用于查找对的压缩矩阵函数

Condensed matrix function to find pairs(用于查找对的压缩矩阵函数)

问题描述

对于一组观察:

[a1,a2,a3,a4,a5]

它们的成对距离

d=[[0,a12,a13,a14,a15]
   [a21,0,a23,a24,a25]
   [a31,a32,0,a34,a35]
   [a41,a42,a43,0,a45]
   [a51,a52,a53,a54,0]]

以压缩矩阵形式给出(上面的上三角，从 scipy.spatial.distance.pdist 计算):

Are given in a condensed matrix form (upper triangular of the above, calculated from scipy.spatial.distance.pdist ):

c=[a12,a13,a14,a15,a23,a24,a25,a34,a35,a45]

问题是，鉴于我在压缩矩阵中有索引，是否有一个函数(最好在 python 中)f 来快速给出使用哪两个观测值来计算它们?

The question is, given that I have the index in the condensed matrix is there a function (in python preferably) f to quickly give which two observations were used to calculate them?

f(c,0)=(1,2)
f(c,5)=(2,4)
f(c,9)=(4,5)
...

我尝试了一些解决方案，但没有一个值得一提:(

I have tried some solutions but none worth mentioning :(

推荐答案

您可能会发现 triu_indices 很有用.喜欢，

You may find triu_indices useful. Like,

In []: ti= triu_indices(5, 1)
In []: r, c= ti[0][5], ti[1][5]
In []: r, c
Out[]: (1, 3)

注意索引从0开始.你可以随意调整它，例如:

Just notice that indices starts from 0. You may adjust it as you like, for example:

In []: def f(n, c):
   ..:     n= ceil(sqrt(2* n))
   ..:     ti= triu_indices(n, 1)
   ..:     return ti[0][c]+ 1, ti[1][c]+ 1
   ..:
In []: f(len(c), 5)
Out[]: (2, 4)

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