I would like to drop the [] for a given df
df=pd.DataFrame(dict(a=[1,2,4,[],5]))
Such that the expected output will be
a
0 1
1 2
2 4
3 5
Edit:
or to make thing more interesting, what if we have two columns and some of the cell is with [] to be dropped.
df=pd.DataFrame(dict(a=[1,2,4,[],5],b=[2,[],1,[],6]))
One way is to get the string repr and filter:
df = df[df['a'].map(repr)!='[]']
Output:
a
0 1
1 2
2 4
4 5
For multiple columns, we could apply the above:
out = df[df.apply(lambda c: c.map(repr)).ne('[]').all(axis=1)]
Output:
a b
0 1 2
2 4 1
4 5 6
You can't use equality directly as pandas will try to align a Series and a list, but you can use isin:
df[~df['a'].isin([[]])]
output:
a
0 1
1 2
2 4
4 5
To act on all columns:
df[~df.isin([[]]).any(1)]
output:
a b
0 1 2
2 4 1
4 5 6