indexes = np.array([[0,1,3],[1,2,4 ]])
data = np.random.rand(2,5)
现在,我想要一个形状为(2,3)的数组,其中
result[0] = data[0,indexes[0]]
result[1] = data[1,indexes[1]]
实现此目标的正确方法是什么?一种麻木的方式,可以推广到更大的数组(也许甚至更高维度)。
请注意与this之类的问题的区别,其中的索引数组包含元组。这不是我要的。
这个问题的更一般的表述是:
比
result[i, j, ..., k] = data[i, j,...,k, indexes[i, j, ..., k]]
其中
len([i, j, ..., k]) == len(data)-1 == len(indexes) - 1
答案 0 :(得分:1)
以下是NumPy和TensorFlow解决方案:
import numpy as np
import tensorflow as tf
def gather_index_np(data, index):
data = np.asarray(data)
index = np.asarray(index)
# Make open grid of all but last dimension indices
grid = np.ogrid[tuple(slice(s) for s in index.shape[:-1])]
# Add extra dimension in grid
grid = [g[..., np.newaxis] for g in grid]
# Complete index
index_full = tuple(grid + [index])
# Index data to get result
result = data[index_full]
return result
def gather_index_tf(data, index):
data = tf.convert_to_tensor(data)
index = tf.convert_to_tensor(index)
index_shape = tf.shape(index)
d = index.shape.ndims
# Make grid of all dimension indices
grid = tf.meshgrid(*(tf.range(index_shape[i]) for i in range(d)), indexing='ij')
# Complete index
index_full = tf.stack(grid[:-1] + [index], axis=-1)
# Index data to get result
result = tf.gather_nd(data, index_full)
return result
示例:
import numpy as np
import tensorflow as tf
data = np.arange(10).reshape((2, 5))
index = np.array([[0, 1, 3], [1, 2, 4]])
print(gather_index_np(data, index))
# [[0 1 3]
# [6 7 9]]
with tf.Session() as sess:
print(sess.run(gather_index_tf(data, index)))
# [[0 1 3]
# [6 7 9]]
答案 1 :(得分:0)
numpy
具有take_along_axis
的功能,它还可以执行您描述的内容,还可以选择轴。
示例:
>>> a = np.arange(24).reshape(2,3,4)
>>> i = np.random.randint(0,4,(2,3,5))
>>> i
array([[[3, 3, 0, 1, 3],
[3, 1, 0, 3, 3],
[3, 2, 0, 3, 3]],
[[2, 3, 0, 0, 0],
[1, 1, 3, 1, 2],
[1, 3, 0, 0, 2]]])
>>> np.take_along_axis(a, i, -1)
array([[[ 3, 3, 0, 1, 3],
[ 7, 5, 4, 7, 7],
[11, 10, 8, 11, 11]],
[[14, 15, 12, 12, 12],
[17, 17, 19, 17, 18],
[21, 23, 20, 20, 22]]])