Numpy矩阵运算

Question

我要计算所有的下列值i和j:

M_ki = Sum[A_ij - A_ik - A_kj + A_kk, 1 <= j <= n]

如何在没有显式循环的情况下使用Numpy(Python)来完成它？

谢谢!

Answer 1

这是解决此类问题的一般策略.

首先,编写一个小脚本,将循环显式写入两个不同的函数,最后进行测试,确保两个函数完全相同:

import numpy as np
from numpy import newaxis

def explicit(a):
    n = a.shape[0]
    m = np.zeros_like(a)
    for k in range(n):
        for i in range(n):
            for j in range(n):
                m[k,i] += a[i,j] - a[i,k] - a[k,j] + a[k,k]
    return m

def implicit(a):
    n = a.shape[0]
    m = np.zeros_like(a)
    for k in range(n):
        for i in range(n):
            for j in range(n):
                m[k,i] += a[i,j] - a[i,k] - a[k,j] + a[k,k]
    return m

a = np.random.randn(10,10)
assert np.allclose(explicit(a), implicit(a), atol=1e-10, rtol=0.)

然后,通过编辑逐步向量化函数implicit,在每一步运行脚本以确保它们继续保持不变:

步骤1

def implicit(a):
    n = a.shape[0]
    m = np.zeros_like(a)
    for k in range(n):
        for i in range(n):
            m[k,i] = (a[i,:] - a[k,:]).sum() - n*a[i,k] + n*a[k,k]
    return m

第2步

def implicit(a):
    n = a.shape[0]
    m = np.zeros_like(a)
    m = - n*a.T + n*np.diag(a)[:,newaxis]
    for k in range(n):
        for i in range(n):
            m[k,i] += (a[i,:] - a[k,:]).sum()
    return m

第3步

def implicit(a):
    n = a.shape[0]
    m = np.zeros_like(a)
    m = - n*a.T + n*np.diag(a)[:,newaxis]
    m += (a.T[newaxis,...] - a[...,newaxis]).sum(1)
    return m

瞧瞧'!最后一个没有循环.为了矢量化这种方程式,广播是要走的路!

警告:确保这explicit是您想要矢量化的等式.我不确定是否j也应该总结不依赖的条款.