正如this issue中正在进行广义切片一样,实现2D张量(矩阵)的op收集列的最佳方法是什么?例如,对于张量t
:
1 2 3 4
5 6 7 8
和指数[1,3],我想得到:
2 4
6 8
相当于numpy t[:, [1,3]]
。
答案 0 :(得分:8)
有一个名为tf.nn.embedding_lookup(params, ind)
的函数可以检索params
张量的行。
为了达到你想要的效果,我们可以首先转置要从中选择某些列的张量t
。然后查找tf.transpose(t)
行(t
列)。选择之后,我们将结果转换回来。
import tensorflow as tf
t = tf.constant([[1, 2, 3],
[4, 5, 6]])
ind = tf.constant([0, 2])
result = tf.transpose(tf.nn.embedding_lookup(tf.transpose(t), ind))
with tf.Session() as sess:
print(sess.run(result))
答案 1 :(得分:5)
到目前为止,我通过展平输入并使用gather
:
def gather_cols(params, indices, name=None):
"""Gather columns of a 2D tensor.
Args:
params: A 2D tensor.
indices: A 1D tensor. Must be one of the following types: ``int32``, ``int64``.
name: A name for the operation (optional).
Returns:
A 2D Tensor. Has the same type as ``params``.
"""
with tf.op_scope([params, indices], name, "gather_cols") as scope:
# Check input
params = tf.convert_to_tensor(params, name="params")
indices = tf.convert_to_tensor(indices, name="indices")
try:
params.get_shape().assert_has_rank(2)
except ValueError:
raise ValueError('\'params\' must be 2D.')
try:
indices.get_shape().assert_has_rank(1)
except ValueError:
raise ValueError('\'indices\' must be 1D.')
# Define op
p_shape = tf.shape(params)
p_flat = tf.reshape(params, [-1])
i_flat = tf.reshape(tf.reshape(tf.range(0, p_shape[0]) * p_shape[1],
[-1, 1]) + indices, [-1])
return tf.reshape(tf.gather(p_flat, i_flat),
[p_shape[0], -1])
适用于:
params = tf.constant([[1, 2, 3],
[4, 5, 6]])
indices = [0, 2]
op = gather_cols(params, indices)
产生预期的输出:
[[1 3]
[4 6]]
答案 2 :(得分:0)
与此同时,gather
方法具有一个axis
参数。
import tensorflow as tf
params = tf.constant([[1,2,3],[4,5,6]])
indices = [0,2]
op = tf.gather(params, indices, axis=1)
产生输出
[[1 3]
[4 6]]