How to filter (or replace) unicode characters that would take more than 3 bytes in UTF-8?


I'm using Python and Django, but I'm having a problem caused by a limitation of MySQL. According to the MySQL 5.1 documentation, their utf8 implementation does not support 4-byte characters. MySQL 5.5 will support 4-byte characters using utf8mb4; and, someday in future, utf8 might support it as well.

But my server is not ready to upgrade to MySQL 5.5, and thus I'm limited to UTF-8 characters that take 3 bytes or less.

My question is: How to filter (or replace) unicode characters that would take more than 3 bytes?

I want to replace all 4-byte characters with the official \ufffd (U+FFFD REPLACEMENT CHARACTER), or with ?.

In other words, I want a behavior quite similar to Python's own str.encode() method (when passing 'replace' parameter). Edit: I want a behavior similar to encode(), but I don't want to actually encode the string. I want to still have an unicode string after filtering.

I DON'T want to escape the character before storing at the MySQL, because that would mean I would need to unescape all strings I get from the database, which is very annoying and unfeasible.

See also:

[EDIT] Added tests about the proposed solutions

So I got good answers so far. Thanks, people! Now, in order to choose one of them, I did a quick testing to find the simplest and fastest one.

#!/usr/bin/env python
# -*- coding: utf-8 -*-
# vi:ts=4 sw=4 et

import cProfile
import random
import re

# How many times to repeat each filtering
repeat_count = 256

# Percentage of "normal" chars, when compared to "large" unicode chars
normal_chars = 90

# Total number of characters in this string
string_size = 8 * 1024

# Generating a random testing string
test_string = u''.join(
            0x10ffff if random.randrange(100) > normal_chars else 0x0fff
        )) for i in xrange(string_size) )

# RegEx to find invalid characters
re_pattern = re.compile(u'[^\u0000-\uD7FF\uE000-\uFFFF]', re.UNICODE)

def filter_using_re(unicode_string):
    return re_pattern.sub(u'\uFFFD', unicode_string)

def filter_using_python(unicode_string):
    return u''.join(
        uc if uc < u'\ud800' or u'\ue000' <= uc <= u'\uffff' else u'\ufffd'
        for uc in unicode_string

def repeat_test(func, unicode_string):
    for i in xrange(repeat_count):
        tmp = func(unicode_string)

print '='*10 + ' filter_using_re() ' + '='*10'repeat_test(filter_using_re, test_string)')
print '='*10 + ' filter_using_python() ' + '='*10'repeat_test(filter_using_python, test_string)')

#print test_string.encode('utf8')
#print filter_using_re(test_string).encode('utf8')
#print filter_using_python(test_string).encode('utf8')

The results:

  • filter_using_re() did 515 function calls in 0.139 CPU seconds (0.138 CPU seconds at the sub() built-in)
  • filter_using_python() did 2097923 function calls in 3.413 CPU seconds (1.511 CPU seconds at the join() call and 1.900 CPU seconds evaluating the generator expression)
  • I did no test using itertools because... well... that solution, although interesting, was quite big and complex.


The RegEx solution was, by far, the fastest one.

5/23/2017 10:31:34 AM

Accepted Answer

Unicode characters in the ranges \u0000-\uD7FF and \uE000-\uFFFF will have 3 byte (or less) encodings in UTF8. The \uD800-\uDFFF range is for multibyte UTF16. I do not know python, but you should be able to set up a regular expression to match outside those ranges.

pattern = re.compile("[\uD800-\uDFFF].", re.UNICODE)
pattern = re.compile("[^\u0000-\uFFFF]", re.UNICODE)

Edit adding Python from Denilson Sá's script in the question body:

re_pattern = re.compile(u'[^\u0000-\uD7FF\uE000-\uFFFF]', re.UNICODE)
filtered_string = re_pattern.sub(u'\uFFFD', unicode_string)    
3/15/2014 12:55:54 PM

You may skip the decoding and encoding steps and directly detect the value of the first byte (8-bit string) of each character. According to UTF-8:

#1-byte characters have the following format: 0xxxxxxx
#2-byte characters have the following format: 110xxxxx 10xxxxxx
#3-byte characters have the following format: 1110xxxx 10xxxxxx 10xxxxxx
#4-byte characters have the following format: 11110xxx 10xxxxxx 10xxxxxx 10xxxxxx

According to that, you only need to check the value of only the first byte of each character to filter out 4-byte characters:

def filter_4byte_chars(s):
    i = 0
    j = len(s)
    # you need to convert
    # the immutable string
    # to a mutable list first
    s = list(s)
    while i < j:
        # get the value of this byte
        k = ord(s[i])
        # this is a 1-byte character, skip to the next byte
        if k <= 127:
            i += 1
        # this is a 2-byte character, skip ahead by 2 bytes
        elif k < 224:
            i += 2
        # this is a 3-byte character, skip ahead by 3 bytes
        elif k < 240:
            i += 3
        # this is a 4-byte character, remove it and update
        # the length of the string we need to check
            s[i:i+4] = []
            j -= 4
    return ''.join(s)

Skipping the decoding and encoding parts will save you some time and for smaller strings that mostly have 1-byte characters this could even be faster than the regular expression filtering.

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