Pandas 数据透视表和小计
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【中文标题】Pandas 数据透视表和小计【英文标题】：Pandas pivot and subtotals 【发布时间】：2021-11-03 00:13:37 【问题描述】：使用这些数据 -
d2 = 'Division': ['DIV1', 'DIV2', 'DIV1', 'DIV3', 'DIV2'],'Region': ['DIV1-South', 'DIV2-North', 'DIV1-North', "DIV3-East", "DIV2-South"]
,'MD': ["Susie", 'Martha', "Jane", "Nichole", "Randall"], 'Month': ['JAN', 'JAN', 'FEB', 'MAR', "APR"]
df2 = pd.DataFrame(d2)
看起来像这样：
Division Region MD Month
0 DIV1 DIV1-South Susie JAN
1 DIV2 DIV2-North Martha JAN
2 DIV1 DIV1-North Jane FEB
3 DIV3 DIV3-East Nichole MAR
4 DIV2 DIV2-South Randall APR
感谢这里的社区，我能够对这些数据进行透视以获得各个月份的总数：使用这行代码
pivoted = df.pivot_table(index=['Division', 'Region', 'NP'], columns='Month', aggfunc=len, fill_value=0)
Month APR FEB JAN MAR
Division Region MD
DIV1 DIV1-North Jane 0 1 0 0
DIV1-South Susie 0 0 1 0
DIV2 DIV2-North Martha 0 0 1 0
DIV2-South Randall 1 0 0 0
DIV3 DIV3-East Nichole 0 0 0 1
因此，这可能是不可能的，但我只在网上找到了一个参考来生成包含各个部分的小计的数据透视结果。不幸的是，这个例子不起作用。
理想的结果是：
Month APR FEB JAN MAR
Division Region MD
DIV1 DIV1-North Jane 0 1 0 0
DIV1-North SubTotal 0 1 0 0
DIV1-South Susie 0 0 1 0
DIV1-South SubTotal 0 0 1 0
DIV1 TOTAL 0 1 1 0
DIV2 DIV2-North Martha 0 0 1 0
DIV2-North SubTotal 0 0 1 0
DIV2-South Randall 1 0 0 0
DIV2-South SubTotal 1 0 0 0
DIV2 TOTAL 1 0 1 0
DIV3 DIV3-East Nichole 0 0 0 1
DIV3-East SubTotal 0 0 0 1
DIV3 TOTAL 0 0 0 1
这有点令人费解，甚至可能不可能，但由于这在 Excel 数据透视表中相当容易，我希望 Pandas 在某个地方启用了此功能，但我找不到它。 （尽管经过数天的搜索和测试，这仍然是正确的。）
【问题讨论】：
【参考方案1】：df = pd.DataFrame("A": ["foo", "foo", "foo", "foo", "foo",
"bar", "bar", "bar", "bar"],
"B": ["one", "one", "one", "two", "two",
"one", "one", "two", "two"],
"C": ["small", "large", "large", "small",
"small", "large", "small", "small",
"large"],
"D": [1, 2, 2, 3, 3, 4, 5, 6, 7],
"E": [2, 4, 5, 5, 6, 6, 8, 9, 9])
输出
A B C D E
0 foo one small 1 2
1 foo one large 2 4
2 foo one large 2 5
3 foo two small 3 5
4 foo two small 3 6
5 bar one large 4 6
6 bar one small 5 8
7 bar two small 6 9
table = pd.pivot_table(df, values='D', index=['A', 'B'],
columns=['C'], aggfunc=np.sum)
输出数据透视表
table
C large small
A B
bar one 4.0 5.0
two 7.0 6.0
foo one 4.0 1.0
two NaN 6.0
【讨论】：
这看起来像一个标准的数据透视表，似乎没有解决关于“小计”的问题 谢谢你的回答，但我的实际数据在上面。为了便于使用，我添加了示例数据框的代码。但是，您的代码是以您提供的一组新数据为中心的，而我正在使用的数据是由原始数据的第一个数据中心产生的多索引 df。 @SeaBean 您的回复非常有效。感谢您花时间添加。我赞成你的回答。绝对是一个复杂的案例，你成功了！【参考方案2】：您可以通过.groupby() 和GroupBy.sum() 与相应级别分组来创建Division Total 和Region SubTotal，如下所示： p>
pivoted2 = pivoted.reset_index()
# Create `Division` Total
df_Div_sum = pivoted2.groupby('Division', as_index=False).sum()
df_Div_sum['Region'] = '_' + df_Div_sum['Division'] + ' Total'
df_Div_sum['MD'] = ''
# Create `Region` SubTotal
df_Reg_sum = pivoted2.groupby(['Division', 'Region'], as_index=False).sum()
df_Reg_sum['MD'] = '_' + df_Reg_sum['Region'] + ' SubTotal'
# Concat results and set index + sort index
df_out = (pd.concat([pivoted2,
df_Reg_sum,
df_Div_sum
])
.set_index(['Division', 'Region', 'MD'])
.sort_index()
)
输入设置
d2 = 'Division': ['DIV1', 'DIV2', 'DIV1', 'DIV3', 'DIV2'],'Region': ['DIV1-South', 'DIV2-North', 'DIV1-North', "DIV3-East", "DIV2-South"]
,'MD': ["Susie", 'Martha', "Jane", "Nichole", "Randall"], 'Month': ['JAN', 'JAN', 'FEB', 'MAR', "APR"]
df = pd.DataFrame(d2)
pivoted = df.pivot_table(index=['Division', 'Region', 'MD'], columns='Month', aggfunc=len, fill_value=0)
输出
print(df_out)
Month APR FEB JAN MAR
Division Region MD
DIV1 DIV1-North Jane 0 1 0 0
_DIV1-North SubTotal 0 1 0 0
DIV1-South Susie 0 0 1 0
_DIV1-South SubTotal 0 0 1 0
_DIV1 Total 0 1 1 0
DIV2 DIV2-North Martha 0 0 1 0
_DIV2-North SubTotal 0 0 1 0
DIV2-South Randall 1 0 0 0
_DIV2-South SubTotal 1 0 0 0
_DIV2 Total 1 0 1 0
DIV3 DIV3-East Nichole 0 0 0 1
_DIV3-East SubTotal 0 0 0 1
_DIV3 Total 0 0 0 1
扩展测试数据
由于您的样本数据每个Region只有一个数据，因此我添加了更多测试数据以进行更完整的测试：
输入设置
data = 'Division': ['DIV1', 'DIV1', 'DIV2', 'DIV2', 'DIV1', 'DIV1', 'DIV3', 'DIV3', 'DIV2', 'DIV2', 'DIV2'],
'Region': ['DIV1-South', 'DIV1-South', 'DIV2-North', 'DIV2-North', 'DIV1-North', 'DIV1-North', 'DIV3-East', 'DIV3-East', 'DIV2-South', 'DIV2-South', 'DIV2-South'],
'MD': ['Susie', 'Susie2', 'Martha', 'Martha2', 'Jane', 'Jane2', 'Nichole', 'Nichole2', 'Randall2', 'Randall3', 'Randall'],
'Month': ['JAN', 'FEB', 'JAN', 'MAR', 'FEB', 'APR', 'MAR', 'APR', 'FEB', 'MAR', 'APR']
df = pd.DataFrame(data)
pivoted = df.pivot_table(index=['Division', 'Region', 'MD'], columns='Month', aggfunc=len, fill_value=0)
print(pivoted)
Month APR FEB JAN MAR
Division Region MD
DIV1 DIV1-North Jane 0 1 0 0
Jane2 1 0 0 0
DIV1-South Susie 0 0 1 0
Susie2 0 1 0 0
DIV2 DIV2-North Martha 0 0 1 0
Martha2 0 0 0 1
DIV2-South Randall 1 0 0 0
Randall2 0 1 0 0
Randall3 0 0 0 1
DIV3 DIV3-East Nichole 0 0 0 1
Nichole2 1 0 0 0
输出
print(df_out)
Month APR FEB JAN MAR
Division Region MD
DIV1 DIV1-North Jane 0 1 0 0
Jane2 1 0 0 0
_DIV1-North SubTotal 1 1 0 0
DIV1-South Susie 0 0 1 0
Susie2 0 1 0 0
_DIV1-South SubTotal 0 1 1 0
_DIV1 Total 1 2 1 0
DIV2 DIV2-North Martha 0 0 1 0
Martha2 0 0 0 1
_DIV2-North SubTotal 0 0 1 1
DIV2-South Randall 1 0 0 0
Randall2 0 1 0 0
Randall3 0 0 0 1
_DIV2-South SubTotal 1 1 0 1
_DIV2 Total 1 1 1 2
DIV3 DIV3-East Nichole 0 0 0 1
Nichole2 1 0 0 0
_DIV3-East SubTotal 1 0 0 1
_DIV3 Total 1 0 0 1
【讨论】：
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