如何获取Tensorflow张量尺寸(形状)作为int值? [英] How to get Tensorflow tensor dimensions (shape) as int values?

问题描述

假设我有一个Tensorflow张量.如何获取张量的尺寸(形状)为整数值?我知道有两种方法，tensor.get_shape()和tf.shape(tensor)，但是我不能将形状值作为整数int32值.

Suppose I have a Tensorflow tensor. How do I get the dimensions (shape) of the tensor as integer values? I know there are two methods, tensor.get_shape() and tf.shape(tensor), but I can't get the shape values as integer int32 values.

例如，下面我创建了一个二维张量，我需要将行数和列数设为int32，以便可以调用reshape()来创建形状为(num_rows * num_cols, 1)的张量.但是，方法tensor.get_shape()返回的值为Dimension类型的值，而不是int32.

For example, below I've created a 2-D tensor, and I need to get the number of rows and columns as int32 so that I can call reshape() to create a tensor of shape (num_rows * num_cols, 1). However, the method tensor.get_shape() returns values as Dimension type, not int32.

import tensorflow as tf
import numpy as np

sess = tf.Session()    
tensor = tf.convert_to_tensor(np.array([[1001,1002,1003],[3,4,5]]), dtype=tf.float32)

sess.run(tensor)    
# array([[ 1001.,  1002.,  1003.],
#        [    3.,     4.,     5.]], dtype=float32)

tensor_shape = tensor.get_shape()    
tensor_shape
# TensorShape([Dimension(2), Dimension(3)])    
print tensor_shape    
# (2, 3)

num_rows = tensor_shape[0] # ???
num_cols = tensor_shape[1] # ???

tensor2 = tf.reshape(tensor, (num_rows*num_cols, 1))    
# Traceback (most recent call last):
#   File "<stdin>", line 1, in <module>
#   File "/usr/local/lib/python2.7/site-packages/tensorflow/python/ops/gen_array_ops.py", line 1750, in reshape
#     name=name)
#   File "/usr/local/lib/python2.7/site-packages/tensorflow/python/framework/op_def_library.py", line 454, in apply_op
#     as_ref=input_arg.is_ref)
#   File "/usr/local/lib/python2.7/site-packages/tensorflow/python/framework/ops.py", line 621, in convert_to_tensor
#     ret = conversion_func(value, dtype=dtype, name=name, as_ref=as_ref)
#   File "/usr/local/lib/python2.7/site-packages/tensorflow/python/framework/constant_op.py", line 180, in _constant_tensor_conversion_function
#     return constant(v, dtype=dtype, name=name)
#   File "/usr/local/lib/python2.7/site-packages/tensorflow/python/framework/constant_op.py", line 163, in constant
#     tensor_util.make_tensor_proto(value, dtype=dtype, shape=shape))
#   File "/usr/local/lib/python2.7/site-packages/tensorflow/python/framework/tensor_util.py", line 353, in make_tensor_proto
#     _AssertCompatible(values, dtype)
#   File "/usr/local/lib/python2.7/site-packages/tensorflow/python/framework/tensor_util.py", line 290, in _AssertCompatible
#     (dtype.name, repr(mismatch), type(mismatch).__name__))
# TypeError: Expected int32, got Dimension(6) of type 'Dimension' instead.

推荐答案

要获取形状为整数列表，请执行tensor.get_shape().as_list().

To get the shape as a list of ints, do tensor.get_shape().as_list().

要完成您的tf.shape()通话，请尝试tensor2 = tf.reshape(tensor, tf.TensorShape([num_rows*num_cols, 1])).或者，您可以直接进行tensor2 = tf.reshape(tensor, tf.TensorShape([-1, 1]))可以推断其第一维度的操作.

To complete your tf.shape() call, try tensor2 = tf.reshape(tensor, tf.TensorShape([num_rows*num_cols, 1])). Or you can directly do tensor2 = tf.reshape(tensor, tf.TensorShape([-1, 1])) where its first dimension can be inferred.

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