Pandas 多索引isn'；当pd包含多个小计行时保留t
当我试图进行第二次小计时，我失去了多重索引。我可以添加第一个小计，但不能添加第二个小计，即B0的总和 这就是我当前的df：Pandas 多索引isn'；当pd包含多个小计行时保留t,pandas,multi-index,Pandas,Multi Index,当我试图进行第二次小计时，我失去了多重索引。我可以添加第一个小计，但不能添加第二个小计，即B0的总和 这就是我当前的df： lvl0 a b lvl1 bar foo bah foo A0 B0 C0 D0 1 0 3 2 D1 5 4 7 6 First Total
lvl0 a b
lvl1 bar foo bah foo
A0 B0 C0 D0 1 0 3 2
D1 5 4 7 6
First Total 6 4 10 8
C1 D0 9 8 11 10
D1 13 12 15 14
First Total 22 20 26 24
C2 D0 17 16 19 18
lvl0 a b
lvl1 bar foo bah foo
(A0, B0, C2, First Total) 38 36 42 40
(A0, B0, C3, D0) 25 24 27 26
(A0, B0, C3, D1) 29 28 31 30
(A0, B0, C3, First Total) 54 52 58 56
(A0, B0, Second Total) 120 112 136 128
(A0, B1, C0, D0) 33 32 35 34
(A0, B1, C0, D1) 37 36 39 38
(A0, B1, C0, First Total) 70 68 74 72
(A0, B1, C1, D0) 41 40 43 42
import pandas as pd
import numpy as np
# creating multiIndex
def mklbl(prefix, n):
return ["%s%s" % (prefix, i) for i in range(n)]
miindex = pd.MultiIndex.from_product([mklbl('A', 4),
mklbl('B', 2),
mklbl('C', 4),
mklbl('D', 2)])
micolumns = pd.MultiIndex.from_tuples([('a', 'foo'), ('a', 'bar'),
('b', 'foo'), ('b', 'bah')],
names=['lvl0', 'lvl1'])
dfmi = pd.DataFrame(np.arange(len(miindex) * len(micolumns))
.reshape((len(miindex), len(micolumns))),
index=miindex,
columns=micolumns).sort_index().sort_index(axis=1)
# My code STARTS HERE
# creating the first subtotal
print(dfmi.index)
df1 = dfmi.groupby(level=[0,1,2]).sum()
df2 = dfmi.groupby(level=[0, 1]).sum()
df1 = df1.set_index(np.array(['First Total'] * len(df1)), append=True)
dfmi = pd.concat([dfmi, df1]).sort_index(level=[0, 1])
print(dfmi)
# this is where the multiIndex is lost
df2 = df2.set_index(np.array(['Second Total'] * len(df2)), append=True)
dfmi = pd.concat([dfmi, df2]).sort_index(level=[1])
print(dfmi)
lvl0 a b
lvl1 bar foo bah foo
A0 B0 C0 D0 1 0 3 2
D1 5 4 7 6
First Total 6 4 10 8
C1 D0 9 8 11 10
D1 13 12 15 14
First Total 22 20 26 24
C2 D0 17 16 19 18
D1 21 20 23 22
First Total 38 36 42 40
C3 D0 25 24 27 26
D1 29 28 31 30
First Total 54 52 58 56
Second Total 120 112 136 128
B1 C0 D0 33 32 35 34
D1 37 36 39 38
First Total 70 68 74 72
C1 D0 41 40 43 42
D1 45 44 47 46
First Total 86 84 90 88
C2 D0 49 48 51 50
D1 53 52 55 54
First Total 102 100 106 104
C3 D0 57 56 59 58
D1 61 60 63 62
First Total 118 116 122 120
Second Total 376 368 392 384
第二个总数是1级dfmi的总和，dfmi有一个4级多索引：
In [208]: dfmi.index.nlevels
Out[208]: 4
df2df2 = df2.set_index([np.array(['Second Total'] * len(df2)), [''] * len(df2)], append=True)
df2dfmidf2pd.concat（[dfmi，df2]）按索引标签排序时可能遇到的一个问题是，它依赖于字符串
'First''Second'顺序列
而按此排序：
dfmi['order'] = range(len(dfmi))
df1['order'] = dfmi.groupby(level=[0,1,2])['order'].last() + 0.1
df2['order'] = dfmi.groupby(level=[0,1])['order'].last() + 0.2
...
dfmi = pd.concat([dfmi, df1, df2])
dfmi = dfmi.sort_values(by='order')
合并后，代码将如下所示：
import pandas as pd
import numpy as np
def mklbl(prefix, n):
return ["%s%s" % (prefix, i) for i in range(n)]
miindex = pd.MultiIndex.from_product([mklbl('A', 4),
mklbl('B', 2),
mklbl('C', 4),
mklbl('Z', 2)])
micolumns = pd.MultiIndex.from_tuples([('a', 'foo'), ('a', 'bar'),
('b', 'foo'), ('b', 'bah')],
names=['lvl0', 'lvl1'])
dfmi = pd.DataFrame(np.arange(len(miindex) * len(micolumns))
.reshape((len(miindex), len(micolumns))),
index=miindex,
columns=micolumns).sort_index().sort_index(axis=1)
df1 = dfmi.groupby(level=[0,1,2]).sum()
df2 = dfmi.groupby(level=[0, 1]).sum()
dfmi['order'] = range(len(dfmi))
df1['order'] = dfmi.groupby(level=[0,1,2])['order'].last() + 0.1
df2['order'] = dfmi.groupby(level=[0,1])['order'].last() + 0.2
df1 = df1.assign(lev4='First').set_index('lev4', append=True)
df2 = df2.assign(lev3='Second', lev4='').set_index(['lev3','lev4'], append=True)
dfmi = pd.concat([dfmi, df1, df2])
dfmi = dfmi.sort_values(by='order')
dfmi = dfmi.drop(['order'], axis=1)
print(dfmi)
产生
lvl0 a b
lvl1 bar foo bah foo
A0 B0 C0 Z0 1 0 3 2
Z1 5 4 7 6
First 6 4 10 8
C1 Z0 9 8 11 10
Z1 13 12 15 14
First 22 20 26 24
C2 Z0 17 16 19 18
Z1 21 20 23 22
First 38 36 42 40
C3 Z0 25 24 27 26
Z1 29 28 31 30
First 54 52 58 56
Second 120 112 136 128
...
@联合国大学指出了问题的本质。df2有三个级别的多索引，您需要第四个级别
我将使用assign
和set\u index
创建第四级：
df2 = df2.assign(lev3='Second Total', lev4='').set_index(['lev3','lev4'], append=True)
这样可以避免计算数据帧的长度