张量流找到到实点的最小距离

Question

我有一个 Bx3 张量，fooB= 批量大小 3D 点。 通过一些幻想，我得到了另一个张量，bar形状为 Bx6x3，其中每个 B 6x3 矩阵对应于foo中的一个点。 该 6x3 矩阵由 6 个复值 3D 点组成。 我想做的是，对于我的每个 B 点，从 6 个bar中找到最接近foo中相应点的实值点，最终得到一个新的 Bx3min_bar，该点由bar中最接近foo中的点组成。

在numpy中，我可以用屏蔽数组完成这一壮举：

foo = np.array([
[1,2,3],
[4,5,6],
[7,8,9]])
# here bar is only Bx2x3 for simplicity, but the solution generalizes
bar = np.array([
[[2,3,4],[1+0.1j,2+0.1j,3+0.1j]],
[[6,5,4],[4,5,7]],
[[1j,1j,1j],[0,0,0]],
])
#mask complex elements of bar
bar_with_masked_imag = np.ma.array(bar)
candidates = bar_with_masked_imag.imag == 0
bar_with_masked_imag.mask = ~candidates
dists = np.sum(bar_with_masked_imag**2, axis=1)
mindists = np.argmin(dists, axis=1)
foo_indices = np.arange(foo.shape[0])
min_bar = np.array(
bar_with_masked_imag[foo_indices,mindists,:], 
dtype=float
)
print(min_bar)
#[[2. 3. 4.]
# [4. 5. 7.]
# [0. 0. 0.]]

但是，张量流没有屏蔽数组等。 如何将其转换为张量流？

Answer 1

这里有一种方法可以做到这一点：

import tensorflow as tf
import math
def solution_tf(foo, bar):
foo = tf.convert_to_tensor(foo)
bar = tf.convert_to_tensor(bar)
# Get real and imaginary parts
bar_r = tf.cast(tf.real(bar), foo.dtype)
bar_i = tf.imag(bar)
# Mask of all real-valued points
m = tf.reduce_all(tf.equal(bar_i, 0), axis=-1)
# Distance to every corresponding point
d = tf.reduce_sum(tf.squared_difference(tf.expand_dims(foo, 1), bar_r), axis=-1)
# Replace distances of complex points with infinity
d2 = tf.where(m, d, tf.fill(tf.shape(d), tf.constant(math.inf, d.dtype)))
# Find smallest distances
idx = tf.argmin(d2, axis=1)
# Get points with smallest distances
b = tf.range(tf.shape(foo, out_type=idx.dtype)[0])
return tf.gather_nd(bar_r, tf.stack([b, idx], axis=1))
# Test
with tf.Graph().as_default(), tf.Session() as sess:
foo = tf.constant([
[1,2,3],
[4,5,6],
[7,8,9]], dtype=tf.float32)
bar = tf.constant([
[[2,3,4],[1+0.1j,2+0.1j,3+0.1j]],
[[6,5,4],[4,5,7]],
[[1j,1j,1j],[0,0,0]]], dtype=tf.complex64)
sol_tf = solution_tf(foo, bar)
print(sess.run(sol_tf))
# [[2. 3. 4.]
#  [4. 5. 7.]
#  [0. 0. 0.]]