# 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.

1
61
6/17/2016 3:08:18 PM

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 : where_are_NaNs
Out:
array([[False, False, False],
[False, False,  True]], dtype=bool)
``````
94
8/15/2014 12:31:24 PM

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.