Applying a mask to an multidimensional array

Question

I want to do this in a proper way:

data = np.array(data)
data =[
[1, 1, 2, 1],
[0, 1, 3, 2],
[0, 2, 3, 2],
[2, 4, 3, 1],
[0, 2, 1, 4],
[3, 1, 4, 1]]

this should become (delete the lines that start with 0):

[1, 1, 2, 1]
[2, 4, 3, 1]
[3, 1, 4, 1]

So far I did it like this:

lines = []
for i in range(0, len(data[0])):
    if data[0,i] != 0:
        lines.append(data[:,i])
lines = np.array(lines)

Then I found this fine method:

mask = 1 <= data[0,:]

and now I want to apply that mask to that array. This Mask reads: [True, False, False, True, False, True]. How do I do that?

you should specify that you have a numpy array

Padraic Cunningham
– Padraic Cunningham

2015-05-24 20:46:30 +00:00
Commented May 24, 2015 at 20:46 — Padraic Cunningham
– Padraic Cunningham, Commented May 24, 2015 at 20:46

Andrey · Accepted Answer · 2015-05-24 20:14:37Z

2

Why not just:

[ar for ar in data if ar[0] != 0]

This assumes that arrays are not empty.

answered May 24, 2015 at 20:14

Andrey

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Padraic Cunningham · Accepted Answer · 2015-05-24 20:41:42Z

1

I presume you have a numpy array based on the data[0,:] and data[0,i] you have in your question and you mean data[:, 0] :

import numpy as np

data = np.array([
    [1, 1, 2, 1],
    [0, 1, 3, 2],
    [0, 2, 3, 2],
    [2, 4, 3, 1],
    [0, 2, 1, 4],
    [3, 1, 4, 1]])

data = data[data[:,0] != 0]
print(data)

Output:

[[1 1 2 1]
 [2 4 3 1]
 [3 1 4 1]]

data[0,:] is the first row [1 1 2 1] not the first column

edited May 24, 2015 at 20:41

answered May 24, 2015 at 20:35

Padraic Cunningham

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2 Comments

xtlc Over a year ago

this is exactly what I was looking for. I was trying something like data[data[:,0].all(1)] but that didn't work :(

Padraic Cunningham Over a year ago

@xtlc, you were not too far off ;)

Ajay · Accepted Answer · 2015-05-24 20:14:23Z

0

Using List comprehension

In [56]: [elem for elem in data if elem[0] !=0]
Out[56]: [[1, 1, 2, 1], [2, 4, 3, 1], [3, 1, 4, 1]]

answered May 24, 2015 at 20:14

Ajay

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