Trying to count words in a string

Question

I\'m trying to analyze the contents of a string. If it has a punctuation mixed in the word I want to replace them with spaces.

For example, If Johnny.Appleseed!is:a*good

Answer 1

Simple loop based solution:

strs = "Johnny.Appleseed!is:a*good&farmer"
lis = []
for c in strs:
    if c.isalnum() or c.isspace():
        lis.append(c)
    else:
        lis.append(' ')

new_strs = "".join(lis)
print new_strs           #print 'Johnny Appleseed is a good farmer'
new_strs.split()         #prints ['Johnny', 'Appleseed', 'is', 'a', 'good', 'farmer']

Better solution:

Using regex:

>>> import re
>>> from string import punctuation
>>> strs = "Johnny.Appleseed!is:a*good&farmer"
>>> r = re.compile(r'[{}]'.format(punctuation))
>>> new_strs = r.sub(' ',strs)
>>> len(new_strs.split())
6
#using `re.split`:
>>> strs = "Johnny.Appleseed!is:a*good&farmer"
>>> re.split(r'[^0-9A-Za-z]+',strs)
['Johnny', 'Appleseed', 'is', 'a', 'good', 'farmer']