假设我有一个包含索引值的元组列表:
mapper= [(0,6),(9,13),(17,27)]
我有一个大的master_df,我想根据上面列表中的元组索引值将其拆分为多个df
映射器[0][0]是起点,映射器[0][1]是终点。我有一个df名字的列表
df_list= ['df_1','df_2,'df_3']
我尝试了下面的代码片段,尝试根据映射器中的索引值填充多个df
for x in range(len(df_list)):
df_list[x] = master_df[mapper[x][0]:mapper[x][1]]
但这并不是我想象的那样。对我来说,理想的解决方案是根据列表中的元组索引值对主函数进行三次单独的df拆分
以下是我试图完成的一个例子:
master_df:
Name Role Location
0 Gina Assistance NY
1 Jake Officer Brooklyn
2 Boyle Detective 99
3 Scully Assistance NY
4 Diaz Officer Brooklyn
5 Hitchcock Detective 99
6 Amy Assistance NY
7 Terry Officer Brooklyn
8 Holt Detective 99
9 Judy Assistance NY
10 Adrian Officer Brooklyn
mapper = [(0,3),(3,6),(6,11)]
df_list = ['df_1','df_2','df_3']
寻求结果
df_1:
Name Role Location
0 Gina Assistance NY
1 Jake Officer Brooklyn
2 Boyle Detective 99
df_2:
Name Role Location
3 Scully Assistance NY
4 Diaz Officer Brooklyn
5 Hitchcock Detective 99
df_3:
Name Role Location
6 Amy Assistance NY
7 Terry Officer Brooklyn
8 Holt Detective 99
9 Judy Assistance NY
10 Adrian Officer Brooklyn
感谢您的帮助/指导
您可以使用
*
解压元组,并将其传递给range函数,然后使用iloc[]
获取这些索引:如果你想把它们分配给名字,你必须把它做成一本字典(见How to create a variable number of variables)
相关问题 更多 >
编程相关推荐