Remove all special characters, punctuation and spaces from string

TLDR

I timed the provided answers.

import re
re.sub('\W+','', string)

is typically 3x faster than the next fastest provided top answer.

Caution should be taken when using this option. Some special characters (e.g. ø) may not be striped using this method.

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After seeing this, I was interested in expanding on the provided answers by finding out which executes in the least amount of time, so I went through and checked some of the proposed answers with timeit against two of the example strings:

• string1 = 'Special $#! characters spaces 888323'
• string2 = 'how much for the maple syrup? $20.99? That s ridiculous!!!'

Example 1

'.join(e for e in string if e.isalnum())

• string1 - Result: 10.7061979771
• string2 - Result: 7.78372597694

Example 2

import re
re.sub('[^A-Za-z0-9]+', '', string)

• string1 - Result: 7.10785102844
• string2 - Result: 4.12814903259

Example 3

import re
re.sub('\W+','', string)

• string1 - Result: 3.11899876595
• string2 - Result: 2.78014397621

The above results are a product of the lowest returned result from an average of: repeat(3, 2000000)

Example 3 can be 3x faster than Example 1.

Shorter way :

import re
cleanString = re.sub('\W+','', string )

If you want spaces between words and numbers substitute '' with ' '

Here is a regex to match a string of characters that are not a letters or numbers:

[^A-Za-z0-9]+

Here is the Python command to do a regex substitution:

re.sub('[^A-Za-z0-9]+', '', mystring)

This can be done without regex:

>>> string = "Special $#! characters   spaces 888323"
>>> ''.join(e for e in string if e.isalnum())
'Specialcharactersspaces888323'

You can use str.isalnum:

S.isalnum() -> bool

Return True if all characters in S are alphanumeric
and there is at least one character in S, False otherwise.

If you insist on using regex, other solutions will do fine. However note that if it can be done without using a regular expression, that's the best way to go about it.