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I need to remove all special characters, punctuation and spaces from a string so that I only have letters and numbers.

19 Answers 19

570

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.

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7 Comments

What is the reason not using regex as a rule of thumb?
@ChrisDutrow regex are slower than python string built-in functions
@DiegoNavarro except that's not true, I benchmarked both the isalnum() and regex versions, and the regex one is 50-75% faster
Tried this in Python3 - it accepts unicode chars so it's useless to me. Try string = "B223323\§§§$3\u445454" as an example. The result? 'B2233233䑔54'
Additionally: "For 8-bit strings, this method is locale-dependent."! Thus the regex alternative is strictly better!
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408

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) 

7 Comments

this also removes the spaces between words, "great place" -> "greatplace". How to avoid it?
@Reihan_amn Simply add a space to the regex, so it becomes: [^A-Za-z0-9 ]+
I guess this doesn't work with modified character in other languages, like á, ö, ñ, etc. Am I right? If so, how would it be the regex for it?
This doesn't work for Spanish, German, Danish and other languages.
just add the special characters of that particular language. For example, to use for german text, re.sub('[^A-Za-z0-9 ,.-_\'äöüÄÖÜß]+', '', sample_text) expression can be used.
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83

Shorter way :

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

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

6 Comments

Except that _ is in \w and is a special character in the context of this question.
Depends on the context - underscore is very useful for filenames and other identifiers, to the point that I don't treat it as a special character but rather a sanitised space.I generally use this method myself.
r'\W+' - slightly off topic (and very pedantic) but I suggest a habit that all regex patterns be raw strings
This procedure does not treat underscore(_) as a special character.
A simple change to remove _ as well: r"[^A-Za-z]+" instead of r"\W+"
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76

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.


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.

9 Comments

@kkurian If you read the beginning of my answer, this is merely a comparison of the previously proposed solutions above. You might want to comment on the originating answer... stackoverflow.com/a/25183802/2560922
Oh, I see where you're going with this. Done!
Must consider Example 3, when dealing with large corpus.
Valid! Thanks for noting.
Third solution didn't remove the "ø" but the second one did remove it.
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34

Python 2.*

I think just filter(str.isalnum, string) works

In [20]: filter(str.isalnum, 'string with special chars like !,#$% etcs.') Out[20]: 'stringwithspecialcharslikeetcs' 

Python 3.*

In Python3, filter( ) function would return an itertable object (instead of string unlike in above). One has to join back to get a string from itertable:

''.join(filter(str.isalnum, string)) 

or to pass list in join use (not sure but can be fast a bit)

''.join([*filter(str.isalnum, string)]) 

note: unpacking in [*args] valid from Python >= 3.5

4 Comments

@Alexey correct, In python3 map, filter, and reduce returns itertable object instead. Still in Python3+ I will prefer ''.join(filter(str.isalnum, string)) (or to pass list in join use ''.join([*filter(str.isalnum, string)])) over accepted answer.
I'm not certain ''.join(filter(str.isalnum, string)) is an improvement on filter(str.isalnum, string), at least to read. Is this really the Pythreenic (yeah, you can use that) way to do this?
@TheProletariat The point is just filter(str.isalnum, string) do not return string in Python3 as filter( ) in Python-3 returns iterator rather than argument type unlike Python-2.+
@GrijeshChauhan, I think you should update your answer to include both your Python2 and Python3 recommendations.
27
#!/usr/bin/python import re strs = "how much for the maple syrup? $20.99? That's ricidulous!!!" print strs nstr = re.sub(r'[?|$|.|!]',r'',strs) print nstr nestr = re.sub(r'[^a-zA-Z0-9 ]',r'',nstr) print nestr 

you can add more special character and that will be replaced by '' means nothing i.e they will be removed.

Comments

21

string.punctuation contains following characters:

'!"#$%&\'()*+,-./:;<=>?@[\]^_`{|}~'

You can use translate and maketrans functions to map punctuations to empty values (replace)

import string 'This, is. A test!'.translate(str.maketrans('', '', string.punctuation)) 

Output:

'This is A test' 

Comments

19

Differently than everyone else did using regex, I would try to exclude every character that is not what I want, instead of enumerating explicitly what I don't want.

For example, if I want only characters from 'a to z' (upper and lower case) and numbers, I would exclude everything else:

import re s = re.sub(r"[^a-zA-Z0-9]","",s) 

This means "substitute every character that is not a number, or a character in the range 'a to z' or 'A to Z' with an empty string".

In fact, if you insert the special character ^ at the first place of your regex, you will get the negation.

Extra tip: if you also need to lowercase the result, you can make the regex even faster and easier, as long as you won't find any uppercase now.

import re s = re.sub(r"[^a-z0-9]","",s.lower()) 

Comments

15
s = re.sub(r"[-()\"#/@;:<>{}`+=~|.!?,]", "", s) 

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10

Assuming you want to use a regex and you want/need Unicode-cognisant 2.x code that is 2to3-ready:

>>> import re >>> rx = re.compile(u'[\W_]+', re.UNICODE) >>> data = u''.join(unichr(i) for i in range(256)) >>> rx.sub(u'', data) u'0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz\xaa\xb2 [snip] \xfe\xff' >>> 

Comments

6

The most generic approach is using the 'categories' of the unicodedata table which classifies every single character. E.g. the following code filters only printable characters based on their category:

import unicodedata # strip of crap characters (based on the Unicode database # categorization: # http://www.sql-und-xml.de/unicode-database/#kategorien PRINTABLE = set(('Lu', 'Ll', 'Nd', 'Zs')) def filter_non_printable(s): result = [] ws_last = False for c in s: c = unicodedata.category(c) in PRINTABLE and c or u'#' result.append(c) return u''.join(result).replace(u'#', u' ') 

Look at the given URL above for all related categories. You also can of course filter by the punctuation categories.

2 Comments

What's with the $ at the end of each line?
If it's copy & paste issue, should you fix it then?
5

For other languages like German, Spanish, Danish, French etc that contain special characters (like German "Umlaute" as ü, ä, ö) simply add these to the regex search string:

Example for German:

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

Comments

5

This will remove all special characters, punctuation, and spaces from a string and only have numbers and letters.

import re sample_str = "Hel&&lo %% Wo$#rl@d" # using isalnum() print("".join(k for k in sample_str if k.isalnum())) # using regex op2 = re.sub("[^A-Za-z]", "", sample_str) print(f"op2 = ", op2) special_char_list = ["$", "@", "#", "&", "%"] # using list comprehension op1 = "".join([k for k in sample_str if k not in special_char_list]) print(f"op1 = ", op1) # using lambda function op3 = "".join(filter(lambda x: x not in special_char_list, sample_str)) print(f"op3 = ", op3) 

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4

Use translate:

import string def clean(instr): return instr.translate(None, string.punctuation + ' ') 

Caveat: Only works on ascii strings.

2 Comments

Version difference? I get TypeError: translate() takes exactly one argument (2 given) with py3.4
It is only working with Python2.7. See below answer for using translate with Python3.
3

This will remove all non-alphanumeric characters except spaces.

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

Special characters spaces 888323

1 Comment

It will not remove double spacings
1
import re my_string = """Strings are amongst the most popular data types in Python. We can create the strings by enclosing characters in quotes. Python treats single quotes the 

same as double quotes."""

# if we need to count the word python that ends with or without ',' or '.' at end count = 0 for i in text: if i.endswith("."): text[count] = re.sub("^([a-z]+)(.)?$", r"\1", i) count += 1 print("The count of Python : ", text.count("python")) 

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1

After 10 Years, below I wrote there is the best solution. You can remove/clean all special characters, punctuation, ASCII characters and spaces from the string.

from clean_text import clean string = 'Special $#! characters spaces 888323' new = clean(string,lower=False,no_currency_symbols=True, no_punct = True,replace_with_currency_symbol='') print(new) Output ==> 'Special characters spaces 888323' you can replace space if you want. update = new.replace(' ','') print(update) Output ==> 'Specialcharactersspaces888323' 

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0
function regexFuntion(st) { const regx = /[^\w\s]/gi; // allow : [a-zA-Z0-9, space] st = st.replace(regx, ''); // remove all data without [a-zA-Z0-9, space] st = st.replace(/\s\s+/g, ' '); // remove multiple space return st; } console.log(regexFuntion('$Hello; # -world--78asdf+-===asdflkj******lkjasdfj67;')); // Output: Hello world78asdfasdflkjlkjasdfj67 

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-4
import re abc = "askhnl#$%askdjalsdk" ddd = abc.replace("#$%","") print (ddd) 

and you shall see your result as

'askhnlaskdjalsdk

1 Comment

wait.... you imported re but never used it. Your replace criteria only works for this specific string. What if your string is abc = "askhnl#$%!askdjalsdk"? I don't think will work on anything other than the #$% pattern. Might wanna tweak it

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