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

  • 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.

     

      December 8, 2020 3:12 PM IST
    0
  • 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.

      December 11, 2020 1:49 PM IST
    0

  • Another simple solution is to use map() as follows:
    tensor_shape = map(int, my_tensor.shape)

    This converts all the Dimension objects to int

      December 14, 2020 7:04 PM IST
    0
  • 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())
    
      January 7, 2022 12:39 PM IST
    0