How to get Tensorflow tensor dimensions (shape) as int values?

Question

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.

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.

martianwars · Accepted Answer · 2017-05-06 19:14:56Z

147

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.

edited May 6, 2017 at 19:14

martianwars

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answered Nov 17, 2016 at 22:41

yuefengz

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6 Comments

stackoverflowuser2010 Over a year ago

Thanks, that lets me call and complete tf.reshape(), but I would really like to get num_rows and num_cols as integers for other operations.

yuefengz Over a year ago

Try tensor.get_shape().as_list()

stackoverflowuser2010 Over a year ago

For completeness, this code works: num_rows, num_cols = x.get_shape().as_list()

SherylHohman Over a year ago

Nice! I was using python int() to cast the results of x.get_shape(). ie num_rows=int(x.get_shape()[1]), num_cols=int(x.get_shape()[2]), etc.Yep, kinda a hacky to get around that pesky error, but it worked. Thanks for enlightening me to a better way :-)

EyesBear Over a year ago

In TF2.0, it seems not work: as_list() is not defined on an unknown TensorShape

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maazza · Accepted Answer · 2017-05-04 08:48:39Z

33

Another way to solve this is like this:

tensor_shape[0].value

This will return the int value of the Dimension object.

edited May 4, 2017 at 8:48

maazza

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answered May 4, 2017 at 8:26

tijmen Verhulsdonck

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Comments

user11530462 · Accepted Answer · 2020-02-03 12:30:19Z

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]

Anna · Accepted Answer · 2017-10-25 14:00:56Z

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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())

edited Oct 25, 2017 at 14:00

answered Oct 25, 2017 at 7:22

Anna

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1 Comment

rayryeng Over a year ago

Very inelegant. How does this add to the already provided answers?

Avandale · Accepted Answer · 2020-07-06 20:15:31Z

Another simple solution is to use map() as follows:

tensor_shape = map(int, my_tensor.shape)

This converts all the Dimension objects to int

thushv89 · Accepted Answer · 2019-10-09 01:25:07Z

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)

Tony · Accepted Answer · 2024-02-03 22:52:03Z

Using this line of code:

tensor_shape[0].value

It will give you the shape of the tensor formatted as an integer.

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