查找数组是否包含有一个不匹配的字符串

Question

我想说列表中有一个字符串只有一个不匹配

def check(list, s):
    n = len(list)
 
    # If the array is empty
    if (n == 0):
        return False
 
    for i in range(0, n, 1):
         
        # If sizes are same
        if (len(list[i]) != len(s)):
            continue
 
        diff = False
        for j in range(0, len(list[i]), 1):
            if (list[i][j] != s[j]):
 
                # If first mismatch
                if (diff == False):
                    diff = True
 
                # Second mismatch
                else:
                    diff = False
                    break
                 
        if (diff):
            return True
 
    return False

这段代码没问题，但运行速度很慢。如何使用字典使其更快？

Answer 1

您可以使用 zip() 将两个字符串的每个字符配对，并使用 sum() 来计算不匹配的数量。 any() 函数将允许您检查列表中每个项目的匹配：

def check(sList,string):
    return any(len(string)==len(s) and sum(a!=b for a,b in zip(s,string))==1 
               for s in sList)

check(["abcd","cdef","efgh"],"xfgh") # True
check(["abcd","cdef","efgh"],"abdc") # False

这将转换为以下基本 for 循环：

def check(sList,string):
    for s in sList:
        if len(string) !=len(s): continue
        if sum(a!=b for a,b in zip(s,string)) == 1:
            return True
    return False

[编辑] 为了使其运行得更快一点，字符串比较可以使用 zip() 作为一个迭代器，以便比较不需要转到每个字符串的末尾：

def check(sList,string):
    for s in sList:
        if len(string) !=len(s): continue
        mismatch = (a!=b for a,b in zip(s,string))  # non-matching iterator
        if not any(mismatch): continue              # at least one
        if any(mismatch): continue                  # no more after that
        return True
    return False