numpy.isposinf() in Python
Last Updated :
08 Mar, 2024
The numpy.isposinf() function tests element-wise whether it is positive infinity or not and returns the result as a boolean array.
Syntax :
numpy.isposinf(array, y = None)
Parameters:
array : [array_like]Input array or object whose elements, we need to test for infinity.
y : [array_like]A boolean array with the same shape and type as x to store the result.
Return:
boolean array containing the result. For scalar input, the result is a new boolean with value
True if the input is positive or negative infinity; otherwise the value is False.
For array input, the result is a boolean array with the same shape as the input and the values
are True where the corresponding element of the input is positive or negative infinity;
elsewhere the values are False.
Code 1:
Python
import numpy as geek
print ( "Positive : " , geek.isposinf( 1 ), "\n" )
print ( "Positive : " , geek.isposinf( 0 ), "\n" )
print ( "Positive : " , geek.isposinf(geek.nan), "\n" )
print ( "Positive : " , geek.isposinf(geek.inf), "\n" )
print ( "Positive : " , geek.isposinf(geek.NINF), "\n" )
x = geek.array([ - geek.inf, 0. , geek.inf])
y = geek.array([ 2 , 2 , 2 ])
print ( "Checking for positivity : " , geek.isposinf(x, y))
|
Output :
Positive : False
Positive : False
Positive : False
Positive : True
Positive : False
Checking for positivity : [0 0 1]
Code 2 :
Python
import numpy as geek
b = geek.arange( 18 ).reshape( 3 , 6 )
print ( "\n" ,b)
print ( "\nIs Positive Infinity : \n" , geek.isposinf(b))
b = [[geek.inf],
[geek.NINF]]
print ( "\nIs Positive Infinity : \n" , geek.isposinf(b))
|
Output :
[[ 0 1 2 3 4 5]
[ 6 7 8 9 10 11]
[12 13 14 15 16 17]]
Is Positive Infinity :
[[False False False False False False]
[False False False False False False]
[False False False False False False]]
Is Positive Infinity :
[[ True]
[False]]
Note :
These codes won’t run on online IDE’s. So please, run them on your systems to explore the working.
Like Article
Suggest improvement
Share your thoughts in the comments
Please Login to comment...