Numpy, sort matrix rows by placing zeros first and not changing the rest of the row

Question

I have a matrix in numpy, i.e.NxM ndarray, which looks like this:

[
  [ 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]
]

I would like to sort it by lines, putting zeros in each line first, without changing the order of the remaining elements along the line.

My desired result is as follows:

[
  [ 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]
]

For efficiency, I have to do this with numpy (therefore, it is not recommended to use regular Python nested lists and do calculations on them). The faster the code, the better.

How can i do this?

Best, Andrea

Answer 1

Is loop through lines allowed?

>>> 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

, , , .

, numpy , .

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

:

>>> 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

, , , , , :

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][:])]))

a .

Answer 4

Python, np.tile np.repeat, - , , :

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]])