从 numpy ndarray 中删除零行

Question

Given mxn nd array of floats, what is the best way to get an m' xn nd array of floats that does not contain all-zero rows?

for example: Given

[ 
  [1.0, 0.0, 2.0], 
  [0.0, 0.0, 0.0], 
  [2.0, 1.0, 0.0] 
]

I want to get

[ 
  [1.0, 0.0, 2.0], 
  [2.0, 1.0, 0.0] 
]

Answer 1

You can exclude those elements as follows:

>>> import numpy as np
>>> x = np.array([ [1.0, 0.0, 2.0], [0.0, 0.0, 0.0], [2.0, 1.0, 0.0] ])
>>> x
array([[1., 0., 2.],
       [0., 0., 0.],
       [2., 1., 0.]])
>>> sumrow = np.abs(x).sum(-1)
>>> x[sumrow>0]
array([[1., 0., 2.],
       [2., 1., 0.]])

Note: @Akavall pointed out correctly that np.abs() would prevent issues with negative values.

Additionally, another more complex approach:

>>> x = np.array([ [1.0, 0.0, 2.0], [0.0, 0.0, 0.0], [2.0, 1.0, 0.0] ])
>>> x[~np.all(x == 0, axis=1)]
array([[1., 0., 2.],
       [2., 1., 0.]])

See: https://www.geeksforgeeks.org/numpy-indexing/

Answer 2

You can index using a boolean array:

a = np.array([[1.0, 0.0, 2.0], [0.0, 0.0, 0.0], [2.0, 1.0, 0.0]])

print(a[a.any(axis=1)])

Here a.any(axis=1) will be True where any elements in the row are non-zero. These are the rows that we want to keep.

Answer 3

A possible solution would be to use the fact that the sum of all zeros is zero. Create a mask using that fact:

>>> bar = np.array ([ [1.0, 0.0, 2.0], [0.0, 0.0, 0.0], [2.0, 1.0, 0.0] ] )
>>> mask = bar.sum(axis=1)==0
>>> bar[mask]
array([[1., 0., 2.],
       [2., 1., 0.]])

Answer 4

Here's one way to do it:

import numpy as np
x    = np.array([ [1.0, 0.0, 2.0], [0.0, 0.0, 0.0], [2.0, 1.0, 0.0] ])
m, n = x.shape
rows = [row for row in range(m) if not all(x[row] == 0)]
x    = x[rows]
print(x)

This works for arrays containing negative data also. If we use sum suppose a row contains [-1, 0, 1] it will be deleted we don't want that.

Answer 5

a=np.array([r for r in a if any(r)])