一维1D阵列和3D阵列的高效乘积-NumPy [英] Efficient product of 1D array and 3D array along one dimension - NumPy

问题描述

我有两个numpy数组:

I have two numpy arrays:

• 一个称为t的形状为(70L，)的一维数组，其元素名为ti.
• 一个名为I的3D数组，形状为(70L，1024L，1024L)，每个元素称为Ii.因此，ii的尺寸为(1024L，1024L)

我想沿第一维制作两个数组的乘积，即:

I would like to make a product of the two array along the first dimension, i.e.:

tI = t1*I1,t2*I2,...,tN*IN

例如，再次获得一个新的维度数组(70L，1024L，1024L)，然后沿第一个维度取和，以获得一个维度数组(1024L，1024L):

such as to obtain again a new array of dimension (70L, 1024L, 1024L) and then take the sum along the first dimension in order to obtain an array of dimension (1024L, 1024L):

tsum = t1*I1 + t2*I2 + ... +tN*IN

目前，我对执行以下操作感到满意:

For the moment I am satisfied with doing the following:

tI = np.asarray([t[i]*I[i,:,:] for i in range(t.shape[0])])
tsum = np.sum(tI,axis=0)

但是随着数组尺寸的增加，它会有点慢. 我想知道是否存在一个numpy或scipy函数，该函数针对该特定任务进行了优化?

But it is going to be a bit slow is the dimensions of my array are increasing. I was wondering if there exist a numpy or scipy function, more optimized for that particular task?

首先感谢任何链接或信息.

Thanks in advance of any link or information.

格雷格

推荐答案

您可以使用

You can use np.tensordot -

np.tensordot(t,I, axes=([0],[0]))

您还可以使用 np.einsum -

You can also use np.einsum -

np.einsum('i,ijk->jk',t,I)

运行时测试和输出验证-

Runtime test and output verification -

In [21]: def original_app(t,I):
    ...:     tI = np.asarray([t[i]*I[i,:,:] for i in range(t.shape[0])])
    ...:     tsum = np.sum(tI,axis=0)
    ...:     return tsum
    ...: 

In [22]: # Inputs with random elements
    ...: t = np.random.rand(70,)
    ...: I = np.random.rand(70,1024,1024)
    ...: 

In [23]: np.allclose(original_app(t,I),np.tensordot(t,I, axes=([0],[0])))
Out[23]: True

In [24]: np.allclose(original_app(t,I),np.einsum('i,ijk->jk',t,I))
Out[24]: True

In [25]: %timeit np.tensordot(t,I, axes=([0],[0]))
1 loops, best of 3: 110 ms per loop

In [26]: %timeit np.einsum('i,ijk->jk',t,I)
1 loops, best of 3: 201 ms per loop

这篇关于一维1D阵列和3D阵列的高效乘积-NumPy的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！