CODEWITHSUNDEEP
tensorflow
YohanRoth
YohanRoth

Reputation: 3253

Variable size mismatch between x.shape and tf.shape(x)?

I am trying to understand tf code and for this I am printing out shapes of tensors. For the following code

print(x.shape)
print(tf.shape(x))

I get output

(?, 32, 32, 3)
Tensor("input/Shape:0", shape=(4,), dtype=int32)

It does not make a lot of sense. Based on what I found online tf.shape(x) can be used to dynamically get the size for the batch. But it gives rather wrong output - 4. I am not sure where this (4,) is coming from and how to get the right value for my tensor.

Upvotes: 0

Views: 165

Answers (1)

giser_yugang
giser_yugang

Reputation: 6166

In fact, the two results are the same. 4 is the shape of (?,32,32,3).

x.shape() returns a tuple, and you can get shape without sess.run(). You can use as_list() to convert it into a list.

tf.shape(x) returns a tensor, and you need to run sess.run() to get the the actual number.

An example:

import tensorflow as tf
import numpy as np

x = tf.placeholder(shape=(None,32,32,3),dtype=tf.float32)
print(x.shape)
print(tf.shape(x))

dim = tf.shape(x)
dim0 = tf.shape(x)[0]

with tf.Session()as sess:
    dim,dim0 = sess.run([dim,dim0],feed_dict={x:np.random.uniform(size=(100,32,32,3))})
    print(dim)
    print(dim0)

#print
(?, 32, 32, 3)
Tensor("Shape:0", shape=(4,), dtype=int32)
[100  32  32   3]
100

Upvotes: 1

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