Question

我有2个张量a和b，它们具有以下形状

>>K.int_shape(a)
(None, 5 , 2)
>>K.int_shape(b)
(None, 5)

我想要的是张量c

>>K.int_shape(c)
(None, 2)

这样，沿着轴0，您可以选择b中最大元素的索引，并使用它来沿轴1索引a。

示例 - 说我有

a = np.array([[[2, 7],
    [6, 5],
    [9, 9],
    [4, 2],
    [5, 9]],

   [[8, 1],
    [8, 8],
    [3, 9],
    [9, 2],
    [9, 1]],

   [[3, 9],
    [6, 4],
    [5, 7],
    [5, 2],
    [5, 6]],

   [[7, 5],
    [9, 9],
    [9, 5],
    [9, 8],
    [5, 7]],

   [[6, 3],
    [1, 7],
    [3, 6],
    [8, 2],
    [3, 2]],

   [[6, 4],
    [5, 9],
    [8, 6],
    [5, 2],
    [5, 2]],

   [[2, 6],
    [6, 5],
    [3, 1],
    [6, 2],
    [6, 4]]])

我有

b = np.array([[ 0.27,  0.25,  0.23,  0.06,  0.19],
[ 0.3 ,  0.13,  0.17,  0.2 ,  0.2 ],
[ 0.08,  0.04,  0.40,  0.36,  0.12],
[ 0.3 ,  0.33,  0.11,  0.07,  0.19],
[ 0.15,  0.21,  0.30,  0.12,  0.22],
[ 0.3 ,  0.13,  0.23,  0.1 ,  0.23],
[ 0.26,  0.35 ,  0.25 ,  0.07,  0.07]])

我期望c成为

c = np.zeros((7,2))
for i in range(7):
    ind = np.argmax(b[i, :])
    c[i, :] = a[i, ind, :]
c
array([[ 2.,  7.],
   [ 8.,  1.],
   [ 5.,  7.],
   [ 9.,  9.],
   [ 3.,  6.],
   [ 6.,  4.],
   [ 6.,  5.]])

Answer 1

使用Tensorflow进行后端（我对Theano知之甚少），使用tf.gather_nd()：

import keras.backend as K
import tensorflow as tf

# `a` and `b` the numpy arrays defined in the question
A = tf.constant(a)
B = tf.constant(b)

# Obtaining your max indices over axis 1, which will be used as indices for axis 1 of A:
col_ind = K.argmax(B, axis=1)

# Creating row range, which will be used as indices for axis 0 of A:
row_ind = K.arange(col_ind.shape[0], dtype='int64')

# Stacking the indices together:
ind = K.stack((row_ind, col_ind), axis=-1)

# Gathering the results:
c = tf.gather_nd(A, ind) # no equivalent I know in K, and no idea about theano...

with tf.Session() as sess:
    print(c.eval())
    # [[2 7]
    #  [8 1]
    #  [5 7]
    #  [9 9]
    #  [3 6]
    #  [6 4]
    #  [6 5]]

Answer 2

找到了解决方案

A = K.constant(a)
B = K.constant(b)
mxidx = K.argmax(B, axis=1)
c = K.map_fn(lambda i: A[i, mxidx[i], :], K.arange(A.shape[0], dtype='int64'))
print K.eval(c)
array([[ 2.,  7.],
   [ 8.,  1.],
   [ 5.,  7.],
   [ 9.,  9.],
   [ 3.,  6.],
   [ 6.,  4.],
   [ 6.,  5.]], dtype=float32)

编辑：添加运行时信息

%timeit K.eval(c)
The slowest run took 9.76 times longer than the fastest. This could mean 
that an intermediate result is being cached.
100000 loops, best of 3: 12.2 µs per loop

如何根据Keras中的另一个数组进行索引

2 个答案: