The question has changed, so too 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']

I think you need dropna for remove NaNs:

incoms=data['int_income'].dropna().unique().tolist()
print (incoms)
[75000.0, 50000.0, 0.0, 200000.0, 100000.0, 25000.0, 10000.0, 175000.0, 150000.0, 125000.0]

And if all values are integers only:

incoms=data['int_income'].dropna().astype(int).unique().tolist()
print (incoms)
[75000, 50000, 0, 200000, 100000, 25000, 10000, 175000, 150000, 125000]

Or remove NaNs by selecting all non NaN values by numpy.isnan:

a = data['int_income'].unique()
incoms= a[~np.isnan(a)].tolist()
print (incoms)
[75000.0, 50000.0, 0.0, 200000.0, 100000.0, 25000.0, 10000.0, 175000.0, 150000.0, 125000.0]

a = data['int_income'].unique()
incoms= a[~np.isnan(a)].astype(int).tolist()
print (incoms)
[75000, 50000, 0, 200000, 100000, 25000, 10000, 175000, 150000, 125000]

Pure python solution - slowier if big DataFrame:

incoms=[x for x in  list(set(data['int_income'])) if pd.notnull(x)]
print (incoms)
[0.0, 100000.0, 200000.0, 25000.0, 125000.0, 50000.0, 10000.0, 150000.0, 175000.0, 75000.0]

incoms=[int(x) for x in  list(set(data['int_income'])) if pd.notnull(x)]
print (incoms)
[0, 100000, 200000, 25000, 125000, 50000, 10000, 150000, 175000, 75000]

It is not working because you are trying to compare it to string 'nan'.

If excel cell is empty it is returned as NaN value in pandas.

You can use numpy library, to compare it with NaN:

import numpy as np

for l in df_list:
    newlist = [x for x in l if x != np.nan]
    print(newlist)

EDIT:

If you want to get all values from the dataframe, which are not NaN, you can just do:

df.stack().tolist()

If you want to print values with the loop (as in your example), you can do:

for l in df.columns:
    print(list(df[l][df[l].notna()]))

To create nested list with a loop:

main = []

for l in df.T.columns:
    new_list = list(df.T[l][df.T[l].notna()])
    main.append(new_list)
    
print(main)

for i in z:
    if not math.isnan(i):
        ztt.append(i)

This is one approach using df.apply.

import pandas as pd
import numpy as np

df = pd.DataFrame({"a": [[np.nan, np.nan, np.nan, "a", np.nan], [np.nan, np.nan], ["a", "b"]]})
df["a"] = df["a"].apply(lambda x: [i for i in x if str(i) != "nan"])
print(df)

Output:

        a
0     [a]
1      []
2  [a, b]

Try iterating over the list and building a new list using numpy's isnan method.

import numpy as np
nan = np.nan
l = [[1,2,3,nan,nan,nan],[4,5,nan],[9,8,7,6,5,4,3, nan, nan, nan, nan, nan]]
[[x for x in y if not np.isnan(x)] for y in l]