python - 比较Python中两个字符串的相似度

Question

我想要以下内容:

Input: ("a#c","abc")
Output: True

Input:("a#c","abd")
Desired Output: False
Real Output: True

因此，如果两个字符串具有相同的长度并且它们仅差字符 #(代表随机字符)，则函数返回 True。如果不是，我希望它返回 False。

我应该在此函数中更改什么？

def checkSolution(problem, solution):

    if len(problem) != len(solution): 
        return False

    if len(problem) == len(solution):
        for i, c in zip(problem, solution):
            if i != c:
                return False

            if i == c or "#" == c:
                return True

print (checkSolution("a#c","abc"))

print (checkSolution("a#c","abd"))

Answer 1

现在您只测试长度和第一个字符是否匹配。

for i, c in zip(problem, solution):
    if i != c:
        # that's the first set of chars, but we're already returning??
        return False

    if i == c or "#" == c:
        # wildcard works here, but already would have failed earlier,
        # and still an early return if it IS true!
        return True

相反，您需要遍历整个字符串并返回结果，或者使用 all 为您完成。

if len(problem) == len(solution):
    for p_ch, s_ch in zip(problem, solution):
        if p_ch == "#":  # wildcard
            continue  # so we skip testing this character
        if p_ch != s_ch:
            return False  # This logic works now that we're skipping testing the "#"
    else:  # if we fall off the bottom here
        return True  # then it must be equal
else:
    return False

或在一行中:

return len(problem) == len(solution) and \
       all(p_ch==s_ch or p_ch=="#" for p_ch, s_ch in zip(problem, solution)

或者如果你真的很疯狂(阅读:你太喜欢正则表达式了)，你可以这样做:

def checkSolution(problem, solution):
    return re.match("^" + ".".join(map(re.escape, problem.split("#"))) + "$",
                    solution)