Logic in pandas apply function not as expected

Question

I have a dataframe in which I construct a list of values as such:

data_df["items"] = data_df.apply(
    lambda x: [x["item_1"], x["item_2"]]
    if x["item_2"]
    else [x["item_1"]],
    axis=1,
)

In my mind this approach should construct a list containing items 1 and 2 if item_2 exist otherwise it should just construct a list containing item_1. However, the check on item_2 doesn't seem to work as expected. I have tried checking for None values as well as empty strings. How do I write this so that the following happens:

if item_1 and item_2 are None or "" return []

if item_1 is not None but item_2 is None or "" return [item_1]

if item_1 and item_2 return [item_1, item_2]

Also, item_1 and item_2 are strings

[How to make good reproducible pandas examples](https://stackoverflow.com/questions/20109391/how-to-make-good-reproducible-pandas-examples) — wwii, May 21 '21 at 13:20

Nk03 · Accepted Answer · 2021-05-21T13:13:34.503

2

use pd.isna():

df['item_list'] = df[['item1', 'item2']].apply(lambda x: [i for i in x if not pd.isna(i)], 1)

update:

df['item_list'] = df[['item1', 'item2']].apply(lambda x: [i for i in x if not (pd.isna(i) or i == '')], 1)

edited May 21 '21 at 13:13

answered May 21 '21 at 13:05

Nk03

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This is close @Nk03. However, I still get [`item_1`, ""]. Is it possible to avoid the empty string? – 00robinette May 21 '21 at 13:12
Add one more condition for an empty string values. – Nk03 May 21 '21 at 13:13

Logic in pandas apply function not as expected

1 Answers1