Question

I have a data frame where the column names share a common element, other columns have been generated with a suffix to this common element. I have a list of these elements that is around ~100 entries. I'd like to iteratively slice the large df using this list, transform the sub-df's by grouping and eventually concatenate them back together.

I was thinking of using a dictionary approach-- using the list as keys, and then defining the columns sharing this element as values. I am not sure how to implement this. I have copied a simplified version to illustrate what I'd like to scale up. In reality there'd be around 100 keys each with 20 associated columns.

   A A_1 A_2 A_3  B B_1 B_2 B_3
0  1   e   f   g  1   x   y   z
1  2   e   f   g  2   x   y   z
2  3   e   f   g  3   x   y   z
3  3   e   f   g  3   x   y   z
4  3   e   f   g  4   x   y   z
5  3   e   f   g  4   x   y   z

df_list = ['A','B']

df_A = df[df.columns[df.columns.to_series().str.contains('A')]]

df_B = df[df.columns[df.columns.to_series().str.contains('B')]]

calc_A = df_A.groupby(['A']).head(1)
print(calc_A)

   A A_1 A_2 A_3
0  1   e   f   g
1  2   e   f   g
2  3   e   f   g


calc_B = df_B.groupby(['B']).head(1)
print(calc_B)

   B B_1 B_2 B_3
0  1   x   y   z
1  2   x   y   z
2  3   x   y   z
4  4   x   y   z

Please advise how to structure this dictionary, iterating through the list to slice the df and assign columns sharing the key as values for the new sub-df. Thank you.

Answer 1

IIUC，您可以对列前缀进行分组，然后初始化字典：

d = {}
for i, g in df.groupby(by=lambda x: x.split('_')[0], axis=1):
    d[i] = g.groupby(i).head(1)

您也可以使用 dict comprehension ：

来完成此操作

d = {
        i : g.groupby(i).head(1) 
        for (i, g) in df.groupby(by=lambda x: x.split('_')[0], axis=1)
}

for k, v in d.items():
    print(v, '\n')

   A A_1 A_2 A_3
0  1   e   f   g
1  2   e   f   g
2  3   e   f   g 

   B B_1 B_2 B_3
0  1   x   y   z
1  2   x   y   z
2  3   x   y   z
4  4   x   y   z 

d.keys()
dict_keys(['A', 'B'])

Split dataFrames based on column header prefix

1 个答案: