I'm new to Python and looking for a way to replace all occurrences of "[A-Z]0" with the [A-Z] portion of the string to get rid of certain numbers that are padded with a zero. I used this snippet to get rid of the whole occurrence from the field I'm processing:
import re
def strip_zeros(s):
return re.sub("[A-Z]0", "", s)
test = strip_zeros(!S_fromManhole!)
How do I perform the same type of procedure but without removing the leading letter of the "[A-Z]0" expression?
Thanks in advance!
Use backreferences.
http://www.regular-expressions.info/refadv.html "\1 through \9 Substituted with the text matched between the 1st through 9th pair of capturing parentheses."
http://docs.python.org/2/library/re.html#re.sub "Backreferences, such as \6, are replaced with the substring matched by group 6 in the pattern."
Untested, but it would look like this:
return re.sub(r"([A-Z])0", r"\1", s)
Placing the first letter inside a capture group and referencing it with \1
r"([A-Z])0".you can try something like
In [47]: s = "ab0"
In [48]: s.translate(None, '0')
Out[48]: 'ab'
In [49]: s = "ab0zy"
In [50]: s.translate(None, '0')
Out[50]: 'abzy'
I like Patashu's answer for this case but for the sake of completeness, passing a function to re.sub instead of a replacement string may be cleaner in more complicated cases. The function should take a single match object and return a string.
>>> def strip_zeros(s):
... def unpadded(m):
... return m.group(1)
... return re.sub("([A-Z])0", unpadded, s)
...
>>> strip_zeros("Q0")
'Q'
