I have a few lists that havethe same IDs that are strings. They are as follows:

我有几个列表，它们的ID与字符串相同。这些建议如下：

list1 = ["1", "2", "3", "4", "5"]
list2 = ["1", "2", "3", "5", "4"]
list3 = ["1", "5", "4", "3", "2"]
list4 = ["4", "2", "5", "3", "1"]

What measure can I use to determine the lists that are closest to each other here in terms of order? Ideally list1 and list2 should be the closest here.

我可以使用什么方法来确定在这里顺序上彼此最接近的列表？理想情况下，list1和list2应该是这里最接近的。

Does the spearman correlation make sense here?

斯皮尔曼的相关性在这里有意义吗？

Edit distance seems to be a good candidate for such a metric.

编辑距离似乎是这样一个指标的一个很好的候选者。

from typing import List


def calcEditDistance(lhs: List[str], rhs: List[str]) -> int:
    '''
    Dynamic programming
    dp[i][j] = minimum number of operations to convert lhs[0:i] to rhs[0:j]
    '''
    m = len(lhs)
    n = len(rhs)
    dp = [[0] * (n + 1) for _ in range(m + 1)]

    for i in range(1, m + 1):
        dp[i][0] = i

    for j in range(1, n + 1):
        dp[0][j] = j

    for i in range(1, m + 1):
        for j in range(1, n + 1):
            if lhs[i - 1] == rhs[j - 1]:
                dp[i][j] = dp[i - 1][j - 1]
            else:
                dp[i][j] = min(dp[i - 1][j - 1], dp[i - 1]
                               [j], dp[i][j - 1]) + 1

    return dp[m][n]


list1 = ["1", "2", "3", "4", "5"]
list2 = ["1", "2", "3", "5", "4"]
list3 = ["1", "5", "4", "3", "2"]
list4 = ["4", "2", "5", "3", "2"]

res = calcEditDistance(list1, list2)
print(f"dis[1, 2] = {res}")

res = calcEditDistance(list1, list3)
print(f"dis[1, 3] = {res}")

res = calcEditDistance(list1, list4)
print(f"dis[1, 4] = {res}")

res = calcEditDistance(list2, list3)
print(f"dis[2, 3] = {res}")

res = calcEditDistance(list2, list4)
print(f"dis[2, 4] = {res}")

res = calcEditDistance(list3, list4)
print(f"dis[3, 4] = {res}")

prints

指纹

dis[1, 2] = 2
dis[1, 3] = 4
dis[1, 4] = 4
dis[2, 3] = 4
dis[2, 4] = 4
dis[3, 4] = 3

which matches your intuition.

这与你的直觉相符。

Note that in the Python code I use the Levenshtein distance where insert, delete, and replace operations are allowed. You can, of course, use other types of edit distance.

请注意，在Python代码中，我使用了允许执行INSERT、DELETE和REPLACE操作的Levenshtein距离。当然，您可以使用其他类型的编辑距离。

The comments have clarified that you're looking for something where two lists are closer together the more elements they have in the same positions. In that case, just count how many elements they have in different positions:

这些评论澄清了你正在寻找的东西，两个列表离得越近，它们在相同位置的元素就越多。在这种情况下，只需数一数它们在不同位置有多少元素：

def distance(l1, l2):
    return sum(1 for i, j in zip(l1, l2) if i != j)