Python strip() multiple characters?


Question

I want to remove any brackets from a string. Why doesn't this work properly?

>>> name = "Barack (of Washington)"
>>> name = name.strip("(){}<>")
>>> print name
Barack (of Washington
1
59
5/7/2014 10:07:37 AM

Accepted Answer

I did a time test here, using each method 100000 times in a loop. The results surprised me. (The results still surprise me after editing them in response to valid criticism in the comments.)

Here's the script:

import timeit

bad_chars = '(){}<>'

setup = """import re
import string
s = 'Barack (of Washington)'
bad_chars = '(){}<>'
rgx = re.compile('[%s]' % bad_chars)"""

timer = timeit.Timer('o = "".join(c for c in s if c not in bad_chars)', setup=setup)
print "List comprehension: ",  timer.timeit(100000)


timer = timeit.Timer("o= rgx.sub('', s)", setup=setup)
print "Regular expression: ", timer.timeit(100000)

timer = timeit.Timer('for c in bad_chars: s = s.replace(c, "")', setup=setup)
print "Replace in loop: ", timer.timeit(100000)

timer = timeit.Timer('s.translate(string.maketrans("", "", ), bad_chars)', setup=setup)
print "string.translate: ", timer.timeit(100000)

Here are the results:

List comprehension:  0.631745100021
Regular expression:  0.155561923981
Replace in loop:  0.235936164856
string.translate:  0.0965719223022

Results on other runs follow a similar pattern. If speed is not the primary concern, however, I still think string.translate is not the most readable; the other three are more obvious, though slower to varying degrees.

46
10/11/2010 3:12:37 AM

Because that's not what strip() does. It removes leading and trailing characters that are present in the argument, but not those characters in the middle of the string.

You could do:

name= name.replace('(', '').replace(')', '').replace ...

or:

name= ''.join(c for c in name if c not in '(){}<>')

or maybe use a regex:

import re
name= re.sub('[(){}<>]', '', name)

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