pandas 提取熊猫列中列表的元素
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Extracting an element of a list in a pandas column
提问by Ignacio Vergara Kausel
I have a DataFrame that contains a list on each column as shown in the example below with only two columns.
我有一个 DataFrame，它在每列上包含一个列表，如下例所示，只有两列。
Gamma Beta
0 [1.4652917656926299, 0.9326935235505321, float] [91, 48.611034768515864, int]
1 [2.6008354611105995, 0.7608529935313189, float] [59, 42.38646954167245, int]
2 [2.6386970166722348, 0.9785848171888037, float] [89, 37.9011122659478, int]
3 [3.49336632573625, 1.0411524946972244, float] [115, 36.211134224288344, int]
4 [2.193991200007534, 0.7955134305428825, float] [128, 50.03563864975485, int]
5 [3.4574527664490997, 0.9399880977511021, float] [120, 41.841146628802875, int]
6 [3.1190582380554863, 1.0839109431114795, float] [148, 55.990072419824514, int]
7 [2.7757359940789916, 0.8889801332053203, float] [142, 51.08885697101243, int]
8 [3.23820908493237, 1.0587479742892683, float] [183, 43.831293356668425, int]
9 [2.2509032790941985, 0.8896196407231622, float] [66, 35.9377662201882, int]
I'd like to extract for every column the first position of the list on each row to get a DataFrame looking as follows.
我想为每一列提取每行列表的第一个位置，以获得如下所示的 DataFrame。
Gamma Beta
0 1.4652917656926299 91
1 2.6008354611105995 59
2 2.6386970166722348 89
...
Up to now, my solution would be like [row[1][0] for row in df_params.itertuples()], which I could iterate for every column index of the row and then compose my new DataFrame.
到目前为止，我的解决方案类似于[row[1][0] for row in df_params.itertuples()]，我可以对行的每个列索引进行迭代，然后组成我的新 DataFrame。
An alternative is new_df = df_params['Gamma'].apply(lambda x: x[0])and then to iterate to go through all the columns.
另一种方法是new_df = df_params['Gamma'].apply(lambda x: x[0])然后迭代以遍历所有列。
My question is, is there a less cumbersome way to perform this operation?
我的问题是，有没有一种不那么麻烦的方法来执行这个操作？
回答by IanS
You can use the straccessor for lists, e.g.:
您可以将str访问器用于列表，例如：
df_params['Gamma'].str[0]
This should work for all columns:
这应该适用于所有列：
df_params.apply(lambda col: col.str[0])
回答by A.Kot
Itertuples would be pretty slow. You could speed this up with the following:
Itertuples 会很慢。您可以通过以下方式加快速度：
for column_name in df_params.columns:
df_params[column_name] = [i[0] for i in df_params[column_name]]
