Tensorflow 中的笛卡尔积 [英] Cartesian Product in Tensorflow

问题描述

有没有像 itertools.product 那样在 Tensorflow 中做笛卡尔积的简单方法?我想获得两个张量元素的组合(a 和 b)，在 Python 中可以通过 itertools 作为 list(product(a, b)).我正在 Tensorflow 中寻找替代方案.

Is there any easy way to do cartesian product in Tensorflow like itertools.product? I want to get combination of elements of two tensors (a and b), in Python it is possible via itertools as list(product(a, b)). I am looking for an alternative in Tensorflow.

推荐答案

这里我将假设 a 和 b 都是一维张量.

I'm going to assume here that both a and b are 1-D tensors.

要获得两者的笛卡尔积，我将使用 tf.expand_dims 和 tf.tile 的组合:

To get the cartesian product of the two, I would use a combination of tf.expand_dims and tf.tile:

a = tf.constant([1,2,3]) 
b = tf.constant([4,5,6,7]) 

tile_a = tf.tile(tf.expand_dims(a, 1), [1, tf.shape(b)[0]])  
tile_a = tf.expand_dims(tile_a, 2) 
tile_b = tf.tile(tf.expand_dims(b, 0), [tf.shape(a)[0], 1]) 
tile_b = tf.expand_dims(tile_b, 2) 

cartesian_product = tf.concat([tile_a, tile_b], axis=2) 

cart = tf.Session().run(cartesian_product) 

print(cart.shape) 
print(cart) 

你最终得到一个 len(a) * len(b) * 2 张量，其中 a 和 b 元素的每个组合在最后一个维度中表示.

You end up with a len(a) * len(b) * 2 tensor where each combination of the elements of a and b is represented in the last dimension.

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