比较字符串python中的字符

Question

I'm trying to compare the characters from 2 separate strings, the idea is that i will return a value corresponding to how many characters the strings both share. for example, if string one was 'mouse' and string 2 was 'house'. They would share 4/5 characters. its important to note that they only share a character if it is in the same 'index position'

def compareWords(word1, word2):
result = 0
if word1[0] in word2[0]:
    result += 1
if word1[1] in word2[1]:
    result += 1
if word1[2] in word2[2]:
    result += 1
if word1[3] in word2[3]:
    result += 1
if word1[4] in word2[4]:
    result += 1
if word1[5] in word2[5]:
    result += 1
    print result, '/5'

Answer 1

zip and sum :

a,b = "house", "mouse"

print(sum(s1 == s2 for s1, s2 in zip(a, b)))
4

zipping will pair the characters at the same index, then summing how many times s1 == s2 will give you the count of matching chars:

In [1]: a,b = "house", "mouse"

In [2]: zip(a, b)
Out[2]: [('h', 'm'), ('o', 'o'), ('u', 'u'), ('s', 's'), ('e', 'e')]

The only thing that is unclear is what you use as the out of if the strings are of different lengths.

If you did want the matches and the sum you can still use the same logic:

def paired(s1, s2):
    sm, index_ch = 0, []
    for ind, (c1, c2) in enumerate(zip(s1, s2)):
        if c1 == c2:
            sm += 1
            index_ch.append((ind, c1))
    return index_ch, sm

index_char, sm = paired("house", "mouse")

print(index_char, sm)

Output:

([(1, 'o'), (2, 'u'), (3, 's'), (4, 'e')], 4)

Answer 2

If you want to preserve the position and character of the matches, you can enumerate the strings, then calculate the intersection of the sets of the resulting tuples. If you don't want to preserve any information about the nature of the matches, I think Padraic's answer is better.

Demo:

>>> s1 = 'hello world'
>>> s2 = 'xelxx worxx'
>>> same = set(enumerate(s1)).intersection(enumerate(s2))
>>> same
set([(7, 'o'), (2, 'l'), (1, 'e'), (8, 'r'), (6, 'w'), (5, ' ')])
>>> len(same)
6