扁平化多索引列的简洁方法 [英] concise way of flattening multiindex columns

问题描述

在groupby-aggregate中使用多个函数会产生一个多索引，然后我希望将其展平.

Using more than 1 function in a groupby-aggregate results in a multi-index which I then want to flatten.

示例:

df = pd.DataFrame(
    {'A': [1,1,1,2,2,2,3,3,3],
     'B': np.random.random(9),
     'C': np.random.random(9)}
)
out = df.groupby('A').agg({'B': [np.mean, np.std], 'C': np.median})

# example output

          B                   C
       mean       std    median
A
1  0.791846  0.091657  0.394167
2  0.156290  0.202142  0.453871
3  0.482282  0.382391  0.892514

目前，我是这样手动完成的

Currently, I do it manually like this

out.columns = ['B_mean', 'B_std', 'C_median']

这给了我想要的结果

     B_mean     B_std  C_median
A
1  0.791846  0.091657  0.394167
2  0.156290  0.202142  0.453871
3  0.482282  0.382391  0.892514

但是我正在寻找一种自动执行此过程的方法，因为这是单调的，耗时的，并且允许我在重命名列时进行拼写错误.

but I'm looking for a way to automate this process, as this is monotonous, time consuming and allows me to make typos as I rename the columns.

在进行groupby-aggregate时，是否有一种方法可以返回扁平索引而不是多索引?

Is there a way to return a flattened index instead of a multi-index when doing a groupby-aggregate?

我需要弄平列以保存到文本文件，然后将由不处理多索引列的其他程序读取该文件.

I need to flatten the columns to save to a text file, which will then be read by a different program that doesn't handle multi-indexed columns.

推荐答案

您可以对列进行map join

out.columns = out.columns.map('_'.join)
out
Out[23]: 
     B_mean     B_std  C_median
A                              
1  0.204825  0.169408  0.926347
2  0.362184  0.404272  0.224119
3  0.533502  0.380614  0.218105

出于某种原因(当列包含int时)我更喜欢这种方式

For some reason (when the column contain int) I like this way better

out.columns.map('{0[0]}_{0[1]}'.format) 
Out[27]: Index(['B_mean', 'B_std', 'C_median'], dtype='object')

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