【发布时间】:2016-09-15 09:45:55
【问题描述】:
我想将形状(..., n * (n - 1) / 2) 的数组打包到形状为(..., n, n) 的张量的下三角形部分,其中... 表示任意形状。在 numpy 中,我会将其实现为
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,我已经能够使用 transpose、reshape 和 scatter_update 的组合在 tensorflow 中实现类似的功能,但感觉很笨拙。
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点?
【问题讨论】:
标签: python indexing tensorflow