remove from the list of missing values None

Question

[['user_1',
  'user_2',
  'user_8',
   None,
   None],
 ['ben',
  'karl',
   None,
   None]]

I try to remove the missing values

for element in df:
     element=[x for x in element if x is not None]

this code leave everything as it was

You could use [**`filter`**](https://docs.python.org/2/library/functions.html#filter), although that would also remove empty strings, zeroes, etc. — Peter Wood, Jan 11 '17 at 09:54
You may find this article helpful: [Facts and myths about Python names and values](http://nedbatchelder.com/text/names.html), which was written by SO veteran Ned Batchelder.. And here's a condensed version of the same info, with cute diagrams: [Other languages have "variables", Python has "names"](http://python.net/~goodger/projects/pycon/2007/idiomatic/handout.html#other-languages-have-variables) — PM 2Ring, Jan 11 '17 at 10:14
FWIW, none of the answers in that dupe target show how to mutate the existing list (or its sublists) rather than replacing it. — PM 2Ring, Jan 11 '17 at 10:20

YOBA · Answer 1 · 2017-01-11T10:00:37.213

1

my_list= [['user_1',
  'user_2',
  'user_8',
   None,
   None],
 ['ben',
  'karl',
   None,
   None]]

print [ [ elt for elt in a_list if elt is not None ] for a_list in my_list ]

[['user_1', 'user_2', 'user_8'], ['ben', 'karl']]

edited Jan 11 '17 at 10:00

answered Jan 11 '17 at 09:36

YOBA

2,759
1
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This would also remove list items with `0` in them... – l'L'l Jan 11 '17 at 09:47
@l'L'l right, fixed. – YOBA Jan 11 '17 at 10:00
This is fine if `len(my_list)` is small, but if it's huge it's probably desirable to save RAM by mutating the existing list rather than replacing it. OTOH, replacing `my_list` _is_ a little faster than mutating it. – PM 2Ring Jan 11 '17 at 10:11

score 1 · Accepted Answer · answered Jan 11 '17 at 09:57

The problem with your code is that the line

element=[x for x in element if x is not None]

creates the new filtered list and binds it to the name element, which replaces the old list object that the name element was bound to. But we can use slice assignment to make it mutate that list instead:

df = [
    [
        'user_1',
        'user_2',
        'user_8',
        None,
        None,
    ],
    [
        'ben',
        'karl',
        None,
        None,
    ]
]

# Remove the `None` values
for element in df:
     element[:] = [x for x in element if x is not None]

for row in df:
    print(row)

output

['user_1', 'user_2', 'user_8']
['ben', 'karl']

score 1 · Answer 3 · answered Jan 11 '17 at 09:58

1

for element in df:
     element[:]=[x for x in element if x is not None]

out:

[['user_1', 'user_2', 'user_8'], ['ben', 'karl']]

Document:

answered Jan 11 '17 at 09:58

宏杰李

11,820
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score 0 · Answer 4 · answered Jan 11 '17 at 09:32

Reassigning element does not alter the list, because element is a separate variable.

With a small change, you can generate a new outer list using a nested comprehension, something like this:

df = [[x for x in element if x is not None] for element in df]

remove from the list of missing values None

4 Answers4