Question

我正在试图找出改变所需的更改次数第一个字符串，s1到s2。我写了一个函数来做到这一点但到目前为止我所拥有的问题是，如果前两个字符是不同的，它会返回1而不是查看字符串的其余部分。

这是我到目前为止所拥有的：

def num_of_changes(s1, s2):
    if not s1:
        return len(s2)
    if not s2:
        return len(s1)
    length = 0
    # if the first char in both the string are the same, then call the function
    # again on the rest of the string after the first char.
    # so if s1 = cat and s2 = cbt since the first char on both strings is
    # the same it call the function again on the rest, the rest in this case
    # being s1 = at and s2 = bt
    if s1[0] == s2[0]:
        num_of_changes(s1[length+1:], s2[length+1:])
    # if the first characters on both the strings aren't the same, then
    # increase the length by 1, and call the function again on the rest of the
    # string after length + 1.
    # so if s1 = cat and s2 = bat, since the first char from both the strings
    # arent the same it calls the function on length+1. Since the length
    # increased to 1 becase the first char isn't the same, the it would be
    # length=1, so 1+1
    # Therefore, it would call the function on s1 = at and s2 = at
    else:
        length += 1
        num_of_changes(s1[length+1:], s2[length+1:])
    # returns the number stored in the length variable.
    return length

Answer 1

您忽略了递归调用的返回值。它们返回一个长度，但这些值被丢弃在地板上。如果我们删除这些递归调用，那么，您的代码等同于：

length = 0
if s1[0] == s2[0]:
    pass
else:
    length += 1
return length

如您所见，它只会返回0或1。

您需要获取递归值，然后在第一个字符匹配时有条件地添加1。

if s1[0] == s2[0]:
    return num_of_changes(s1[length+1:], s2[length+1:])
else:
    return num_of_changes(s1[length+1:], s2[length+1:]) + 1

通过将length+1更改为1并合并两个return语句，可以将其简化为：

return num_of_changes(s1[1:], s2[1:]) + (1 if s1[0] == s2[0] else 0)

它可能有点密集，但这使得你很清楚你要返回的是什么：它是递归调用的值，加上s1[0] == s2[0]时的值。

Answer 2

您要查找的功能称为两个字符串之间的编辑距离或Levenshtein距离。您可以在python-Levenshtein包中找到它。通过以下方式安装：

$ sudo pip install python-Levenshtein

并像这样使用它：

>>> import Levenshtein
>>> Levenshtein.distance('abba', 'baba')
2

将字符串转换为另一个字符串所需的更改次数

2 个答案: