简明矩阵功能找到对的 [英] Condensed matrix function to find pairs

问题描述

有关的一组观察：

[a1,a2,a3,a4,a5]

他们的成对距离

their pairwise distances

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]]

在一个浓缩的矩阵形式给定（以上的上三角，从计算的 scipy.spatial.distance.pdist ）：

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

现在的问题是，因为我有指数在凝聚基体有一个函数（在Python preferably）的 F 即可快速给这两个观察来计算呢？

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)
...

我已经尝试了一些解决方案，但没有任何值得一提的：（

I have tried some solutions but none worth mentioning :(

推荐答案

您可能会发现<一href="http://docs.scipy.org/doc/numpy-1.5.x/reference/generated/numpy.triu_indices.html#numpy.triu_indices"相对=nofollow> triu_indices 的有用。像，

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)

只需注意指数从0开始，你喜欢，你可以调整它，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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