张量的访问元素 [英] Access elements of a Tensor

问题描述

我有以下TensorFlow张量.

I have the following TensorFlow tensors.

tensor1 = tf.constant(np.random.randint(0,255, (2,512,512,1)), dtype='int32') #All elements in range [0,255]
tensor2 = tf.constant(np.random.randint(0,255, (2,512,512,1)), dtype='int32') #All elements in range [0,255]

tensor3 = tf.keras.backend.flatten(tensor1)
tensor4 = tf.keras.backend.flatten(tensor2)

tensor5 = tf.constant(np.random.randint(0,255, (255,255)), dtype='int32') #All elements in range [0,255]

我希望使用张量3和张量4中存储的值创建一个元组，并查询张量5中该元组给定位置的元素.例如，假设张量3中的第0个元素，即tensor3 [0 ] = 5和tensor4 [0] = 99. 因此，元组变为(5,99).我希望在张量5中查找元素(5,99)的值.我希望以批量处理的方式对Tensor3和Tensor4中的所有元素进行处理.那就是我不想遍历(len(Tensor3))范围内的所有值.为了达到这个目的，我做了以下工作.

I wish to use the values stored in tensor 3 and tensor 4 to make a tuple and query the element at position given by the tuple in tensor 5. For example, let's say 0th element in tensor 3, that is tensor3[0]=5 and tensor4[0]=99. So the tuple becomes (5,99). I wish to look up the value of element (5,99) in tensor 5. I wish to do it for all elements in Tensor3 and Tensor4 in a batch processing manner. That is I do not want to loop over all values in the range of (len(Tensor3)). I did the following to achieve this.

tensor6 = tensor5[tensor3[0],tensor4[0]]

但是tensor6的形状为(255,255)，正如我希望得到的形状的张量(len(tensor3)，len(tensor3))一样.我想在len(tensor3)中所有可能的位置评估tensor5.那是在(0,0),...(1000,1000),....(2000,2000),....我正在使用TensorFlow 1.12.0版本.我该如何实现?

But tensor6 has the shape (255,255) where as I was hoping to get a tensor of shape (len(tensor3),len(tensor3)). I wanted to evaluate tensor5 at all possible locations in len(tensor3). That is at (0,0),...(1000,1000),....(2000,2000),.... I am using TensorFlow version 1.12.0. How can I achieve this?

推荐答案

我设法在Tensorflow v 1.12中工作了一些，但请告诉我它是否是预期的代码:

I have managed to get something working in Tensorflow v 1.12, but do let me know if it is the expected code:

import tensorflow as tf
print(tf.__version__)
import numpy as np

tensor1 = tf.constant(np.random.randint(0,255, (2,512,512,1)), dtype='int32') #All elements in range [0,255]
tensor2 = tf.constant(np.random.randint(0,255, (2,512,512,1)), dtype='int32') #All elements in range [0,255]

tensor3 = tf.keras.backend.flatten(tensor1)
tensor4 = tf.keras.backend.flatten(tensor2)

tensor5 = tf.constant(np.random.randint(0,255, (255,255)), dtype='int32') #All elements in range [0,255]

elems = (tensor3, tensor4)
a = tf.map_fn(lambda x: tensor5[x[0], x[1]], elems, dtype=tf.int32)

print(tf.Session().run(a))

基于下面的注释，我想对代码中使用的map_fn添加一个说明.由于没有eager_execution，不支持for循环，因此map_fn等效于for循环.

Based on the comment below I'd like to add an explanation for the map_fn used in the code. Since for loops are not supported without eager_execution, map_fn is (sort of) equivalent to for loops.

A map_fn具有以下参数:operation_performed，input_arguments，optional_dtype.实际情况是，沿着input_arguments中的值的长度(必须包含一个可迭代的对象)运行for循环，然后对每个获得的值执行operation_performed.如需进一步的说明，请参阅 docs .

A map_fn has the following parameters: operation_performed, input_arguments, optional_dtype. What happens under the hood is that a for loop is run along the length of the values in input_arguments (which must contain an iterable object) and then for each value obtained operation_performed is performed. For further clarification please refer docs.

函数的参数名称是我解释它们的方式，正如我想了解的那样，官方文档中未给出. :)

The names given to the arguments of the function is my way of interpreting them, as I'd like understand it, and is not given in the official docs. :)

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