How can I remove Nan from list Python/NumPy

Question

I have a list that countain values, one of the values I got is \'nan\'

countries= [nan, \'USA\', \'UK\', \'France\']

I tried to remove it,

Answer 1

Using your example where...

countries= [nan, 'USA', 'UK', 'France']

Since nan is not equal to nan (nan != nan) and countries[0] = nan, you should observe the following:

countries[0] == countries[0]
False

However,

countries[1] == countries[1]
True
countries[2] == countries[2]
True
countries[3] == countries[3]
True

Therefore, the following should work:

cleanedList = [x for x in countries if x == x]

Answer 2

The question has changed, so to has the answer:

Strings can't be tested using math.isnan as this expects a float argument. In your countries list, you have floats and strings.

In your case the following should suffice:

cleanedList = [x for x in countries if str(x) != 'nan']

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Old answer

In your countries list, the literal 'nan' is a string not the Python float nan which is equivalent to:

float('NaN')

In your case the following should suffice:

cleanedList = [x for x in countries if x != 'nan']

Answer 3

if you check for the element type

type(countries[1])

the result will be <class float> so you can use the following code:

[i for i in countries if type(i) is not float]

Answer 4

In your example 'nan' is a string so instead of using isnan() just check for the string

like this:

cleanedList = [x for x in countries if x != 'nan']

Answer 5

use numpy fancy indexing:

In [29]: countries=np.asarray(countries)

In [30]: countries[countries!='nan']
Out[30]: 
array(['USA', 'UK', 'France'], 
      dtype='|S6')

Answer 6

import numpy as np

mylist = [3, 4, 5, np.nan]
l = [x for x in mylist if ~np.isnan(x)]

This should remove all NaN. Of course, I assume that it is not a string here but actual NaN (np.nan).