如何将Tensorflow张量的维度(shape)作为整数值获取?
如何将Tensorflow张量的维度(shape)作为整数值获取?
假设我有一个Tensorflow张量。我如何将张量的维度(形状)作为整数值获取?我知道有两种方法,tensor.get_shape()和tf.shape(tensor),但我无法将形状值作为整数int32值获取。
例如,下面我创建了一个二维张量,我需要将行数和列数作为int32,以便我可以调用reshape()创建形状为(num_rows * num_cols, 1)的张量。然而,方法tensor.get_shape()返回的是Dimension类型的值,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 "
# 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.