访问包含列表的 Pandas DataFrame 列的每个第一个元素
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Accessing every 1st element of Pandas DataFrame column containing lists
提问by mkoala
I have a Pandas DataFrame with a column containing lists objects
我有一个包含列表对象的列的 Pandas DataFrame
A
0 [1,2]
1 [3,4]
2 [8,9]
3 [2,6]
How can I access the first element of each list and save it into a new column of the DataFrame? To get a result like this:
如何访问每个列表的第一个元素并将其保存到 DataFrame 的新列中？要得到这样的结果：
A new_col
0 [1,2] 1
1 [3,4] 3
2 [8,9] 8
3 [2,6] 2
I know this could be done via iterating over each row, but is there any "pythonic" way?
我知道这可以通过迭代每一行来完成，但是有什么“pythonic”的方式吗？
回答by DSM
As always, remember that storing non-scalar objects in frames is generally disfavoured, and should really only be used as a temporary intermediate step.
与往常一样，请记住，在帧中存储非标量对象通常是不受欢迎的，并且实际上应该仅用作临时的中间步骤。
That said, you can use the .straccessor even though it's not a column of strings:
也就是说，您可以使用.str访问器，即使它不是一列字符串：
>>> 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
回答by dmb
You can use mapand a lambdafunction
您可以使用map和一个lambda功能
df.loc[:, 'new_col'] = df.A.map(lambda x: x[0])
回答by jezrael
回答by Alexander
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.
您可以只使用条件列表推导式，它采用任何可迭代对象的第一个值，或者为该项目使用 None 。列表推导式非常 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
时间安排
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.
删除安全检查以确保可交互。
%timeit df['new_col'] = [val[0] for val in df["A"]]
100 loops, best of 3: 7.38 ms per loop
