Finding all numbers in string and replacing with ''

[Bouzy]

I have a list of words and am trying to replace all the numbers in my list with whitespace.

for word in words:
      numbers = re.search('[0-9]+', word)
      word = clearup(word)
      if word in dictionary:
         pass
      else:
         print word     

def clearup(tor):
 
   numbers = re.search('[0-9]+', tor)
   return tor.replace('%s' % numbers, '')\
   .replace('.', '')\
   .replace(',', '')\
   .replace(',', '')\
   .replace(':', '')\
   .replace('!', '')\
   .replace('@', '')\
   .replace('#', '')\
   .replace('$', '')\
   .replace('%', '')\
   .replace('(', '')\
   .replace(')', '')\
   .replace('?', '')\
   .replace('-', ' ')\
   .replace(';', '')\
   .lower()

I am not getting any errors, but when the words are printed, all of the numbers in in list words are printed. So it doesn't give me any errors but it doesn't work. What would I need to do to accomplish my task?

[bvdet]

Try this:

import re, string
 
def clearup(s, chars):
    return re.sub('[%s]' % chars, '', s).lower()
 
s = 'This is %a t1e22st !st4ring6 w.it6h 87embed766ded punct,:ua-tion and nu=mbe]rS6.'
 
print clearup(s, string.punctuation+string.digits)
 

Output:

>>> this is a test string with embedded punctuation and numbers

[Bouzy]

Thanks! That worked exactly as needed.

[markus314]

Or if you use python 2.6 and s is not a unicode string:

s.translate(None, string.punctuation+string.digits).lower()
 

which is approx. 12 times faster than a regex.