比较Python

Question

我希望得到以下内容：

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，并且它们仅由字符＃（它代表随机字符）不同，则函数返回True。 如果没有，我希望它返回False。

我应该在这个功能中改变什么？

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

您只检查第一个字符。如果第一个字符相同或者是True，则不应返回#，但是您应该继续查找第一个不匹配并仅在for循环之外返回True。 / p>

第二个问题是，在您的测试用例中，变量c永远不会是'#'，因为i是problem的字符，而c是solution。

的字符

def checkSolution(problem, solution):
    if len(problem) != len(solution): 
        return False
    for i, c in zip(problem, solution):
        if i != '#' and c != '#' and i != c :
            return False
    return True

Answer 2

现在你只测试长度和第一个字符匹配。

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，或使用all为您执行此操作。

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

或一行：

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

或者如果你真的疯了（读：你喜欢正则表达式），你可以这样做：

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

Answer 3

正如评论中所指出的那样，你的缩进是愚蠢的，应该修复。

if len(problem) == len(solution):
    # in the line below, 
    # 'i' will contain the next char from problem
    # 'c' will contain the next char from solution
    for i, c in zip(problem, solution):
        # in this line, if they're not equal, you return False
        # before you have a chance to look for the wildcard character
        if i != c:
            return False
        # ...and here you'd fail anyway, because you're testing 
        # the character from the solution string against the wildcard...
        if i == c or "#" == c:
            return True
# ...while in your test, you pass the wildcard in as part of the problem string.
print (checkSolution("a#c","abc"))

Answer 4

您的函数的一行版本：

def check_solution(problem, solution):
    return (len(problem) == len(solution) and
            all(ch==solution[i] for i, ch in enumerate(problem) if ch != '#'))

测试：

>>> check_solution("a#c", "abc")
True
>>> check_solution("a#c", "abd")
False