Question

我有一个三维数组，我想使用二维数组进行掩码，该数组的尺寸与三维数组中最右边的两个相同。有没有办法在不编写以下循环的情况下执行此操作？

import numpy as np

nx = 2
nt = 4

field3d = np.random.rand(nt, nx, nx)
field2d = np.random.rand(nx, nx)

field3d_mask = np.zeros(field3d.shape, dtype=bool)

for t in range(nt):
    field3d_mask[t,:,:] = field2d > 0.3

field3d = np.ma.array(field3d, mask=field3d_mask)

print field2d
print field3d

Answer 1

有numpy.broadcast_to（Numpy 1.10.0中的新内容）：

field3d_mask = np.broadcast_to(field2d > 0.3, field3d.shape)

Answer 2

没有循环，你可以把它写成：

field3d_mask[:,:,:] = field2d[np.newaxis,:,:] > 0.3

例如：

field3d_mask_1 = np.zeros(field3d.shape, dtype=bool)
field3d_mask_2 = np.zeros(field3d.shape, dtype=bool)

for t in range(nt):
    field3d_mask_1[t,:,:] = field2d > 0.3

field3d_mask_2[:,:,:] = field2d[np.newaxis,:,:] > 0.3

print((field3d_mask_1 == field3d_mask_2).all())

给出：

真

在numpy中使用2d掩码掩盖3d数组

2 个答案: