Delete "nan" in python list

Question

I'm new in python and I have a simple question how could I delete the first "nan" of this list but maintain the second in case the value is duplicated.

for example:

my_list = ['experiencia: 4',
           'nan',
          'experiencia: 2',
           'nan',
           'nan',
          'experiencia: 10',
           'nan',
          'experiencia: 2',
           'nan',
          'experiencia: 3',
           'nan',
           'nan',
           'nan',
           'experiencia: 9',
           'nan',
           'experiencia: 10']

and the output I want to obtain is the following:

 my_list = ['experiencia: 4',
          'experiencia: 2',
           'nan',
          'experiencia: 10',
          'experiencia: 2',
          'experiencia: 3',
           'nan',
           'nan',
           'experiencia: 9',
           'experiencia: 10']

Have you tried coding it up yet? sometimes when it's tricky to figure out how to go about implementing something, it helps to break it up into a flow chart of individual steps. — Aaron, Sep 05 '20 at 22:08
I also would point out, it is often easier to build a new list rather than trying to modify an existing one in-place. — Aaron, Sep 05 '20 at 22:10

score 2 · Answer 1 · answered Sep 05 '20 at 22:18

You can ... get creative:

Group the whole list by its contents and disassemle the grouping again. If the key is a "nan" you need to use one less then the groupings value has elements - for non-"nan" use the grouping value as is. Lastly, remove empty "nan" - groupings (key = "nan", value = ["nan"] gets reduced to [] due to subtracting one):

my_list = ['experiencia: 4',
           'nan',
          'experiencia: 2',
           'nan',
           'nan',
          'experiencia: 10',
           'nan',
          'experiencia: 2',
           'nan',
          'experiencia: 3',
           'nan',
           'nan',
           'nan',
           'experiencia: 9',
           'nan',
           'experiencia: 10']

from itertools import groupby


result = [x for y in (list(amount) if value != "nan" else ["nan"] * (len(list(amount))-1)
                  for value, amount in groupby(my_list)) for x in y if x]
print(result)

Output (reformatted):

['experiencia: 4', 
 'experiencia: 2', 
 'nan', 
 'experiencia: 10', 
 'experiencia: 2',
 'experiencia: 3', 
 'nan', 
 'nan', 
 'experiencia: 9', 
 'experiencia: 10']

Readup:

Explanation of how nested list comprehension works? (as to how the get rid of [] and unstacking of inner lists works)

thanks a lot! i was using the Pandas Series and dropping the index using a similar criteria, but your way to do it is better than mine, thanks! — cristobalvch, Sep 05 '20 at 22:32

Seyi Daniel · Answer 2 · 2020-09-05T22:35:25.093

Create a list for the indices you wish to do delete and delete them using the pop method

delete_list = [1,2,4,5,6,9]
newlist = ''
for i in delete_list:
   my_list.pop(i)
print (my_list)

Result

['experiencia: 4', 'experiencia: 2', 'nan', 'experiencia: 10', 'experiencia: 2', 'experiencia: 3', 'nan', 'nan', 'experiencia: 9', 'experiencia: 10']

Note also, that i have shifted my deletion pattern to suit the index adjustment after every deletion

score -1 · Answer 3 · answered Sep 05 '20 at 22:12

-1

my_list.remove('nan')

will delete only the first 'nan'

answered Sep 05 '20 at 22:12

Hadrian

917
5
10

but not the first nan of every nan occurences – Patrick Artner Sep 05 '20 at 22:15

Brian Larsen · Answer 4 · 2020-09-06T18:05:55.167

-1

This will also give you the desired output:

from itertools import groupby
from itertools import chain
my_list = [*chain.from_iterable([[k] * (len([*v]) - 1) if k == "nan" else [k] for k, v in groupby(my_list)])]

edited Sep 06 '20 at 18:05

answered Sep 05 '20 at 22:32

Brian Larsen

612
8
9

Delete "nan" in python list

4 Answers4