从 NumPy 矩阵中满足条件的每一行中取 N 个第一个值

Question

我有一个numpy vector，还有一个numpy array。

我需要从矩阵中的每一行中取出小于（或等于）向量中相应行的前 N 个（比如说 3 个）值。

如果这是我的矢量：

7,
9,
22,
38,
6,
15

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.]]

输出应该是：

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

有没有什么有效的方法可以用面具之类的东西做到这一点，而不是丑陋的 for 循环？

任何帮助将不胜感激！

Answer 1

方法 #1

这是 broadcasting -

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]

方法 #2

受, we can leverage np.lib.stride_tricks.as_strided启发，用于高效的补丁提取，就像这样-

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]

从 NumPy 矩阵中满足条件的每一行中取 N 个第一个值

Take N first values from every row in NumPy matrix that fulfill condition

python

arrays

numpy

mask

matrix