How to fill dataframe Nan values with empty list [] in pandas?

My approach is similar to @hellpanderrr's, but instead tests for list-ness rather than using isnan:

df['ids'] = df['ids'].apply(lambda d: d if isinstance(d, list) else [])

I originally tried using pd.isnull (or pd.notnull) but, when given a list, that returns the null-ness of each element.

After a lot of head-scratching I found this method that should be the most efficient (no looping, no apply), just assigning to a slice:

isnull = df.ids.isnull()

df.loc[isnull, 'ids'] = [ [[]] * isnull.sum() ]

The trick was to construct your list of [] of the right size (isnull.sum()), and then enclose it in a list: the value you are assigning is a 2D array (1 column, isnull.sum() rows) containing empty lists as elements.

You can first use loc to locate all rows that have a nan in the ids column, and then loop through these rows using at to set their values to an empty list:

for row in df.loc[df.ids.isnull(), 'ids'].index:
    df.at[row, 'ids'] = []

>>> df
        date                                             ids
0 2011-04-23  [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13]
1 2011-04-24  [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13]
2 2011-04-25  [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13]
3 2011-04-26                                              []
4 2011-04-27  [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13]
5 2011-04-28  [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13]

A simple solution would be:

df['ids'].fillna("").apply(list)

As noted by @timgeb, this requires df['ids'] to contain lists or nan only.