Stripping everything but alphanumeric chars from a string in Python

Question

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.

Answer 1

Regular expressions to the rescue:

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

Answer 2

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.

Answer 3

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

Answer 4

You could try:

print ''.join(ch for ch in some_string if ch.isalnum())

Answer 5

I just timed some functions out of curiosity

$ 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

Answer 6

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>\python26\python -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>\python26\python -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>\python26\python -c "import string; s = string.printable; print len(s),repr(s)"
100 '0123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ!"#$%&\'()*+,-./:;=>?@[\\]^_`{|}~ \t\n\r\x0b\x0c'

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

C:\junk>\python26\python -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>\python26\python -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