Python，从字符串中删除所有非字母字符

Question

I am writing a python MapReduce word count program. Problem is that there are many non-alphabet chars strewn about in the data, I have found this post Stripping everything but alphanumeric chars from a string in Python which shows a nice solution using regex, but I am not sure how to implement it

def mapfn(k, v):
    print v
    import re, string 
    pattern = re.compile('[\W_]+')
    v = pattern.match(v)
    print v
    for w in v.split():
        yield w, 1

I'm afraid I am not sure how to use the library re or even regex for that matter. I am not sure how to apply the regex pattern to the incoming string (line of a book) v properly to retrieve the new line without any non-alphanumeric chars.

Suggestions?

Answer 1

Use re.sub

import re

regex = re.compile('[^a-zA-Z]')
#First parameter is the replacement, second parameter is your input string
regex.sub('', 'ab3d*E')
#Out: 'abdE'

Alternatively, if you only want to remove a certain set of characters (as an apostrophe might be okay in your input...)

regex = re.compile('[,\.!?]') #etc.

Answer 2

如果您不想使用正则表达式，您可以尝试

''.join([i for i in s if i.isalpha()])

Answer 3

You can use the re.sub() function to remove these characters:

>>> import re
>>> re.sub("[^a-zA-Z]+", "", "ABC12abc345def")
'ABCabcdef'

re.sub(MATCH PATTERN, REPLACE STRING, STRING TO SEARCH)

• "[^a-zA-Z]+" - look for any group of characters that are NOT a-zA-z.
• "" - Replace the matched characters with ""

Answer 4

Try:

s = ''.join(filter(str.isalnum, s))

This will take every char from the string, keep only alphanumeric ones and build a string back from them.

Answer 5

The fastest method is regex

#Try with regex first
t0 = timeit.timeit("""
s = r2.sub('', st)

""", setup = """
import re
r2 = re.compile(r'[^a-zA-Z0-9]', re.MULTILINE)
st = 'abcdefghijklmnopqrstuvwxyz123456789!@#$%^&*()-=_+'
""", number = 1000000)
print(t0)

#Try with join method on filter
t0 = timeit.timeit("""
s = ''.join(filter(str.isalnum, st))

""", setup = """
st = 'abcdefghijklmnopqrstuvwxyz123456789!@#$%^&*()-=_+'
""",
number = 1000000)
print(t0)

#Try with only join
t0 = timeit.timeit("""
s = ''.join(c for c in st if c.isalnum())

""", setup = """
st = 'abcdefghijklmnopqrstuvwxyz123456789!@#$%^&*()-=_+'
""", number = 1000000)
print(t0)


2.6002226710006653 Method 1 Regex
5.739747313000407 Method 2 Filter + Join
6.540099570000166 Method 3 Join

Answer 6

It is advisable to use PyPi regex module if you plan to match specific Unicode property classes. This library has also proven to be more stable, especially handling large texts, and yields consistent results across various Python versions. All you need to do is to keep it up-to-date.

If you install it (using pip intall regex or pip3 install regex ), you may use

import regex
print ( regex.sub(r'\P{L}+', '', 'ABCŁąć1-2!Абв3§4“5def”') )
// => ABCŁąćАбвdef

to remove all chunks of 1 or more characters other than Unicode letters from text . See an online Python demo . You may also use "".join(regex.findall(r'\\p{L}+', 'ABCŁąć1-2!Абв3§4“5def”')) to get the same result.

In Python re , in order to match any Unicode letter, one may use the [^\\W\\d_] construct ( Match any unicode letter? ).

So, to remove all non-letter characters, you may either match all letters and join the results:

result = "".join(re.findall(r'[^\W\d_]', text))

Or, remove all chars other than those matched with [^\\W\\d_] :

result = re.sub(r'([^\W\d_])|.', r'\1', text, re.DOTALL)

See the regex demo online . However , you may get inconsistent results across various Python versions because the Unicode standard is evolving, and the set of chars matched with \\w will depend on the Python version. Using PyPi regex library is highly recommended to get consistent results.

Answer 7

Here's yet another callable function that removes every that is not in plain english:

import re
remove_non_english = lambda s: re.sub(r'[^a-zA-Z\s\n\.]', ' ', s)

Usage:

remove_non_english('a€bñcá`` something. 2323')
> 'a b c    something     '