Removing control characters from a string in python

I currently have the following code

def removeControlCharacters(line):
    i = 0
    for c in line:
        if (c < chr(32)):
            line = line[:i - 1] + line[i+1:]
            i += 1
    return line

This is just does not work if there are more than one character to be deleted.

Answers:

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Method 1

There are hundreds of control characters in unicode. If you are sanitizing data from the web or some other source that might contain non-ascii characters, you will need Python’s unicodedata module. The unicodedata.category(…) function returns the unicode category code (e.g., control character, whitespace, letter, etc.) of any character. For control characters, the category always starts with “C”.

This snippet removes all control characters from a string.

import unicodedata
def remove_control_characters(s):
    return "".join(ch for ch in s if unicodedata.category(ch)[0]!="C")

Examples of unicode categories:

>>> from unicodedata import category
>>> category('r')      # carriage return --> Cc : control character
'Cc'
>>> category('')      # null character ---> Cc : control character
'Cc'
>>> category('t')      # tab --------------> Cc : control character
'Cc'
>>> category(' ')       # space ------------> Zs : separator, space
'Zs'
>>> category(u'u200A') # hair space -------> Zs : separator, space
'Zs'
>>> category(u'u200b') # zero width space -> Cf : control character, formatting
'Cf'
>>> category('A')       # letter "A" -------> Lu : letter, uppercase
'Lu'
>>> category(u'u4e21') # 両 ---------------> Lo : letter, other
'Lo'
>>> category(',')       # comma  -----------> Po : punctuation
'Po'
>>>

Method 2

You could use str.translate with the appropriate map, for example like this:

>>> mpa = dict.fromkeys(range(32))
>>> 'abc2de'.translate(mpa)
'abcde'

Method 3

Anyone interested in a regex character class that matches any Unicode control character may use [x00-x1fx7f-x9f].

You may test it like this:

>>> import unicodedata, re, sys
>>> all_chars = [chr(i) for i in range(sys.maxunicode)]
>>> control_chars = ''.join(c for c in all_chars if unicodedata.category(c) == 'Cc')
>>> expanded_class = ''.join(c for c in all_chars if re.match(r'[x00-x1fx7f-x9f]', c))
>>> control_chars == expanded_class
True

So to remove the control characters using re just use the following:

>>> re.sub(r'[x00-x1fx7f-x9f]', '', 'abc2de')
'abcde'

Method 4

This is the easiest, most complete, and most robust way I am aware of. It does require an external dependency, however. I consider it to be worth it for most projects.

pip install regex

import regex as rx
def remove_control_characters(str):
    return rx.sub(r'p{C}', '', 'my-string')

p{C} is the unicode character property for control characters, so you can leave it up to the unicode consortium which ones of the millions of unicode characters available should be considered control. There are also other extremely useful character properties I frequently use, for example p{Z} for any kind of whitespace.

Method 5

Your implementation is wrong because the value of i is incorrect. However that’s not the only problem: it also repeatedly uses slow string operations, meaning that it runs in O(n²) instead of O(n). Try this instead:

return ''.join(c for c in line if ord(c) >= 32)

Method 6

And for Python 2, with the builtin translate:

import string
all_bytes = string.maketrans('', '')  # String of 256 characters with (byte) value 0 to 255

line.translate(all_bytes, all_bytes[:32])  # All bytes < 32 are deleted (the second argument lists the bytes to delete)

Method 7

You modify the line during iterating over it. Something like ''.join([x for x in line if ord(x) >= 32])

Method 8

filter(string.printable[:-5].__contains__,line)

Method 9

I’ve tried all the above and it didn’t help. In my case, I had to remove Unicode ‘LRM’ chars:

Finally I found this solution that did the job:

df["AMOUNT"] = df["AMOUNT"].str.encode("ascii", "ignore")
df["AMOUNT"] = df["AMOUNT"].str.decode('UTF-8')

Reference here.

All methods was sourced from stackoverflow.com or stackexchange.com, is licensed under cc by-sa 2.5, cc by-sa 3.0 and cc by-sa 4.0