Replace non-ASCII characters with a single space

Question

Your ''.join() expression is filtering, removing anything non-ASCII; you could use a conditional expression instead:

return ''.join([i if ord(i) < 128 else ' ' for i in text])

This handles characters one by one and would still use one space per character replaced.

Your regular expression should just replace consecutive non-ASCII characters with a space:

re.sub(r'[^\x00-\x7F]+',' ', text)

Note the + there.

For you the get the most alike representation of your original string I recommend the unidecode module:

# python 2.x:
from unidecode import unidecode
def remove_non_ascii(text):
    return unidecode(unicode(text, encoding = "utf-8"))

Then you can use it in a string:

remove_non_ascii("Ceñía")
Cenia

For character processing, use Unicode strings:

PythonWin 3.3.0 (v3.3.0:bd8afb90ebf2, Sep 29 2012, 10:57:17) [MSC v.1600 64 bit (AMD64)] on win32.
>>> s='ABC马克def'
>>> import re
>>> re.sub(r'[^\x00-\x7f]',r' ',s)   # Each char is a Unicode codepoint.
'ABC  def'
>>> b = s.encode('utf8')
>>> re.sub(rb'[^\x00-\x7f]',rb' ',b) # Each char is a 3-byte UTF-8 sequence.
b'ABC      def'

But note you will still have a problem if your string contains decomposed Unicode characters (separate character and combining accent marks, for example):

>>> s = 'mañana'
>>> len(s)
6
>>> import unicodedata as ud
>>> n=ud.normalize('NFD',s)
>>> n
'mañana'
>>> len(n)
7
>>> re.sub(r'[^\x00-\x7f]',r' ',s) # single codepoint
'ma ana'
>>> re.sub(r'[^\x00-\x7f]',r' ',n) # only combining mark replaced
'man ana'

If the replacement character can be '?' instead of a space, then I'd suggest result = text.encode('ascii', 'replace').decode():

"""Test the performance of different non-ASCII replacement methods."""


import re
from timeit import timeit


# 10_000 is typical in the project that I'm working on and most of the text
# is going to be non-ASCII.
text = 'Æ' * 10_000


print(timeit(
    """
result = ''.join([c if ord(c) < 128 else '?' for c in text])
    """,
    number=1000,
    globals=globals(),
))

print(timeit(
    """
result = text.encode('ascii', 'replace').decode()
    """,
    number=1000,
    globals=globals(),
))

Results:

0.7208260721400134
0.009975979187503592

What about this one?

def replace_trash(unicode_string):
     for i in range(0, len(unicode_string)):
         try:
             unicode_string[i].encode("ascii")
         except:
              #means it's non-ASCII
              unicode_string=unicode_string[i].replace(" ") #replacing it with a single space
     return unicode_string

As a native and efficient approach, you don't need to use ord or any loop over the characters. Just encode with ascii and ignore the errors.

The following will just remove the non-ascii characters:

new_string = old_string.encode('ascii',errors='ignore')

Now if you want to replace the deleted characters just do the following:

final_string = new_string + b' ' * (len(old_string) - len(new_string))