有没有办法找出将字符串值列表转换为第二个字符串值列表所需的更改/编辑总数？

Question

我有两个列表：list1 和 list2（等长），由由字母“a”到“o”组成的相同长度的字符串值组成。

list1= ["gdmnbl","cfakdd","lkbhad",....]
list2= ["gamndl","bfakdd","lkbhad",....]

我需要 output 作为总编辑/更改的列表，需要将值从 list1 转换为 list2。 示例如下所示：

Taking into consideration the first value from list1 and list2 and comparing.
"gdmnbl" and "gamndl"

First character matches, so move on to next character.
Second character not a match, it takes 3 edits/changes to make "d" to "a". EXPLANATION:( "d" to "c", "c" to "b" and "b" to "a").
Third character, fourth character  are a match, so ignore.
Fifth character not a match, it takes 2 edits/changes to make "b" to "d". ("b" to c" and then "c" to "d".
Sixth character matches, so move on.

I want the output to be a list consisting of the sum of edits/changes. For the above example it is 3+2= 5.

所以 output 列表应该是这样的：

list3=[5,1,0.....]

有没有办法做到这一点？ 非常感谢您的帮助。 谢谢

Answer 1

如果您的意思是对同一 position 的字符之间的距离求和，您可以执行以下操作：

list3=[sum([abs(ord(i)-ord(j)) for i,j in zip(a,b)]) for a,b in zip(list1,list2)  ] 

list1和list2必须具有相同数量的字符串，并且字符串对的长度必须相等。

Answer 2

你可以尝试这样的事情：

def get_diff(s1,s2):
    d = 0
    s1 = s1.lower()
    s2 = s2.lower()
    for i in range(len(s1)):
        d = d+abs(ord(s1[i])-ord(s2[i]))
    return d

list1= ["gdmnbl","cfakdd","lkbhad"]
list2= ["gamndl","bfakdd","lkbhad"]
final_out = []
for i in range(len(list1)):
    d = get_diff(list1[i],list2[i])
    final_out.append(d)
print(final_out)