python: remove substring only at the end of string

Question

i have a bunch of strings

some of them have ' rec'

i want to remove that only if those are the last 4 characters

so another words

somestring='this is some string rec'

i want it to be:

somestring='this is some string'

what is the python way to approach this?

Answer 1

def rchop(thestring, ending):
  if thestring.endswith(ending):
    return thestring[:-len(ending)]
  return thestring

somestring = rchop(somestring, ' rec')

Answer 2

You could use a regular expression as well:

from re import sub

str = r"this is some string rec"
regex = r"(.*)\srec$"
print sub(regex, r"\1", str)

Answer 3

Since you have to get len(trailing) anyway (where trailing is the string you want to remove IF it's trailing), I'd recommend avoiding the slight duplication of work that .endswith would cause in this case. Of course, the proof of the code is in the timing, so, let's do some measurement (naming the functions after the respondents proposing them):

import re

astring = 'this is some string rec'
trailing = ' rec'

def andrew(astring=astring, trailing=trailing):
    regex = r'(.*)%s$' % re.escape(trailing)
    return re.sub(regex, r'\1', astring)

def jack0(astring=astring, trailing=trailing):
    if astring.endswith(trailing):
        return astring[:-len(trailing)]
    return astring

def jack1(astring=astring, trailing=trailing):
    regex = r'%s$' % re.escape(trailing)
    return re.sub(regex, '', astring)

def alex(astring=astring, trailing=trailing):
    thelen = len(trailing)
    if astring[-thelen:] == trailing:
        return astring[:-thelen]
    return astring

Say we've named this python file a.py and it's in the current directory; now, ...:

$ python2.6 -mtimeit -s'import a' 'a.andrew()'
100000 loops, best of 3: 19 usec per loop
$ python2.6 -mtimeit -s'import a' 'a.jack0()'
1000000 loops, best of 3: 0.564 usec per loop
$ python2.6 -mtimeit -s'import a' 'a.jack1()'
100000 loops, best of 3: 9.83 usec per loop
$ python2.6 -mtimeit -s'import a' 'a.alex()'
1000000 loops, best of 3: 0.479 usec per loop

As you see, the RE-based solutions are "hopelessly outclassed" (as often happens when one "overkills" a problem -- possibly one of the reasons REs have such a bad rep in the Python community!-), though the suggestion in @Jack's comment is way better than @Andrew's original. The string-based solutions, as expected, shing, with my endswith-avoiding one having a miniscule advantage over @Jack's (being just 15% faster). So, both pure-string ideas are good (as well as both being concise and clear) -- I prefer my variant a little bit only because I am, by character, a frugal (some might say, stingy;-) person... "waste not, want not"!-)

Answer 4

As kind of one liner generator joined:

test = """somestring='this is some string rec'
this is some string in the end word rec
This has not the word."""
match = 'rec'
print('\n'.join((line[:-len(match)] if line.endswith(match) else line)
      for line in test.splitlines()))
""" Output:
somestring='this is some string rec'
this is some string in the end word 
This has not the word.
"""