CODEWITHSUNDEEP
python
AP257
AP257

Reputation: 93973

Python strip() multiple characters?

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

Upvotes: 98

Views: 277502

Answers (7)

Litz
Litz

Reputation: 1

Since strip only removes characters from start and end, one idea could be to break the string into list of words, then remove chars, and then join:

s = 'Barack (of Washington)'
x = [j.strip('(){}<>') for j in s.split()]
ans = ' '.join(j for j in x)
print(ans)

Upvotes: 0

Siva Kumar
Siva Kumar

Reputation: 479

For example string s="(U+007c)"

To remove only the parentheses from s, try the below one:

import re
a=re.sub("\\(","",s)
b=re.sub("\\)","",a)
print(b)

Upvotes: -3

cherish
cherish

Reputation: 1400

string.translate with table=None works fine.

>>> name = "Barack (of Washington)"
>>> name = name.translate(None, "(){}<>")
>>> print name
Barack of Washington

Upvotes: 17

JasonFruit
JasonFruit

Reputation: 7875

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.

Upvotes: 81

Ruel
Ruel

Reputation: 15780

Because strip() only strips trailing and leading characters, based on what you provided. I suggest:

>>> import re
>>> name = "Barack (of Washington)"
>>> name = re.sub('[\(\)\{\}<>]', '', name)
>>> print(name)
Barack of Washington

Upvotes: 16

bobince
bobince

Reputation: 536715

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)

Upvotes: 125

unutbu
unutbu

Reputation: 880797

strip only strips characters from the very front and back of the string.

To delete a list of characters, you could use the string's translate method:

import string
name = "Barack (of Washington)"
table = string.maketrans( '', '', )
print name.translate(table,"(){}<>")
# Barack of Washington

Upvotes: 10

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