将数组打包成张量的下三角

Question

我想包形状的阵列(..., n * (n - 1) / 2)与形状的张量的下三角部分(..., n, n)，其中...表示一个任意的形状。在 numpy 中，我会将它实现为

import numpy as np

# Create the array to store data in
arbitrary_shape = (10, 11, 12)
n = 5
target = np.zeros(arbitrary_shape + (n, n))
# Create the source array
source = np.random.normal(0, 1, arbitrary_shape + (n * (n - 1) / 2,))
# Create indices and set values
u, v = np.tril_indices(n, -1)
target[..., u, v] = source
# Check that everything went ok
print target[0, 0, 0]

到目前为止，我已经能够实现在使用的组合tensorflow类似的东西transpose，reshape和scatter_update但感觉笨拙。

import tensorflow as tf

# Create the source array
source = np.random.normal(0, 1, (n * (n - 1) / 2,) + arbitrary_shape)

sess = tf.InteractiveSession()

# Create a flattened representation
target = tf.Variable(np.zeros((n * n,) + arbitrary_shape))
# Assign the values
target = tf.scatter_update(target, u * n + v, source)
# Reorder the axes and reshape into a square matrix along the last dimension
target = tf.transpose(target, (1, 2, 3, 0))
target = tf.reshape(target, arbitrary_shape + (n, n))

# Initialise variables and check results
sess.run(tf.initialize_all_variables())
print target.eval()[0, 0, 0]

sess.close()

有没有更好的方法来实现这一目标？

Answer 1

我意识到这有点晚了，但我一直在尝试加载下三角矩阵，并且我使用稀疏到密集使其工作：

import tensorflow as tf
import numpy as np

session = tf.InteractiveSession()

n = 4 # Number of dimensions of matrix

# Get pairs of indices of positions
indices = list(zip(*np.tril_indices(n)))
indices = tf.constant([list(i) for i in indices], dtype=tf.int64)

# Test values to load into matrix
test = tf.constant(np.random.normal(0, 1, int(n*(n+1)/2)), dtype=tf.float64)

# Can pass in list of values and indices to tf.sparse_to_dense 
# and it will return a dense matrix
dense = tf.sparse_to_dense(sparse_indices=indices, output_shape=[n, n], \
                           sparse_values=test, default_value=0, \
                           validate_indices=True)

sess.close()