tensorflow : How to outer product when the tensor's first dim is None?

Question

I have a tensor whose shape is None,10 . I want to get a outer product result whose shape is None,100 or None,10,10 . Here is my code :

# output'shape is None,10
output = tf.keras.layers.Concatenate()(encoded_feature_list)
# wrong
cross_output = tf.keras.layers.Lambda(lambda x:tf.linalg.matmul(x,x,transpose_a=True))(output)
cross_output = tf.keras.layers.Flatten()(cross_output)

Does this answer your question? [Outer product in tensorflow](https://stackoverflow.com/questions/33858021/outer-product-in-tensorflow) — David, Nov 30 '20 at 06:03
@DavidS no , it's not . its first dim isn't ```None``` ...... — DachuanZhao, Nov 30 '20 at 06:12

David · Accepted Answer · 2020-11-30T06:56:50.980

1

The answer I provided does have the answer, but not the immediate one!

Given the example you gave in your comment to @Meow Cat 2012:

import tensorflow as tf
import numpy as np

a = np.array([[1,2,3.0],[4,5,6.0]]
res = tf.einsum('ki,kj->kij',a,a)
print(res.shape)  # TensorShape([2, 3, 3])

tf.einsum() function will compute the outer product of two tensors.

Another solution using tf.linalg.matmul is (@DachuanZhao pointed it out):

import tensorflow as tf
import numpy as np
    
a = np.array([[1,2,3.0],[4,5,6.0]]
res = tf.linalg.matmul(tf.expand_dims(a, axis=-1),tf.expand_dims(a, axis=1))
print(res.shape)  # (2, 3, 3)

edited Nov 30 '20 at 06:56

answered Nov 30 '20 at 06:40

David

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Yes , I have found anther solution : – DachuanZhao Nov 30 '20 at 06:53
```tf.linalg.matmul(tf.expand_dims(a, axis=-1),tf.expand_dims(a, axis=1))``` – DachuanZhao Nov 30 '20 at 06:54
@DachuanZhao Cool, happy you managed. I will add this solution to my answer for future users – David Nov 30 '20 at 06:55

score 0 · Answer 2 · answered Nov 30 '20 at 05:58

0

Already answered: https://stackoverflow.com/a/51323886/9399618

According to the documentation:

Additionally outer product is supported by passing axes=0

Tested as follow:

a = np.array([1,2,3,4,5,6,7,8,9,10.0])
tf.tensordot(a, a, axes=0)
# Got <tf.Tensor: shape=(10, 10), dtype=float64, numpy=xxx>

answered Nov 30 '20 at 05:58

Meow Cat 2012

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This is coeerct dim=1 , but get wrong result when dim=2 . – DachuanZhao Nov 30 '20 at 06:05
When ```a = np.array([[1,2,3.0],[4,5,6.0]])``` , ```tf.tensordot(a, a, axes=0)``` get the shape ```(2,3,2,3)``` , not ```(2,3,3)``` or ```(2,9)``` – DachuanZhao Nov 30 '20 at 06:07

tensorflow : How to outer product when the tensor's first dim is None?

2 Answers2