[FIXED] different outputs for checking if a nan value is in the list or not

Issue

I have a list and I want to check if there is a missing value or not! I used two different ways to check it but didn't get the same outputs!

np.nan in lst1

output for this one is "False"

n = 0
for i in range(len(lst1)):
    if lst1[i] is np.nan :
        n+=1
print(n)

and for this one is 1!

the second answer is actually true but I don't know why in the first code I got "False"

Solution

To use math.isnan(x) is preferred way of NaN finding. As I understand np.nan and pure python NaN are not the same objects. And you get False at checking.

import math
import numpy as np

lst1 = [1,2,3, float('nan'), 2, np.nan, 4]
for ix, vl in enumerate(lst1):
    if math.isnan(vl):
        print(ix, vl)

for ix, vl in enumerate(lst1):
    if vl is np.nan:
        print(ix, vl)

Output for math.isnan()

3 nan
5 nan

Output for np.nan check

5 nan

Answered By - Andy Pavlov