在theano中使用索引矩阵索引张量?

问题描述

我有一个theano张量A，使得A.shape =(40，20，5)，一个theano矩阵B，使得B.shape =(40，20).我可以执行单行运算来获得矩阵C，其中C.shape =(40，20)并且C(i，j)= A [i，j，B [i，j]]用theano语法?

本质上，我想使用B作为索引矩阵；使用theano最有效/最优雅的方法是什么?

解决方案

您可以在numpy中执行以下操作:

import numpy as np

A = np.arange(4 * 2 * 5).reshape(4, 2, 5)
B = np.arange(4 * 2).reshape(4, 2) % 5

C = A[np.arange(A.shape[0])[:, np.newaxis], np.arange(A.shape[1]), B]

因此，您可以在theano中执行相同的操作:

import theano
import theano.tensor as T

AA = T.tensor3()
BB = T.imatrix()

CC = AA[T.arange(AA.shape[0]).reshape((-1, 1)), T.arange(AA.shape[1]), BB]

f = theano.function([AA, BB], CC)

f(A.astype(theano.config.floatX), B)

I have a theano tensor A such that A.shape = (40, 20, 5) and a theano matrix B such that B.shape = (40, 20). Is there a one-line operation I can perform to get a matrix C, where C.shape = (40, 20) and C(i,j) = A[i, j, B[i,j]] with theano syntax?

Essentially, I want to use B as an indexing matrix; what is the most efficient/elegant to do this using theano?

解决方案

You can do the following in numpy:

import numpy as np

A = np.arange(4 * 2 * 5).reshape(4, 2, 5)
B = np.arange(4 * 2).reshape(4, 2) % 5

C = A[np.arange(A.shape[0])[:, np.newaxis], np.arange(A.shape[1]), B]

So you can do the same thing in theano:

import theano
import theano.tensor as T

AA = T.tensor3()
BB = T.imatrix()

CC = AA[T.arange(AA.shape[0]).reshape((-1, 1)), T.arange(AA.shape[1]), BB]

f = theano.function([AA, BB], CC)

f(A.astype(theano.config.floatX), B)

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