I have a Pandas DataFrame with a column containing lists objects
A
0 [1,2]
1 [3,4]
2 [8,9]
3 [2,6]
How can I access the firs
As always, remember that storing non-scalar objects in frames is generally disfavoured, and should really only be used as a temporary intermediate step.
That said, you can use the .str
accessor even though it's not a column of strings:
>>> df = pd.DataFrame({"A": [[1,2],[3,4],[8,9],[2,6]]})
>>> df["new_col"] = df["A"].str[0]
>>> df
A new_col
0 [1, 2] 1
1 [3, 4] 3
2 [8, 9] 8
3 [2, 6] 2
>>> df["new_col"]
0 1
1 3
2 8
3 2
Name: new_col, dtype: int64
Use apply with x[0]
:
df['new_col'] = df.A.apply(lambda x: x[0])
print df
A new_col
0 [1, 2] 1
1 [3, 4] 3
2 [8, 9] 8
3 [2, 6] 2
You can use map
and a lambda
function
df.loc[:, 'new_col'] = df.A.map(lambda x: x[0])
You can just use a conditional list comprehension which takes the first value of any iterable or else uses None for that item. List comprehensions are very Pythonic.
df['new_col'] = [val[0] if hasattr(val, '__iter__') else None for val in df["A"]]
>>> df
A new_col
0 [1, 2] 1
1 [3, 4] 3
2 [8, 9] 8
3 [2, 6] 2
Timings
df = pd.concat([df] * 10000)
%timeit df['new_col'] = [val[0] if hasattr(val, '__iter__') else None for val in df["A"]]
100 loops, best of 3: 13.2 ms per loop
%timeit df["new_col"] = df["A"].str[0]
100 loops, best of 3: 15.3 ms per loop
%timeit df['new_col'] = df.A.apply(lambda x: x[0])
100 loops, best of 3: 12.1 ms per loop
%timeit df.A.map(lambda x: x[0])
100 loops, best of 3: 11.1 ms per loop
Removing the safety check ensuring an interable.
%timeit df['new_col'] = [val[0] for val in df["A"]]
100 loops, best of 3: 7.38 ms per loop