如何有效使用strip（）函数

Question

Can you please tell me why the strip() function does not work?

str1= 'aaaadfffdswefoijeowji'

def char_freq():
    for x in range (0, len(str1)):
        sub = str1[x]
        print 'the letter',str1[x],'appearence in the sentence=', str1.count(sub, 0,len(str1))
        str1.strip(str1[x])

def main():
    char_freq()

main()

Answer 1

.strip() is working just fine, but strings are immutable. str.strip() returns the new stripped string:

>>> str1 = 'foofoof'
>>> str1.strip('f')
'oofoo'
>>> str1
'foofoof'

You are ignoring the return value. If you do store the altered string, however, your for loop will run into an IndexError , as the string will be shorter the next iteration:

>>> for x in range (0, len(str1)):
...     str1 = str1.strip(str1[x])
... 
Traceback (most recent call last):
  File "<stdin>", line 2, in <module>
IndexError: string index out of range

To count strings, don't str.strip() ; that just removes characters from the start and end of a string, not in the middle. You could use str.replace(character, '') but that would be inefficient too; but combined with a while loop to avoid the IndexError problem that'd look like:

while str1:
    c = str1[0]
    print 'the letter {} appearence in the sentence={}'.format(c, str1.count(c))
    str1 = str1.replace(c, '')

Much easier would be to just use a collections.Counter() object :

from collections import Counter

freq = Counter(str1)
for character, count in freq.most_common():
    print '{} appears {} times'.format(character, count)

Without a dedicated Counter object, you could use a dictionary to count characters instead:

freq = {}
for c in str1:
    if c not in freq:
        freq[c] = 0
    freq[c] += 1

for character, count in freq.items():
    print '{} appears {} times'.format(character, count)

where freq then holds character counts after the loop.