BTW, when creating regexes, you may find it much easier if you use raw
strings to avoid needing to escape backslashes:

'\n' in a string is a newline escape. To get a literal backslash followed
by an n, you can write '\\n' or r'\n'.