将上三角条目的平面列表复制到完整矩阵? [英] Copy flat list of upper triangle entries to full matrix?

问题描述

我在平面列表(连接的行)中具有对称矩阵的上三角条目(包括对角线)，并且我想使用它们来填充完整的矩阵，包括下三角.最快的方法是什么?

I have the upper triangle entries (including diagonal) of a symmetric matrix in a flat list (concatenated rows), and I want to use them to fill in the full matrix, including the lower triangle. What's the fastest way to do this?

这是我目前的做法.如此简单的操作似乎需要做很多工作.

Here is my current approach. Seems like a lot of work for such a simple operation.

import numpy as np
def utri2mat(utri,ntotal):
    iu1 = np.triu_indices(ntotal)
    ret = np.zeros([ntotal,ntotal])
    ret[iu1] = utri
    ret = ret + ret.transpose() - np.diag(ret.diagonal())
    return ret

推荐答案

这是我提名的一种更快，甚至可能更好的方法，用于根据平面值制作对称矩阵:

Here's my nomination for a faster, and possibly better, way to make a symmetric matrix from flat values:

def make_sym(val, n):
    # uses boolean mask
    # uses the same lower tri as np.triu
    mask = ~np.tri(5,k=-1,dtype=bool)
    out = np.zeros((n,n),dtype=val.dtype)
    out[mask] = val
    out.T[mask] = val
    return out

测试:

In [939]: val=np.arange(1,16)
In [940]: make_sym(val, 5)
Out[940]: 
array([[ 1,  2,  3,  4,  5],
       [ 2,  6,  7,  8,  9],
       [ 3,  7, 10, 11, 12],
       [ 4,  8, 11, 13, 14],
       [ 5,  9, 12, 14, 15]])

与其他答案一样，它使用out.T[]分配下部三角形.

Like the other answers it uses out.T[] to assign the lower triangle.

Warren的答案使用np.triu_indices，即where值.这种类型的索引比布尔掩码慢一点.

Warren's answer uses np.triu_indices, which are the where values. This type of indexing is a bit slower than boolean masking.

但是正如我提到的，Divakar使用的np.triu在早期的numpy版本(例如1.9)中不会返回布尔掩码.这就是促使我深入研究问题的原因.

But as I noted the np.triu that Divakar uses does not return a boolean mask in earlier numpy versions (e.g. 1.9). This is what prompted me to dig into the issue.

在1.10中，此函数被重写为:

In 1.10 this function was rewritten as:

mask = np.tri(*m.shape[-2:], k=k-1, dtype=bool)
return np.where(mask, np.zeros(1, m.dtype), m)

通过将where替换为~mask，我获得了一些速度.结果相同，但只是删除了一个中间步骤.

I gain a bit of speed by replacing the where with ~mask. Same result, but just cutting out an intermediate step.

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