拆分后如何根据索引过滤熊猫中的列表[重复]

Question

假设我有一个看起来像这样的数据框。

import pandas as pd

df = {'col1':["A,1,a,text,stack,over,flow","B,2,b,text,stack,over,flow","C,3,c,text,stack,over,flow","D,4,d,text,stack,over,flow"]}
df = pd.DataFrame(df)

给#

                         col1
0  A,1,a,text,stack,over,flow
1  B,2,b,text,stack,over,flow
2  C,3,c,text,stack,over,flow
3  D,4,d,text,stack,over,flow

我正在尝试拆分 columnby,` 并相应地访问元素。

new = df["col1"].str.split(pat = ",", expand=True)
print(new)

给我 ＃

   0  1  2     3      4     5     6
0  A  1  a  text  stack  over  flow
1  B  2  b  text  stack  over  flow
2  C  3  c  text  stack  over  flow
3  D  4  d  text  stack  over  flow

我的问题来了

我如何在str.split()? 之后访问多个手工挑选的索引。如果我想访问1 索引，我可以轻松地通过

new = df["col1"].str.split(pat = ",", expand=True)[1]

给我 ＃

0    1
1    2
2    3
3    4
Name: 1, dtype: object

从上面的语法我怎么能一次提到多个索引？就像如果我想过滤1st,4th,6th我怎么能提到它？

new = df["col1"].str.split(pat = ",", expand=True)[1][4][6] ?

肯定会给我一个错误。

raise KeyError(key) from err
KeyError: 4

Answer 1

使用清单：

new = df["col1"].str.split(pat = ",", expand=True)[[1,4,6]]