Python 具有多索引列的 Pandas 数据框 - 合并级别
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Pandas dataframe with multiindex column - merge levels
提问by Zhubarb
I have a dataframe, grouped, with multiindex columns as below:
我有一个数据框，grouped具有多索引列，如下所示：
import pandas as pd
codes = ["one","two","three"];
colours = ["black", "white"];
textures = ["soft", "hard"];
N= 100 # length of the dataframe
df = pd.DataFrame({ 'id' : range(1,N+1),
'weeks_elapsed' : [random.choice(range(1,25)) for i in range(1,N+1)],
'code' : [random.choice(codes) for i in range(1,N+1)],
'colour': [random.choice(colours) for i in range(1,N+1)],
'texture': [random.choice(textures) for i in range(1,N+1)],
'size': [random.randint(1,100) for i in range(1,N+1)],
'scaled_size': [random.randint(100,1000) for i in range(1,N+1)]
}, columns= ['id', 'weeks_elapsed', 'code','colour', 'texture', 'size', 'scaled_size'])
grouped = df.groupby(['code', 'colour']).agg( {'size': [np.sum, np.average, np.size, pd.Series.idxmax],'scaled_size': [np.sum, np.average, np.size, pd.Series.idxmax]}).reset_index()
>> grouped
code colour size scaled_size
sum average size idxmax sum average size idxmax
0 one black 1031 60.647059 17 81 185.153944 10.891408 17 47
1 one white 481 37.000000 13 53 204.139249 15.703019 13 53
2 three black 822 48.352941 17 6 123.269405 7.251141 17 31
3 three white 1614 57.642857 28 50 285.638337 10.201369 28 37
4 two black 523 58.111111 9 85 80.908912 8.989879 9 88
5 two white 669 41.812500 16 78 82.098870 5.131179 16 78
[6 rows x 10 columns]
How can I flatten/merge the column index levels as: "Level1|Level2", e.g. size|sum, scaled_size|sum. etc? If this is not possible, is there a way to groupby()as I did above without creating multi-index columns?
如何将列索引级别展平/合并为：“Level1|Level2”，例如size|sum，scaled_size|sum。等等？如果这是不可能的，有没有办法groupby()像我上面所做的那样不创建多索引列？
采纳答案by acushner
you could always change the columns:
您可以随时更改列：
grouped.columns = ['%s%s' % (a, '|%s' % b if b else '') for a, b in grouped.columns]
回答by Scott Boston
There is potentially a better way, more pythonic way to flatten multiindex columns.
可能有更好的方法，更pythonic 的方法来展平多索引列。
1. Use map and join with string column headers:
1. 使用 map 并连接字符串列标题：
grouped.columns = grouped.columns.map('|'.join).str.strip('|')
print(grouped)
Output:
输出：
code colour size|sum size|average size|size size|idxmax \
0 one black 862 53.875000 16 14
1 one white 554 46.166667 12 18
2 three black 842 49.529412 17 90
3 three white 740 56.923077 13 97
4 two black 1541 61.640000 25 50
scaled_size|sum scaled_size|average scaled_size|size scaled_size|idxmax
0 6980 436.250000 16 77
1 6101 508.416667 12 13
2 7889 464.058824 17 64
3 6329 486.846154 13 73
4 12809 512.360000 25 23
2. Use map with format for column headers that has numeric data types.
2. 对具有数字数据类型的列标题使用带格式的映射。
grouped.columns = grouped.columns.map('{0[0]}|{0[1]}'.format)
Output:
输出：
code| colour| size|sum size|average size|size size|idxmax \
0 one black 734 52.428571 14 30
1 one white 1110 65.294118 17 88
2 three black 930 51.666667 18 3
3 three white 1140 51.818182 22 20
4 two black 656 38.588235 17 77
5 two white 704 58.666667 12 17
scaled_size|sum scaled_size|average scaled_size|size scaled_size|idxmax
0 8229 587.785714 14 57
1 8781 516.529412 17 73
2 10743 596.833333 18 21
3 10240 465.454545 22 26
4 9982 587.176471 17 16
5 6537 544.750000 12 49
3. Use list comprehension with f-string for Pytnon 3.6+:
3. 对 Pytnon 3.6+ 使用带有 f-string 的列表理解：
grouped.columns = [f'{i}|{j}' if j != '' else f'{i}' for i,j in grouped.columns]
Output:
输出：
code colour size|sum size|average size|size size|idxmax \
0 one black 1003 43.608696 23 76
1 one white 1255 59.761905 21 66
2 three black 777 45.705882 17 39
3 three white 630 52.500000 12 23
4 two black 823 54.866667 15 33
5 two white 491 40.916667 12 64
scaled_size|sum scaled_size|average scaled_size|size scaled_size|idxmax
0 12532 544.869565 23 27
1 13223 629.666667 21 13
2 8615 506.764706 17 92
3 6101 508.416667 12 43
4 7661 510.733333 15 42
5 6143 511.916667 12 49
回答by Ningrong Ye
Based on Scott Boston's answer, little update(it will be work for 2 or more levels column):
根据 Scott Boston 的回答，几乎没有更新（它将适用于 2 个或更多级别的列）：
temp.columns.map(lambda x: '|'.join([str(i) for i in x]))
Thank you, Boston!
谢谢你，波士顿！
