Given m x n nd array of floats, what is the best way to get an m' x n 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]
]
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.
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.]])
[-1., 0., 1.]?np.abs(x).sum(-1) will solve the issue that I pointed out.
~np.all(x == 0, axis=1) you can just do x.any(axis=1). This saves some computation steps and temporary arrays, although by De Morgan's Laws we can see that it arrives at the same answer.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.]])
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.
a=np.array([r for r in a if any(r)])
