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This is my dataframe:

          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  NaN
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,...

I want to replace Nan with [[],[],[],[]]. How to do that?

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

Is working well for a 1 element list, but how can we use it with 4 element list?

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  • Is it a string or np.nan? Commented May 9, 2022 at 17:43
  • 1
    it is a np.nan. Commented May 9, 2022 at 17:46

3 Answers 3

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2

You can't use fillna with lists, but you can create a Series containing your list repeated for the length of the dataframe, and assign that to the b where b is NaN:

df.loc[df['b'].isna(), 'b'] = pd.Series([ [[]]*4 ] * len(df))
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2

You can try

df['ids'] = df['ids'].apply(lambda x: [[],[],[],[]] if x!=x else x)

This uses the feature that np.nan is not equal with itself.

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1 Comment

Cool to take advantage of that property of NaN. However, apply is necessarily going to be slow.
1

You can use .apply() checking if the value is np.nan or not. If yes, then fill the nested list otherwise the normal value.

Try this:

df['ids'] = df['ids'].apply(lambda d: d if d is not np.nan else [[],[],[],[]])
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