如何在 numpy 中快速实现这样的操作 - V2EX
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Deepseafish
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如何在 numpy 中快速实现这样的操作

  Deepseafish 2020-07-31 11:51:21 +08:00 1929 次点击
这是一个创建于 1901 天前的主题,其中的信息可能已经有所发展或是发生改变。

通过一个 mask 得到数组中的某几位后,再通过 mask 恢复原来的位置,并在其余地方填充 0 。具体实例代码如下。

除了遍历有没有更好方法?

https://gist.github.com/SungYK/4a23c94c9eb6b275d3d2bd90a0b2a5dd

2 条回复    2020-07-31 13:20:38 +08:00
Xs0ul
    1
Xs0ul  
   2020-07-31 12:17:35 +08:00
lz 的 mask = np.array([1,3,5,7]) 一般不叫 mask,叫 indices 比较合理

mask 应该是
mask = np.zeros_like(arr)
mask[indices] = 1

然后
res = np.where(mask, arr, 0)
ytterbium
    2
ytterbium  
   2020-07-31 13:20:38 +08:00 via Android
和 numpy 关系比较近的 pytorch 里有个 masked_scatter

https://pytorch.org/docs/stable/tensors.html#torch.Tensor.masked_scatter

比如,mask 和 val 都是 np.array 类型

mask = [

[0, 1, 0],

[1, 0, 0],

[1, 1, 0]

]

val = [1, 2, 3, 4]

x = torch.Tensor.masked_scatter(torch.from_numpy(mask), torch.from_numpy(val)).data.numpy()

输出 x 为 np.array 类型

x = [

[0, 1, 0],

[2, 0, 0],

[3, 4, 0]

]
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