Question

我有一个numpy矩阵，这是一个NxM ndarray，看起来像下面的那个：

[
  [ 0, 5, 11, 22, 0, 0, 11, 22], 
  [ 1, 4, 11, 20, 0, 4, 11, 20], 
  [ 1, 6, 11, 22, 0, 1, 11, 22], 
  [ 4, 7, 12, 21, 0, 4, 12, 21], 
  [ 5, 7, 12, 22, 0, 7, 12, 22], 
  [ 5, 7, 12, 22, 0, 5, 12, 22]
]

我想按行排序，先将每行放零，而不改变行中其他元素的顺序。

我想要的输出如下：

[
  [ 0, 0, 0, 5, 11, 22, 11, 22], 
  [ 0, 1, 4, 11, 20, 4, 11, 20], 
  [ 0, 1, 6, 11, 22, 1, 11, 22], 
  [ 0, 4, 7, 12, 21, 4, 12, 21], 
  [ 0, 5, 7, 12, 22, 7, 12, 22], 
  [ 0, 5, 7, 12, 22, 5, 12, 22]
]

为了提高效率，我需要使用numpy（因此不建议切换到Python的常规嵌套列表并对它们进行计算）。代码越快越好。

我怎么能这样做？

最佳，安德烈

Answer 1

是否允许循环行？

>>> a
array([[ 0,  5, 11, 22,  0,  0, 11, 22],
       [ 1,  4, 11, 20,  0,  4, 11, 20],
       [ 1,  6, 11, 22,  0,  1, 11, 22],
       [ 4,  7, 12, 21,  0,  4, 12, 21],
       [ 5,  7, 12, 22,  0,  7, 12, 22],
       [ 5,  7, 12, 22,  0,  5, 12, 22]])
>>> for row in a:
...     row[:] = np.r_[row[row == 0], row[row != 0]]
...     
>>> a
array([[ 0,  0,  0,  5, 11, 22, 11, 22],
       [ 0,  1,  4, 11, 20,  4, 11, 20],
       [ 0,  1,  6, 11, 22,  1, 11, 22],
       [ 0,  4,  7, 12, 21,  4, 12, 21],
       [ 0,  5,  7, 12, 22,  7, 12, 22],
       [ 0,  5,  7, 12, 22,  5, 12, 22]])

Answer 2

此方法获取二进制数组，其中数组为零且非零，然后获取该数组的排序索引，然后将其应用于原始数组。

你需要一个与你的待排序数组一样大的数组来保存索引，但由于它是所有numpy操作，它可能比循环更快。

ind = (a>0).astype(int)
ind = ind.argsort(axis=1)
a[np.arange(ind.shape[0])[:,None], ind]

输出：

>>> a
array([[ 0,  0,  0,  5, 11, 22, 11, 22],
       [ 0,  1,  4, 11, 20,  4, 11, 20],
       [ 0,  1,  6, 11, 22,  1, 11, 22],
       [ 0,  4,  7, 12, 21,  4, 12, 21],
       [ 0,  5,  7, 12, 22,  7, 12, 22],
       [ 0,  5,  7, 12, 22,  5, 12, 22]])

Answer 3

可能不是最有效的，因为它在线上循环，但可能是一个很好的起点：

import numpy as np

a = np.array([[ 0,  5, 11, 22,  0,  0, 11, 22],
             [ 1,  4, 11, 20,  0,  4, 11, 20],
             [ 1,  6, 11, 22,  0,  1, 11, 22],
             [ 4,  7, 12, 21,  0,  4, 12, 21],
             [ 5,  7, 12, 22,  0,  7, 12, 22],
             [ 5,  7, 12, 22,  0,  5, 12, 22]])

size = a.shape[1]

for i, line in enumerate(a):
    nz = np.nonzero(a[i][:])[0]
    z = np.zeros(size - nz.shape[0])
    a[i][:] = np.concatenate((z,a[i][:][np.nonzero(a[i][:])]))

对于a中的每一行，您会找到非零索引并添加一些零以匹配大小。

Answer 4

有可能摆脱所有的Python循环，在np.tile和np.repeat的帮助下构建一个布尔掩码，尽管你需要在一些更大的例子上计时，看它是否值得额外的复杂性：

rows, cols = a.shape
mask = a != 0
nonzeros_per_row = mask.sum(axis=1)
repeats = np.column_stack((cols-nonzeros_per_row, nonzeros_per_row)).ravel()
new_mask = np.repeat(np.tile([False, True], rows), repeats).reshape(rows, cols)
out = np.zeros_like(a)
out[new_mask] = a[mask]

>>> a
array([[ 0,  5, 11, 22,  0,  0, 11, 22],
       [ 1,  4, 11, 20,  0,  4, 11, 20],
       [ 1,  6, 11, 22,  0,  1, 11, 22],
       [ 4,  7, 12, 21,  0,  4, 12, 21],
       [ 5,  7, 12, 22,  0,  7, 12, 22],
       [ 5,  7, 12, 22,  0,  5, 12, 22]])
>>> out
array([[ 0,  0,  0,  5, 11, 22, 11, 22],
       [ 0,  1,  4, 11, 20,  4, 11, 20],
       [ 0,  1,  6, 11, 22,  1, 11, 22],
       [ 0,  4,  7, 12, 21,  4, 12, 21],
       [ 0,  5,  7, 12, 22,  7, 12, 22],
       [ 0,  5,  7, 12, 22,  5, 12, 22]])

Numpy，排序矩阵的行首先放零，而不是修改行的其余部分

4 个答案: