我有一個 Bx3 張量，fooB= 批量大小的 3D 點。通過某種幻想，我得到了另一個張量，bar形狀為 Bx6x3，其中每個 B 6x3 矩陣對應于foo. 該 6x3 矩陣由 6 個復值 3D 點組成。我想做的是，對于我的每個 B 點，從6 in 中找到最接近對應點 in的實值點，最終得到一個新的 Bx3 ，其中包含與 in點的最近點。barfoomin_barbarfoo在numpy中，我可以使用屏蔽數組來完成這一壯舉：foo = np.array([ [1,2,3], [4,5,6], [7,8,9]])# here bar is only Bx2x3 for simplicity, but the solution generalizesbar = 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 barbar_with_masked_imag = np.ma.array(bar)candidates = bar_with_masked_imag.imag == 0bar_with_masked_imag.mask = ~candidatesdists = 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，tensorflow 沒有掩碼數組等。我如何將其翻譯成張量流？
1 回答
幕布斯7119047
TA貢獻1794條經驗 獲得超8個贊
這是一種方法：
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.]]
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