[python] How can I remove Nan from list Python/NumPy

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, but I everytime get an error

cleanedList = [x for x in countries if (math.isnan(x) == True)]
TypeError: a float is required

When I tried this one :

cleanedList = cities[np.logical_not(np.isnan(countries))]
cleanedList = cities[~np.isnan(countries)]

TypeError: ufunc 'isnan' not supported for the input types, and the inputs could not be safely coerced to any supported types according to the casting rule ''safe''

use numpy fancy indexing:

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

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

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]

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).

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]

The problem comes from the fact that np.isnan() does not handle string values correctly. For example, if you do:

np.isnan("A")
TypeError: ufunc 'isnan' not supported for the input types, and the inputs could not be safely coerced to any supported types according to the casting rule ''safe''

However the pandas version pd.isnull() works for numeric and string values:

pd.isnull("A")
> False

pd.isnull(3)
> False

pd.isnull(np.nan)
> True

pd.isnull(None)
> True

I like to remove missing values from a list like this:

list_no_nan = [x for x in list_with_nan if pd.notnull(x)]

Another way to do it would include using filter like this:

countries = list(filter(lambda x: str(x) != 'nan', countries))

I noticed that Pandas for example will return 'nan' for blank values. Since it's not a string you need to convert it to one in order to match it. For example:

ulist = df.column1.unique() #create a list from a column with Pandas which 
for loc in ulist:
    loc = str(loc)   #here 'nan' is converted to a string to compare with if
    if loc != 'nan':
        print(loc)

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']

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']

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']