使用列表理解从索引列表列表生成 pandas DataFrame 列

Question

这是我编写的应用于 df 的每一行的函数

def use_list_of_list_of_indexes(column1, column2):
"""
Column that contains list of lists of indices is used to generate a new column based on indices
_
Args:
column1 - list of elements
column2 - list of lists of indexes
_
Returns:
pandas.core.series.Series
"""
list_of_lists = column1.apply(lambda row: [[row[i] for i in x] for x in column2])
return list_of_lists

这里是df

col1                     col2
[55, 475,243,60]         [[0,1,2], [0,1,3], [0,2,3], [1,2,3]]

使用时

df['col3'] = use_list_of_list_of_indexes(df.col1, df.col2)

我想得到

col3
[[55, 475, 243], [55, 475, 60], [55, 243, 60], [475, 243, 60]]

但是，我得到错误

TypeError: list indices must be integers or slices, not list

将 [0] 添加到 df.col2 解决了问题，但是我应该如何重写 use_list_of_list_of_indexes 函数才能将其应用于整个列？

df['col3'] = use_list_of_list_of_indexes(df.col1, df.col2[0])

Answer 1

如果你必须这样做，你就错过了一个循环：

column1.apply(lambda row: [[[row[z] for z in y] for y in x] for x in column2])

x 给你 [[0,1,2], [0,1,3], [0,2,3], [1,2,3]]

y 给你 [0,1,2]

所以你必须使用 z 来遍历 y

稍微好一点的做法：

df['col3'] = df[['col1', 'col2']].apply(lambda r: [[r['col1'][y] for y in x] for x in r['col2']], axis = 1)

使用列表理解从索引列表列表生成 pandas DataFrame 列

Generate pandas DataFrame column from list of lists of indexes using list comprehension

python

list-comprehension

dataframe

pandas