前言：我建立了this answer。
考虑以下代码：
import numpy as np
# Create an array to work with.
np.random.seed(123)
full = np.random.randint(1, 99, size=(8, 8))
# Get the indices for the largest `num_largest` values.
num_largest = 8
indices = (-full).argpartition(num_largest, axis=None)[:num_largest]
# OR, if you want to avoid the temporary array created by `-full`:
# indices = full.argpartition(full.size - num_largest, axis=None)[-num_largest:]
x, y = np.unravel_index(indices, full.shape)
print "full:"
print full
print "x =", x
print "y =", y
print "Largest values:", full[x, y]
输出：
full:
[[67 93 18 84 58 87 98 97]
[48 74 33 47 97 26 84 79]
[37 97 81 69 50 56 68 3]
[85 40 67 85 48 62 49 8]
[93 53 98 86 95 28 35 98]
[77 41 4 70 65 76 35 59]
[11 23 78 19 16 28 31 53]
[71 27 81 7 15 76 55 72]]
x = [0 2 4 4 0 1 4 0]
y = [6 1 7 2 7 4 4 1]
Largest values: [98 97 98 98 97 97 95 93]
如何找到高于的最大值？
换句话说，如果我有兴趣只返回上面的值，例如，我该如何做？
期望的输出：
Largest values: [98 97 98 98 97 97] # 95 and 93 are missing because it's below the threshold
除了过滤，我还想保留索引信息;也就是说，最大值的x和y坐标。
因此，下面的解决方案不会起作用，因为它只是在最后过滤值并删除索引信息。
full[full>95]