Spearman的脚掌距离(以R为底) [英] Spearman's footrule distance with base R

问题描述

给出两个排列:

v1
[1] 4 3 1 5 2

v2
[1] 2 3 4 5 1

如何计算基数为R的Spearman的尺距(所有元素的总位移)? (可灵活更改大小为n的任意两个)

How do you compute the Spearman's footrule distance (total displacement of all elements) with base R? (flexible for any two permutations of size n)

例如，对于这两个向量，如下所示:

For example, for these two vectors, it's as follows:

1从v1移到v2
2个位置 2从v1移到v2
的4个位置 3从v1移到v2
的0个位置 4从v1移到v2
的2个位置 5从v1移到v2

1 is moved 2 places from v1 to v2
2 is moved 4 places from v1 to v2
3 is moved 0 places from v1 to v2
4 is moved 2 places from v1 to v2
5 is moved 0 places from v1 to v2

总距离为:2+4+0+2+0 = 8

推荐答案

以下是使用sapply，which和sum的方法:

Here is a method using sapply, which, and sum:

sum(sapply(seq_along(v1), function(i) abs(i - (which(v2 == v1[i])))))

在这里，我们沿v1的索引移动，并计算当前索引中元素的索引距其在v2中的位置的距离.然后将它们汇总在一起.

Here, we move along the indices of v1 and calculate the distance of the index of the element in the current index from its position in v2. These are then summed together.

我怀疑评论中类似@alexis_laz的解决方案可能会提高计算效率.

I suspect something along the lines of @alexis_laz's solution in the comments may have greater computational efficiency.

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