[英]Split dataFrames based on column header prefix
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. 我有一个数据框,其中的列名称共享一个公共元素,而其他列的生成都带有此公共元素的后缀。 I have a list of these elements that is around ~100 entries.
我有这些元素的列表,大约有100个条目。 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.
我想使用此列表来迭代切片大型df,通过分组来转换子df,最后将它们重新连接在一起。
I was thinking of using a dictionary approach-- using the list as keys, and then defining the columns sharing this element as values. 我当时正在考虑使用字典方法-将列表用作键,然后将共享该元素的列定义为值。 I am not sure how to implement this.
我不确定如何执行此操作。 I have copied a simplified version to illustrate what I'd like to scale up.
我复制了一个简化版本,以说明我想要扩展的内容。 In reality there'd be around 100 keys each with 20 associated columns.
实际上,大约有100个键,每个键具有20个关联的列。
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. 请建议如何构造该字典,遍历列表以对df进行切片,并将共享密钥的列分配为新sub-df的值。 Thank you.
谢谢。
IIUC, you can group on column prefixes, and then initialise a dictionary: 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)
You could also do this using a dict comprehension : 您也可以使用dict理解来做到这一点:
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'])
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