[FIXED] Replace Numpy NaN with string in list with strings

Issue

I have a list containing string elements, and several NaN numpy floats. E.g.

l=['foo', 'bar', 'baz', 'nan']

How do I replace the float nan to the string missing?

Most answers I found regard this issue in a pandas DataFrame.

---

Try 1:

for x in l:
    x=x.replace('nan', 'missing')

gives AttributeError: 'float' object has no attribute 'replace'

Try 2:

for x in l:
   if str(x)=='nan':
       x=str(x)

Command executes, but nothing changes.

---

Advised by comments:

1. ['missing' if x is 'nan' else x for x in l]
2. ['missing' if x is np.isnan else x for x in l]
3. ['missing' if x is np.nan else x for x in l]

Commands execute, but nothing changes.

Solution

I think you have a bad format for your NaN's (notice the nan is outputed nan and not 'nan'). The answers from the comment should work:

>>> import numpy as np
>>> l=['foo', 'bar', 'baz', np.nan]
>>> print l
['foo', 'bar', 'baz', nan]
>>> l_new=['missing' if x is np.nan else x for x in l]
>>> print l_new
['foo', 'bar', 'baz', 'missing']

For your current problem maybe make the following solution:

my_nan=l['some_index_with_nan']
l_new=['missing' if x is my_nan else x for x in l]

Answered By - Peter Mølgaard Pallesen