13

I have a pandas DataFrame where one column contains lists and would like to select the rows where the lists are not empty.

Example data:

df = pd.DataFrame({'letter': ["a", "b", "c", "d", "e"], 
               'my_list':[[0,1,2],[1,2],[],[],[0,1]]})

df
    letter   my_list
0   a   [0, 1, 2]
1   b   [1, 2]
2   c   []
3   d   []
4   e   [0, 1]

What I'd like:

df
    letter  my_list
0   a   [0, 1, 2]
1   b   [1, 2]
4   e   [0, 1]

What I'm trying:

df[df.my_list.map(lambda x: if len(x) !=0)]

... which returns an invalid syntax error. Any suggestions?

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3 Answers 3

Reset to default
24

Empty lists evaluate to False in a boolean context

df[df.my_list.astype(bool)]

  letter    my_list
0      a  [0, 1, 2]
1      b     [1, 2]
4      e     [0, 1]
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1 Comment

Thanks! I just got this to work df[df.my_list.map(lambda x: len(x) !=0)]. But your solution is much better.
5

Or you can follow your own logic using length of the list to determine keep or not .

df[df.my_list.str.len().gt(0)]
Out[1707]: 
  letter    my_list
0      a  [0, 1, 2]
1      b     [1, 2]
4      e     [0, 1]
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Comments

0

Another option I found helpful is to 'filter' the dataframe like this:

df = df [df ['my_list'] != '']

The != '' is the operation what you want to filter. In this case: no rows where the cell value is None / empty. In other cases it could be simply changed, e.g. list must be longer then 3 df[ len(df ['my_list']) > 3, or first value has to be bigger than 2 df [df ['my_list'][0] > 2], or many other conditions.

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