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在2d numpy数组的给定索引之间填入值

[英]Fill in values between given indices of 2d numpy array

Given a numpy array, 给定一个numpy数组,

a = np.zeros((10,10))

[[0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
 [0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
 [0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
 [0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
 [0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
 [0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
 [0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
 [0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
 [0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
 [0, 0, 0, 0, 0, 0, 0, 0, 0, 0]]

And for a set of indices, eg: 对于一组指数,例如:

start = [0,1,2,3,4,4,3,2,1,0]
end   = [9,8,7,6,5,5,6,7,8,9]

how do you get the "select" all the values/range between the start and end index and get the following: 如何获得“选择”开始和结束索引之间的所有值/范围并获得以下内容:

result = [[1, 0, 0, 0, 0, 0, 0, 0, 0, 1],
          [1, 1, 0, 0, 0, 0, 0, 0, 1, 1],
          [1, 1, 1, 0, 0, 0, 0, 1, 1, 1],
          [1, 1, 1, 1, 0, 0, 1, 1, 1, 1],
          [1, 1, 1, 1, 1, 1, 1, 1, 1, 1],
          [1, 1, 1, 1, 1, 1, 1, 1, 1, 1],
          [1, 1, 1, 1, 0, 0, 1, 1, 1, 1],
          [1, 1, 1, 0, 0, 0, 0, 1, 1, 1],
          [1, 1, 0, 0, 0, 0, 0, 0, 1, 1],
          [1, 0, 0, 0, 0, 0, 0, 0, 0, 1]]

My goal is to 'select' the all the values between the each given indices of the columns. 我的目标是“选择”列的每个给定索引之间的所有值。

I know that using apply_along_axis can do the trick, but is there a better or more elegant solution? 我知道使用apply_along_axis可以解决问题,但是有更好或更优雅的解决方案吗?

Any inputs are welcomed!! 欢迎任何投入!!

You can use broadcasting - 你可以使用broadcasting -

r = np.arange(10)[:,None]
out = ((start  <= r) & (r <= end)).astype(int)

This would create an array of shape (10,len(start) . Thus, if you need to actually fill some already initialized array filled_arr , do - 这将创建一个形状数组(10,len(start) 。因此,如果你需要实际填充一些已初始化的数组filled_arr ,请执行 -

m,n = out.shape
filled_arr[:m,:n] = out

Sample run - 样品运行 -

In [325]: start = [0,1,2,3,4,4,3,2,1,0]
     ...: end   = [9,8,7,6,5,5,6,7,8,9]
     ...: 

In [326]: r = np.arange(10)[:,None]

In [327]: ((start  <= r) & (r <= end)).astype(int)
Out[327]: 
array([[1, 0, 0, 0, 0, 0, 0, 0, 0, 1],
       [1, 1, 0, 0, 0, 0, 0, 0, 1, 1],
       [1, 1, 1, 0, 0, 0, 0, 1, 1, 1],
       [1, 1, 1, 1, 0, 0, 1, 1, 1, 1],
       [1, 1, 1, 1, 1, 1, 1, 1, 1, 1],
       [1, 1, 1, 1, 1, 1, 1, 1, 1, 1],
       [1, 1, 1, 1, 0, 0, 1, 1, 1, 1],
       [1, 1, 1, 0, 0, 0, 0, 1, 1, 1],
       [1, 1, 0, 0, 0, 0, 0, 0, 1, 1],
       [1, 0, 0, 0, 0, 0, 0, 0, 0, 1]])

If you meant to use this as a mask with 1s as the True ones, skip the conversion to int . 如果您打算将此作为掩码使用1s作为True ,请跳过转换为int Thus, (start <= r) & (r <= end) would be the mask. 因此, (start <= r) & (r <= end)将是掩码。

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