convert nan value to zero

Question:

I have a 2D numpy array. Some of the values in this array are NaN. I want to perform certain operations using this array. For example consider the array:

[[   0.   43.   67.    0.   38.]
 [ 100.   86.   96.  100.   94.]
 [  76.   79.   83.   89.   56.]
 [  88.   NaN   67.   89.   81.]
 [  94.   79.   67.   89.   69.]
 [  88.   79.   58.   72.   63.]
 [  76.   79.   71.   67.   56.]
 [  71.   71.   NaN   56.  100.]]

I am trying to take each row, one at a time, sort it in reversed order to get max 3 values from the row and take their average. The code I tried is:

# nparr is a 2D numpy array
for entry in nparr:
    sortedentry = sorted(entry, reverse=True)
    highest_3_values = sortedentry[:3]
    avg_highest_3 = float(sum(highest_3_values)) / 3

This does not work for rows containing NaN. My question is, is there a quick way to convert all NaN values to zero in the 2D numpy array so that I have no problems with sorting and other things I am trying to do.

Asked By: Curious2learn

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Answers:

Where A is your 2D array:

import numpy as np
A[np.isnan(A)] = 0

The function isnan produces a bool array indicating where the NaN values are. A boolean array can by used to index an array of the same shape. Think of it like a mask.

Answered By: Paul

This should work:

from numpy import *

a = array([[1, 2, 3], [0, 3, NaN]])
where_are_NaNs = isnan(a)
a[where_are_NaNs] = 0

In the above case where_are_NaNs is:

In [12]: where_are_NaNs
Out[12]: 
array([[False, False, False],
       [False, False,  True]], dtype=bool)
Answered By: Marcin

For your purposes, if all the items are stored as str and you just use sorted as you are using and then check for the first element and replace it with ‘0’

>>> l1 = ['88','NaN','67','89','81']
>>> n = sorted(l1,reverse=True)
['NaN', '89', '88', '81', '67']
>>> import math
>>> if math.isnan(float(n[0])):
...     n[0] = '0'
... 
>>> n
['0', '89', '88', '81', '67']
Answered By: Senthil Kumaran

How about nan_to_num()?

Answered By: Drake Guan

nan is never equal to nan

if z!=z:z=0

so for a 2D array

for entry in nparr:
    if entry!=entry:entry=0
Answered By: litepresence

A code example for drake’s answer to use nan_to_num:

>>> import numpy as np
>>> A = np.array([[1, 2, 3], [0, 3, np.NaN]])
>>> A = np.nan_to_num(A)
>>> A
array([[ 1.,  2.,  3.],
       [ 0.,  3.,  0.]])
Answered By: serv-inc

You can use numpy.nan_to_num :

numpy.nan_to_num(x) : Replace nan with zero and inf with finite numbers.

Example (see doc) :

>>> np.set_printoptions(precision=8)
>>> x = np.array([np.inf, -np.inf, np.nan, -128, 128])
>>> np.nan_to_num(x)
array([  1.79769313e+308,  -1.79769313e+308,   0.00000000e+000,
        -1.28000000e+002,   1.28000000e+002])
Answered By: Bilal

You could use np.where to find where you have NaN:

import numpy as np

a = np.array([[   0,   43,   67,    0,   38],
              [ 100,   86,   96,  100,   94],
              [  76,   79,   83,   89,   56],
              [  88,   np.nan,   67,   89,   81],
              [  94,   79,   67,   89,   69],
              [  88,   79,   58,   72,   63],
              [  76,   79,   71,   67,   56],
              [  71,   71,   np.nan,   56,  100]])

b = np.where(np.isnan(a), 0, a)

In [20]: b
Out[20]: 
array([[   0.,   43.,   67.,    0.,   38.],
       [ 100.,   86.,   96.,  100.,   94.],
       [  76.,   79.,   83.,   89.,   56.],
       [  88.,    0.,   67.,   89.,   81.],
       [  94.,   79.,   67.,   89.,   69.],
       [  88.,   79.,   58.,   72.,   63.],
       [  76.,   79.,   71.,   67.,   56.],
       [  71.,   71.,    0.,   56.,  100.]])
Answered By: Anton Protopopov

You can use lambda function, an example for 1D array:

import numpy as np
a = [np.nan, 2, 3]
map(lambda v:0 if np.isnan(v) == True else v, a)

This will give you the result:

[0, 2, 3]
Answered By: Mohanad Kaleia
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