Stripping everything but alphanumeric chars from a string in Python


What is the best way to strip all non alphanumeric characters from a string, using Python?

The solutions presented in the PHP variant of this question will probably work with some minor adjustments, but don’t seem very ‘pythonic’ to me.

For the record, I don’t just want to strip periods and commas (and other punctuation), but also quotes, brackets, etc.

Asked By: Mark van Lent



Regular expressions to the rescue:

import re
re.sub(r'W+', '', your_string)

By Python definition 'W == [^a-zA-Z0-9_], which excludes all numbers, letters and _

Answered By: Ants Aasma

How about:

def ExtractAlphanumeric(InputString):
    from string import ascii_letters, digits
    return "".join([ch for ch in InputString if ch in (ascii_letters + digits)])

This works by using list comprehension to produce a list of the characters in InputString if they are present in the combined ascii_letters and digits strings. It then joins the list together into a string.

Answered By: DrAl
>>> import re
>>> string = "Kl13@£$%[};'""
>>> pattern = re.compile('W')
>>> string = re.sub(pattern, '', string)
>>> print string
Answered By: DisplacedAussie

You could try:

print ''.join(ch for ch in some_string if ch.isalnum())
Answered By: ars

I just timed some functions out of curiosity. In these tests I’m removing non-alphanumeric characters from the string string.printable (part of the built-in string module). The use of compiled '[W_]+' and pattern.sub('', str) was found to be fastest.

$ python -m timeit -s 
     "import string" 
     "''.join(ch for ch in string.printable if ch.isalnum())" 
10000 loops, best of 3: 57.6 usec per loop

$ python -m timeit -s 
    "import string" 
    "filter(str.isalnum, string.printable)"                 
10000 loops, best of 3: 37.9 usec per loop

$ python -m timeit -s 
    "import re, string" 
    "re.sub('[W_]', '', string.printable)"
10000 loops, best of 3: 27.5 usec per loop

$ python -m timeit -s 
    "import re, string" 
    "re.sub('[W_]+', '', string.printable)"                
100000 loops, best of 3: 15 usec per loop

$ python -m timeit -s 
    "import re, string; pattern = re.compile('[W_]+')" 
    "pattern.sub('', string.printable)" 
100000 loops, best of 3: 11.2 usec per loop
Answered By: Otto Allmendinger

Use the str.translate() method.

Presuming you will be doing this often:

  1. Once, create a string containing all the characters you wish to delete:

    delchars = ''.join(c for c in map(chr, range(256)) if not c.isalnum())
  2. Whenever you want to scrunch a string:

    scrunched = s.translate(None, delchars)

The setup cost probably compares favourably with re.compile; the marginal cost is way lower:

C:junk>python26python -mtimeit -s"import string;d=''.join(c for c in map(chr,range(256)) if not c.isalnum());s=string.printable" "s.translate(None,d)"
100000 loops, best of 3: 2.04 usec per loop

C:junk>python26python -mtimeit -s"import re,string;s=string.printable;r=re.compile(r'[W_]+')" "r.sub('',s)"
100000 loops, best of 3: 7.34 usec per loop

Note: Using string.printable as benchmark data gives the pattern '[W_]+' an unfair advantage; all the non-alphanumeric characters are in one bunch … in typical data there would be more than one substitution to do:

C:junk>python26python -c "import string; s = string.printable; print len(s),repr(s)"
100 '0123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ!"#$%&'()*+,-./:;=>?@[\]^_`{|}~ tnrx0bx0c'

Here’s what happens if you give re.sub a bit more work to do:

C:junk>python26python -mtimeit -s"d=''.join(c for c in map(chr,range(256)) if not c.isalnum());s='foo-'*25" "s.translate(None,d)"
1000000 loops, best of 3: 1.97 usec per loop

C:junk>python26python -mtimeit -s"import re;s='foo-'*25;r=re.compile(r'[W_]+')" "r.sub('',s)"
10000 loops, best of 3: 26.4 usec per loop
Answered By: John Machin

As a spin off from some other answers here, I offer a really simple and flexible way to define a set of characters that you want to limit a string’s content to. In this case, I’m allowing alphanumerics PLUS dash and underscore. Just add or remove characters from my PERMITTED_CHARS as suits your use case.

PERMITTED_CHARS = "0123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ_-" 
someString = "".join(c for c in someString if c in PERMITTED_CHARS)
Answered By: BuvinJ
for char in my_string:
    if not char.isalnum():
        my_string = my_string.replace(char,"")
Answered By: Junior Ogun
sent = "".join(e for e in sent if e.isalpha())
Answered By: Tom Kalvijn

Timing with random strings of ASCII printables:

from inspect import getsource
from random import sample
import re
from string import printable
from timeit import timeit

pattern_single = re.compile(r'[W]')
pattern_repeat = re.compile(r'[W]+')
translation_tb = str.maketrans('', '', ''.join(c for c in map(chr, range(256)) if not c.isalnum()))

def generate_test_string(length):
    return ''.join(sample(printable, length))

def main():
    for i in range(0, 60, 10):
        for test in [
            lambda: ''.join(c for c in generate_test_string(i) if c.isalnum()),
            lambda: ''.join(filter(str.isalnum, generate_test_string(i))),
            lambda: re.sub(r'[W]', '', generate_test_string(i)),
            lambda: re.sub(r'[W]+', '', generate_test_string(i)),
            lambda: pattern_single.sub('', generate_test_string(i)),
            lambda: pattern_repeat.sub('', generate_test_string(i)),
            lambda: generate_test_string(i).translate(translation_tb),

            print(timeit(test), i, getsource(test).lstrip('            lambda: ').rstrip(',n'), sep='t')

if __name__ == '__main__':

Result (Python 3.7):

       Time       Length                           Code                           
6.3716264850008880  00  ''.join(c for c in generate_test_string(i) if c.isalnum())
5.7285426190064750  00  ''.join(filter(str.isalnum, generate_test_string(i)))
8.1875841680011940  00  re.sub(r'[W]', '', generate_test_string(i))
8.0002205439959650  00  re.sub(r'[W]+', '', generate_test_string(i))
5.5290945199958510  00  pattern_single.sub('', generate_test_string(i))
5.4417179649972240  00  pattern_repeat.sub('', generate_test_string(i))
4.6772285089973590  00  generate_test_string(i).translate(translation_tb)
23.574712151996210  10  ''.join(c for c in generate_test_string(i) if c.isalnum())
22.829975890002970  10  ''.join(filter(str.isalnum, generate_test_string(i)))
27.210196289997840  10  re.sub(r'[W]', '', generate_test_string(i))
27.203713296003116  10  re.sub(r'[W]+', '', generate_test_string(i))
24.008979928999906  10  pattern_single.sub('', generate_test_string(i))
23.945240008994006  10  pattern_repeat.sub('', generate_test_string(i))
21.830899796994345  10  generate_test_string(i).translate(translation_tb)
38.731336012999236  20  ''.join(c for c in generate_test_string(i) if c.isalnum())
37.942474347000825  20  ''.join(filter(str.isalnum, generate_test_string(i)))
42.169366310001350  20  re.sub(r'[W]', '', generate_test_string(i))
41.933375883003464  20  re.sub(r'[W]+', '', generate_test_string(i))
38.899814646996674  20  pattern_single.sub('', generate_test_string(i))
38.636144253003295  20  pattern_repeat.sub('', generate_test_string(i))
36.201238164998360  20  generate_test_string(i).translate(translation_tb)
49.377356811004574  30  ''.join(c for c in generate_test_string(i) if c.isalnum())
48.408927293996385  30  ''.join(filter(str.isalnum, generate_test_string(i)))
53.901889764994850  30  re.sub(r'[W]', '', generate_test_string(i))
52.130339455994545  30  re.sub(r'[W]+', '', generate_test_string(i))
50.061149017004940  30  pattern_single.sub('', generate_test_string(i))
49.366573111998150  30  pattern_repeat.sub('', generate_test_string(i))
46.649754120997386  30  generate_test_string(i).translate(translation_tb)
63.107938601999194  40  ''.join(c for c in generate_test_string(i) if c.isalnum())
65.116287978999030  40  ''.join(filter(str.isalnum, generate_test_string(i)))
71.477421126997800  40  re.sub(r'[W]', '', generate_test_string(i))
66.027950693998720  40  re.sub(r'[W]+', '', generate_test_string(i))
63.315361931003280  40  pattern_single.sub('', generate_test_string(i))
62.342320287003530  40  pattern_repeat.sub('', generate_test_string(i))
58.249303059004890  40  generate_test_string(i).translate(translation_tb)
73.810345625002810  50  ''.join(c for c in generate_test_string(i) if c.isalnum())
72.593953348005020  50  ''.join(filter(str.isalnum, generate_test_string(i)))
76.048324580995540  50  re.sub(r'[W]', '', generate_test_string(i))
75.106637657001560  50  re.sub(r'[W]+', '', generate_test_string(i))
74.681338128997600  50  pattern_single.sub('', generate_test_string(i))
72.430461594005460  50  pattern_repeat.sub('', generate_test_string(i))
69.394243567003290  50  generate_test_string(i).translate(translation_tb)

str.maketrans & str.translate is fastest, but includes all non-ASCII characters.
re.compile & pattern.sub is slower, but is somehow faster than ''.join & filter.

Answered By: Solomon Ucko

If i understood correctly the easiest way is to use regular expression as it provides you lots of flexibility but the other simple method is to use for loop following is the code with example I also counted the occurrence of word and stored in dictionary..

s = """An... essay is, generally, a piece of writing that gives the author's own 
argument — but the definition is vague, 
overlapping with those of a paper, an article, a pamphlet, and a short story. Essays 
have traditionally been 
sub-classified as formal and informal. Formal essays are characterized by "serious 
purpose, dignity, logical 
organization, length," whereas the informal essay is characterized by "the personal 
element (self-revelation, 
individual tastes and experiences, confidential manner), humor, graceful style, 
rambling structure, unconventionality 
or novelty of theme," etc.[1]"""

d = {}      # creating empty dic      
words = s.split() # spliting string and stroing in list
for word in words:
    new_word = ''
    for c in word:
        if c.isalnum(): # checking if indiviual chr is alphanumeric or not
            new_word = new_word + c
    print(new_word, end=' ')
    # if new_word not in d:
    #     d[new_word] = 1
    # else:
    #     d[new_word] = d[new_word] +1

please rate this if this answer is useful!

For a simple one-liner (Python 3.0):

''.join(filter( lambda x: x in '0123456789abcdefghijklmnopqrstuvwxyz', the_string_you_want_stripped ))

For Python < 3.0:

filter( lambda x: x in '0123456789abcdefghijklmnopqrstuvwxyz', the_string_you_want_stripped )

Note: you could add other characters to the allowed characters list if desired (e.g. ‘0123456789abcdefghijklmnopqrstuvwxyz.,_’).

Answered By: Ben Pritchard

Python 3

Uses the same method as @John Machin’s answer but updated for Python 3:

  • larger character set
  • slight changes to how translate works.

Python code is now assumed to be encoded in UTF-8
(source: PEP 3120)

This means the string containing all the characters you wish to delete gets much larger:

del_chars = ''.join(c for c in map(chr, range(1114111)) if not c.isalnum())

And the translate method now needs to consume a translation table which we can create with maketrans():

del_map = str.maketrans('', '', del_chars)

Now, as before, any string s you want to "scrunch":

scrunched = s.translate(del_map)

Using the last timing example from @Joe Machin, we can see it still beats re by an order of magnitude:

> python -mtimeit -s"d=''.join(c for c in map(chr,range(1114111)) if not c.isalnum());m=str.maketrans('','',d);s='foo-'*25" "s.translate(m)"
1000000 loops, best of 5: 255 nsec per loop
> python -mtimeit -s"import re;s='foo-'*25;r=re.compile(r'[W_]+')" "r.sub('',s)"
50000 loops, best of 5: 4.8 usec per loop
Answered By: jslatane

A simple solution because all answers here are complicated

filtered = ''
for c in unfiltered:
    if str.isalnum(c):
        filtered += c
Answered By: Ahmed Tremo

I checked the results with perfplot (a project of mine) and found that for short strings,

"".join(filter(str.isalnum, s))

is fastest. For long strings (200+ chars)

re.sub("[W_]", "", s)

is fastest.

enter image description here

Code to reproduce the plot:

import perfplot
import random
import re
import string

pattern = re.compile("[W_]+")

def setup(n):
    return "".join(random.choices(string.ascii_letters + string.digits, k=n))

def string_alphanum(s):
    return "".join(ch for ch in s if ch.isalnum())

def filter_str(s):
    return "".join(filter(str.isalnum, s))

def re_sub1(s):
    return re.sub("[W_]", "", s)

def re_sub2(s):
    return re.sub("[W_]+", "", s)

def re_sub3(s):
    return pattern.sub("", s)

b = perfplot.bench(
    kernels=[string_alphanum, filter_str, re_sub1, re_sub2, re_sub3],
    n_range=[2**k for k in range(10)],
Answered By: Nico Schlömer

If you’d like to preserve characters like áéíóúãẽĩõũ for example, use this:

import re
re.sub('[Wd_]+', '', your_string)
Answered By: Diogo de Toledo
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