How to delete columns in numpy.array

Question

I would like to delete selected columns in a numpy.array . This is what I do:

n [397]: a = array([[ NaN,   2.,   3., NaN],
   .....:        [  1.,   2.,   3., 9]])

In [398]: print a
[[ NaN   2.   3.  NaN]
 [  1.   2.   3.   9.]]

In [399]: z = any(isnan(a), axis=0)

In [400]: print z
[ True False False  True]

In [401]: delete(a, z, axis = 1)
Out[401]:
 array([[  3.,  NaN],
       [  3.,   9.]])

In this example my goal is to delete all the columns that contain NaN's. I expect the last command to result in:

array([[2., 3.],
       [2., 3.]])

How can I do that?

score 167 · Accepted Answer · edited Apr 10 '19 at 07:30

167

Given its name, I think the standard way should be delete:

import numpy as np

A = np.delete(A, 1, 0)  # delete second row of A
B = np.delete(B, 2, 0)  # delete third row of B
C = np.delete(C, 1, 1)  # delete second column of C

According to numpy's documentation page, the parameters for numpy.delete are as follow:

numpy.delete(arr, obj, axis=None)

arr refers to the input array,
obj refers to which sub-arrays (e.g. column/row no. or slice of the array) and
axis refers to either column wise (axis = 1) or row-wise (axis = 0) delete operation.

edited Apr 10 '19 at 07:30

Ph03nIX

173
1
10

answered Feb 17 '11 at 20:57

Steve Tjoa

59,122
18
90
101

14

I believe you should be referencing `numpy`, not `scipy`. http://docs.scipy.org/doc/numpy/reference/generated/numpy.delete.html – hlin117 Mar 26 '15 at 03:37
6

just add explanation: array, index and axis as parameters – maximus Jun 23 '18 at 06:49

score 32 · Answer 2 · edited Feb 18 '17 at 13:29

32

Example from the numpy documentation:

>>> a = numpy.array([[ 0,  1,  2,  3],
               [ 4,  5,  6,  7],
               [ 8,  9, 10, 11],
               [12, 13, 14, 15]])

>>> numpy.delete(a, numpy.s_[1:3], axis=0)                       # remove rows 1 and 2

array([[ 0,  1,  2,  3],
       [12, 13, 14, 15]])

>>> numpy.delete(a, numpy.s_[1:3], axis=1)                       # remove columns 1 and 2

array([[ 0,  3],
       [ 4,  7],
       [ 8, 11],
       [12, 15]])

edited Feb 18 '17 at 13:29

qwerty_so

35,448
8
62
86

answered Jul 05 '11 at 17:42

Nikolay Frick

2,145
22
17

@alvas here is a well-organized explanation! http://stackoverflow.com/questions/32682754/np-delete-and-np-s-whats-so-special-about-np-s – Daeyoung Sep 17 '16 at 08:50
1

@Alvas , s_ is: `A nicer way to build up index tuples for arrays.`: https://docs.scipy.org/doc/numpy/reference/generated/numpy.s_.html – Minions Jul 16 '19 at 14:54

score 16 · Answer 3 · answered Oct 29 '09 at 12:44

Another way is to use masked arrays:

import numpy as np
a = np.array([[ np.nan,   2.,   3., np.nan], [  1.,   2.,   3., 9]])
print(a)
# [[ NaN   2.   3.  NaN]
#  [  1.   2.   3.   9.]]

The np.ma.masked_invalid method returns a masked array with nans and infs masked out:

print(np.ma.masked_invalid(a))
[[-- 2.0 3.0 --]
 [1.0 2.0 3.0 9.0]]

The np.ma.compress_cols method returns a 2-D array with any column containing a masked value suppressed:

a=np.ma.compress_cols(np.ma.masked_invalid(a))
print(a)
# [[ 2.  3.]
#  [ 2.  3.]]

See manipulating-a-maskedarray

score 9 · Answer 4 · answered Oct 29 '09 at 11:19

9

This creates another array without those columns:

  b = a.compress(logical_not(z), axis=1)

answered Oct 29 '09 at 11:19

Paul

42,322
15
106
123

2

cool. I wish matlab's syntax worked here: "a(:,z) = []" is much simpler – Boris Gorelik Oct 29 '09 at 11:22
1

@bpowah: indeed. the more general way would be b = a[:,z]. You might want to update your answer accordingly – Boris Gorelik Oct 29 '09 at 12:12

score 8 · Answer 5 · answered Nov 10 '13 at 05:55

From Numpy Documentation

np.delete(arr, obj, axis=None) Return a new array with sub-arrays along an axis deleted.

>>> arr
array([[ 1,  2,  3,  4],
       [ 5,  6,  7,  8],
       [ 9, 10, 11, 12]])
>>> np.delete(arr, 1, 0)
array([[ 1,  2,  3,  4],
       [ 9, 10, 11, 12]])

>>> np.delete(arr, np.s_[::2], 1)
array([[ 2,  4],
       [ 6,  8],
       [10, 12]])
>>> np.delete(arr, [1,3,5], None)
array([ 1,  3,  5,  7,  8,  9, 10, 11, 12])

score 6 · Answer 6 · answered Oct 16 '11 at 19:36

6

In your situation, you can extract the desired data with:

a[:, -z]

"-z" is the logical negation of the boolean array "z". This is the same as:

a[:, logical_not(z)]

answered Oct 16 '11 at 19:36

Olivier

393
3
11

score 2 · Answer 7 · answered Mar 23 '15 at 23:55

2

>>> A = array([[ 1,  2,  3,  4],
               [ 5,  6,  7,  8],
               [ 9, 10, 11, 12]])

>>> A = A.transpose()

>>> A = A[1:].transpose()

answered Mar 23 '15 at 23:55

yonglin

416
1
5
8

score -1 · Answer 8 · edited May 05 '17 at 12:11

Removing Matrix columns that contain NaN. This is a lengthy answer, but hopefully easy to follow.

def column_to_vector(matrix, i):
    return [row[i] for row in matrix]
import numpy
def remove_NaN_columns(matrix):
    import scipy
    import math
    from numpy import column_stack, vstack

    columns = A.shape[1]
    #print("columns", columns)
    result = []
    skip_column = True
    for column in range(0, columns):
        vector = column_to_vector(A, column)
        skip_column = False
        for value in vector:
            # print(column, vector, value, math.isnan(value) )
            if math.isnan(value):
                skip_column = True
        if skip_column == False:
            result.append(vector)
    return column_stack(result)

### test it
A = vstack(([ float('NaN'), 2., 3., float('NaN')], [ 1., 2., 3., 9]))
print("A shape", A.shape, "\n", A)
B = remove_NaN_columns(A)
print("B shape", B.shape, "\n", B)

A shape (2, 4) 
 [[ nan   2.   3.  nan]
 [  1.   2.   3.   9.]]
B shape (2, 2) 
 [[ 2.  3.]
 [ 2.  3.]]

I actually don't follow you. How does this code work? – RamenChef Nov 13 '16 at 23:49 — RamenChef, Nov 13 '16 at 23:49

How to delete columns in numpy.array

8 Answers8

Linked