如何得到张量张量尺寸(形状)作为整数值？

假设我有张量流张量。如何得到张量的尺寸(形状)作为整数值？我知道有两个方法，tensor.get_shape()和 tf.shape(tensor)，但是我不能得到整数 int32的形状值。

例如，下面我已经创建了一个2-D 张量，我需要得到行数和列数作为 int32，这样我就可以调用 reshape()来创建一个形状为 (num_rows * num_cols, 1)的张量。但是，方法 tensor.get_shape()返回的值为 Dimension类型，而不是 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.
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最佳答案

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

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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Another way to solve this is like this:

tensor_shape[0].value

This will return the int value of the Dimension object.

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for a 2-D tensor, you can get the number of rows and columns as int32 using the following code:

rows, columns = map(lambda i: i.value, tensor.get_shape())
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In later versions (tested with TensorFlow 1.14) there's a more numpy-like way to get the shape of a tensor. You can use tensor.shape to get the shape of the tensor.

tensor_shape = tensor.shape
print(tensor_shape)
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2.0 Compatible Answer: In Tensorflow 2.x (2.1), you can get the dimensions (shape) of the tensor as integer values, as shown in the Code below:

Method 1 (using tf.shape):

import tensorflow as tf
c = tf.constant([[1.0, 2.0, 3.0], [4.0, 5.0, 6.0]])
Shape = c.shape.as_list()
print(Shape)   # [2,3]

Method 2 (using tf.get_shape()):

import tensorflow as tf
c = tf.constant([[1.0, 2.0, 3.0], [4.0, 5.0, 6.0]])
Shape = c.get_shape().as_list()
print(Shape)   # [2,3]
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Another simple solution is to use map() as follows:

tensor_shape = map(int, my_tensor.shape)

This converts all the Dimension objects to int


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