How do I delete a row in a numpy array which contains a zero?

Question

I'm trying to write a function to delete all rows in which have a zero value in. This is not from my code, but an example of the idea I am using:

import numpy as np
a=np.array(([7,1,2,8],[4,0,3,2],[5,8,3,6],[4,3,2,0]))
b=[]

for i in range(len(a)):
    for j in range (len(a[i])):
        if a[i][j]==0:
            b.append(i)

print 'b=', b
for zero_row in b:
    x=np.delete(a,zero_row, 0)

print 'a=',a

and this is my output:

b= [1, 3]
a= [[7 1 2 8]
 [4 0 3 2]
 [5 8 3 6]
 [4 3 2 0]]

How do I get rid of the rows with the index in b? Sorry, I'm fairly new to this any help would be really appreciated.

Just one comment. That piece of example code you posted loops over `b` to delete individual rows. That's not needed at all. `a = np.delete(a, b, axis=0)` works like a charm (`axis` specified to make it more evident what we're doing) — Ricardo Cárdenes, Feb 06 '14 at 11:24

score 16 · Answer 1 · answered Aug 23 '13 at 09:07

16

I'm trying to write a function to delete all rows in which have a zero value in.

You don't need to write a function for that, it can be done in a single expression:

>>> a[np.all(a != 0, axis=1)]
array([[7, 1, 2, 8],
       [5, 8, 3, 6]])

Read as: select from a all rows that are entirely non-zero.

answered Aug 23 '13 at 09:07

Fred Foo

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is there a way to do this for a 1D array? would you just leave out the axis specification? i.e. **b[np.all(b != 0, )]** where b is a 1d array – Ashleigh Clayton Aug 23 '13 at 09:39
2

`b = b[b!=0]` removes the `0`'s in `b`. It is called boolean indexing. – J. Martinot-Lagarde Aug 23 '13 at 11:33
What if one wants to remove all columns containing "2"? `a[np.all(a !=2, axis=0)]` does not work as expected. I had to resort to manipulating the transpose instead. Any suggestion? – neither-nor Apr 10 '14 at 13:51
@user1211129 I think transposing and later transposing back is the sensible thing to do. – Fred Foo Apr 10 '14 at 14:15

score 3 · Answer 2 · answered Aug 23 '13 at 08:09

3

Looks like np.delete does't change the array, just returns a new array, so

Instead of

x = np.delete(a,zero_row, 0)

try

a = np.delete(a,zero_row, 0)

answered Aug 23 '13 at 08:09

tuxcanfly

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ah that was stupid, probably to early in the morning. However I did change this and got **ValueError:invalid entry** thanks for the quick response – Ashleigh Clayton Aug 23 '13 at 08:12

score 1 · Answer 3 · answered Aug 23 '13 at 08:22

I think I have found the answer:

as @tuxcanfly said I changed x to a. Also I have now removed the for loop as it removed the row with index 2 for some reason.

Instead I now just chose to delete the rows using b as the delete function with use the elements in the list to remove the row with that index.

the new code:

import numpy as np
a=np.array(([7,1,2,8],[4,0,3,2],[5,8,3,6],[4,3,2,0]))
b=[]

for i in range(len(a)):
    for j in range (len(a[i])):
        if a[i][j]==0:
            b.append(i)
print 'b=',b
for zero_row in b:
    a=np.delete(a,b, 0)

print 'a=',a

and the output:

b= [1, 3]
a= [[7 1 2 8]
 [5 8 3 6]]

score 1 · Answer 4 · edited Apr 18 '19 at 19:53

I think this helps readability (and allows you to loop once, not twice):

#!/usr/bin/env python

import numpy as np
a = np.array(([7,1,2,8], [4,0,3,2], [5,8,3,6], [4,3,2,0]))
b = None

for row in a:
    if 0 not in row:
        b = np.vstack((b, row)) if b is not None else row

a = b
print 'a = ', a

In this version, you loop over each row and test for 0's membership in the row. If the row does not contain a zero, you attempt to use np.vstack to append the row to an array called b. If b has not yet been assigned to, it is initialized to the current row.

How do I delete a row in a numpy array which contains a zero?

4 Answers4

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