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numpy.nan_to_num() in Python

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numpy.nan_to_num() function is used when we want to replace nan(Not A Number) with zero and inf with finite numbers in an array. It returns (positive) infinity with a very large number and negative infinity with a very small (or negative) number.

Syntax : numpy.nan_to_num(arr, copy=True)

Parameters :
arr : [array_like] Input data.
copy : [bool, optional] Whether to create a copy of arr (True) or to replace values in-place (False). The in-place operation only occurs if casting to an array does not require a copy. Default is True.

Return : [ndarray] New Array with the same shape as arr and dtype of the element in arr with the greatest precision. If arr is inexact, then NaN is replaced by zero, and infinity (-infinity) is replaced by the largest (smallest or most negative) floating point value that fits in the output dtype. If arr is not inexact, then a copy of arr is returned.

Code #1 : Working




# Python program explaining
# numpy.nan_to_num() function
  
import numpy as geek
in_num = geek.nan
  
print ("Input  number : ", in_num)
    
out_num = geek.nan_to_num(in_num) 
print ("output  number : ", out_num) 


Output :

Input  number :  nan
output  number :  0.0

 
Code #2 :




# Python program explaining
# numpy.nan_to_num function
  
import numpy as geek
  
in_arr = geek.array([[2, geek.inf, 2], [2, 2, geek.nan]])
   
print ("Input array : ", in_arr) 
    
out_arr = geek.nan_to_num(in_arr) 
print ("output array: ", out_arr) 


Output :

Input array :  [[  2.  inf   2.]
 [  2.   2.  nan]]
output array:  [[  2.00000000e+000   1.79769313e+308   2.00000000e+000]
 [  2.00000000e+000   2.00000000e+000   0.00000000e+000]]

 
Code #3 :




# Python program explaining
# numpy.nan_to_num function
  
import numpy as geek
  
in_arr = geek.array([[2, 2, 2], [2, 2, 6]])
   
print ("Input array : ", in_arr) 
    
out_arr = geek.nan_to_num(in_arr) 
print ("Output array: ", out_arr) 


Output :

Input array :  Input array :  [[2 2 2]
 [2 2 6]]
Output array:  [[2 2 2]
 [2 2 6]]


Last Updated : 28 Nov, 2018
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