从满足条件的NumPy矩阵的每一行中取N个第一个值

I have a numpy vector, and a numpy array.

我有一个numpy向量和一个numpy数组。

I need to take from every row in the matrix the first N (lets say 3) values that are smaller than (or equal to) the corresponding line in the vector.

我需要从矩阵中的每一行获取小于(或等于)向量中相应行的第一个N(比如3)值。

so if this is my vector:

所以如果这是我的载体:

7,
9,
22,
38,
6,
15

and this is my matrix:

这是我的矩阵:

[[ 20.,   9.,   7.,   5.,   None,   None],
 [ 33.,  21.,  18.,   9.,   8.,   7.],
 [ 31.,  21.,  13.,  12.,   4.,   0.],
 [ 36.,  18.,  11.,   7.,   7.,   2.],
 [ 20.,  14.,  10.,   6.,   6.,   3.],
 [ 14.,  14.,  13.,  11.,   5.,   5.]]

the output should be:

输出应该是:

[[7,5,None],
 [9,8,7],
 [21,13,12],
 [36,18,11],
 [6,6,3],
 14,14,13]]

Is there any efficient way to do that with masks or something, without an ugly for loop?

是否有任何有效的方法来做掩码或其他东西,没有丑陋的for循环?

Any help will be appreciated!

任何帮助将不胜感激!

1 个解决方案

#1

Approach #1

Here's one with broadcasting -

这是一个广播 -

def takeN_le_per_row_broadcasting(a, b, N=3): # a, b : 1D, 2D arrays respectively
    # First col indices in each row of b with <= corresponding one in a
    idx = (b <= a[:,None]).argmax(1)

    # Get all N ranged column indices
    all_idx = idx[:,None] + np.arange(N)

    # Finally advanced-index with those indices into b for desired output
    return b[np.arange(len(all_idx))[:,None], all_idx]

Approach #2

Inspired by NumPy Fancy Indexing - Crop different ROIs from different channels's solution, we can leverage np.lib.stride_tricks.as_strided for efficient patch extraction, like so -

受NumPy花式索引的启发 - 从不同渠道的解决方案中获取不同的投资回报率,我们可以利用np.lib.stride_tricks.as_strided进行有效的补丁提取,如下所示 -

from skimage.util.shape import view_as_windows

def takeN_le_per_row_strides(a, b, N=3): # a, b : 1D, 2D arrays respectively
    # First col indices in each row of b with <= corresponding one in a
    idx = (b <= a[:,None]).argmax(1)

    # Get 1D sliding windows for each element off data
    w = view_as_windows(b, (1,N))[:,:,0]

    # Use fancy/advanced indexing to select the required ones
    return w[np.arange(len(idx)), idx]

#1