pandas group分组与agg聚合的实例

import pandas as pd
 
df = pd.DataFrame({'Country':['China','China', 'India', 'India', 'America', 'Japan', 'China', 'India'], 
     'Income':[10000, 10000, 5000, 5002, 40000, 50000, 8000, 5000],
     'Age':[5000, 4321, 1234, 4010, 250, 250, 4500, 4321]})

 Age Country Income
0 5000 China 10000
1 4321 China 10000
2 1234 India 5000
3 4010 India 5002
4 250 America 40000
5 250 Japan 50000
6 4500 China 8000
7 4321 India 5000

df_gb = df.groupby('Country')
for index, data in df_gb:
 print(index)
 print(data)

America
 Age Country Income
4 250 America 40000
China
 Age Country Income
0 5000 China 10000
1 4321 China 10000
6 4500 China 8000
India
 Age Country Income
2 1234 India 5000
3 4010 India 5002
7 4321 India 5000
Japan
 Age Country Income
5 250 Japan 50000

df_gb = df.groupby(['Country', 'Income'])
for (index1, index2), data in df_gb:
 print((index1, index2))
 print(data)

('America', 40000)
 Age Country Income
4 250 America 40000
('China', 8000)
 Age Country Income
6 4500 China 8000
('China', 10000)
 Age Country Income
0 5000 China 10000
1 4321 China 10000
('India', 5000)
 Age Country Income
2 1234 India 5000
7 4321 India 5000
('India', 5002)
 Age Country Income
3 4010 India 5002
('Japan', 50000)
 Age Country Income
5 250 Japan 50000

df_agg = df.groupby('Country').agg(['min', 'mean', 'max'])
print(df_agg)

 Age     Income      
   min   mean max min   mean max
Country              
America 250 250.000000 250 40000 40000.000000 40000
China 4321 4607.000000 5000 8000 9333.333333 10000
India 1234 3188.333333 4321 5000 5000.666667 5002
Japan  250 250.000000 250 50000 50000.000000 50000

num_agg = {'Age':['min', 'mean', 'max']}
print(df.groupby('Country').agg(num_agg))

 Age     
   min   mean max
Country       
America 250 250.000000 250
China 4321 4607.000000 5000
India 1234 3188.333333 4321
Japan  250 250.000000 250
num_agg = {'Age':['min', 'mean', 'max'], 'Income':['min', 'max']}
print(df.groupby('Country').agg(num_agg))

  Age     Income  
   min   mean max min max
Country          
America 250 250.000000 250 40000 40000
China 4321 4607.000000 5000 8000 10000
India 1234 3188.333333 4321 5000 5002
Japan  250 250.000000 250 50000 50000

DataFrame
import pandas as pd
import numpy as np

 df = pd.DataFrame({'key1':['a', 'a', 'b', 'b', 'a'],
     'key2':['one', 'two', 'one', 'two', 'one'],
     'data1':np.random.randn(5),
     'data2':np.random.randn(5)})
print(df)

  data1  data2 key1 key2
0 -0.410122 0.247895 a one
1 -0.627470 -0.989268 a two
2 0.179488 -0.054570 b one
3 -0.299878 -1.640494 b two
4 -0.297191 0.954447 a one

list(df.groupby(['key1']))#list后得到：[(group1),(group2),......]

[('a',  data1  data2 key1 key2
 0 -0.410122 0.247895 a one
 1 -0.627470 -0.989268 a two
 4 -0.297191 0.954447 a one), ('b',  data1  data2 key1 key2
 2 0.179488 -0.054570 b one
 3 -0.299878 -1.640494 b two)]

for name,group in df.groupby(['key1']):
 print(name)
 print(group)

a
  data1  data2 key1 key2
0 -0.410122 0.247895 a one
1 -0.627470 -0.989268 a two
4 -0.297191 0.954447 a one
b
  data1  data2 key1 key2
2 0.179488 -0.054570 b one
3 -0.299878 -1.640494 b two

for name,group in df.groupby(['key1','key2']):
 print(name) #name=(k1,k2)
 print(group)

('a', 'one')
  data1  data2 key1 key2
0 -0.410122 0.247895 a one
4 -0.297191 0.954447 a one
('a', 'two')
  data1  data2 key1 key2
1 -0.62747 -0.989268 a two
('b', 'one')
  data1 data2 key1 key2
2 0.179488 -0.05457 b one
('b', 'two')
  data1  data2 key1 key2
3 -0.299878 -1.640494 b two

dict(list(df.groupby(['key1'])))#dict(list())

{'a':  data1  data2 key1 key2
 0 -0.410122 0.247895 a one
 1 -0.627470 -0.989268 a two
 4 -0.297191 0.954447 a one, 'b':  data1  data2 key1 key2
 2 0.179488 -0.054570 b one
 3 -0.299878 -1.640494 b two}

df.groupby(['key1']).size()

key1
a 3
b 2
dtype: int64

dict(['a1','x2','e3'])



{'a': '1', 'e': '3', 'x': '2'}

df.groupby(['key1','key2']).size()

key1 key2
a  one  2
  two  1
b  one  1
  two  1
dtype: int64

df['data1'].groupby(df['key1']).mean()
#groupby没有进行任何的计算。它只是进行了一个分组

key1
a -0.444928
b -0.060195
Name: data1, dtype: float64

df.groupby(['key1'])['data1'].mean()#理解：对df按key1分组，并计算分组后df['data1']的均值
#等价于：df.groupby(['key1']).data1.mean()

key1
a -0.444928
b -0.060195
Name: data1, dtype: float64

df.groupby(['key1'])['data1'].size()

key1
a 3
b 2
Name: data1, dtype: int64

df['data1'].groupby([df['key1'],df['key2']]).mean()

key1 key2
a  one -0.353657
  two -0.627470
b  one  0.179488
  two -0.299878
Name: data1, dtype: float64

df.groupby(['key1','key2'])['data1'].mean()
#等价于：df.groupby(['key1','key2']).data1'.mean()

key1 key2
a  one -0.353657
  two -0.627470
b  one  0.179488
  two -0.299878
Name: data1, dtype: float64

df.groupby(['key1','key2'])['data1'].mean().unstack()

list(df.groupby(df.dtypes, axis=1))

[(dtype('float64'),  data1  data2
 0 -0.410122 0.247895
 1 -0.627470 -0.989268
 2 0.179488 -0.054570
 3 -0.299878 -1.640494
 4 -0.297191 0.954447), (dtype('O'), key1 key2
 0 a one
 1 a two
 2 b one
 3 b two
 4 a one)]

aggregate(self, func_or_funcs, * args, ** kwargs)
func: function, string, dictionary, or list of string/functions

s= pd.Series([10,20,30,40])
s

0 10
1 20
2 30
3 40
dtype: int64

for n,g in s.groupby([1,1,2,2]):
 print(n)
 print(g)

0 10
1 20
dtype: int64
2
2 30
3 40
dtype: int64

s.groupby([1,1,2,2]).min()
1
1 10
2 30
dtype: int64

#等价于这个：
s.groupby([1,1,2,2]).agg('min')

1 10
2 30
dtype: int64

s.groupby([1,1,2,2]).agg(['min','max'])#加[]，func仅接受一个参数

df.groupby(['key1'])['data1'].min()

key1
a -0.627470
b -0.299878
Name: data1, dtype: float64

df.groupby(['key1'])['data1'].agg({'min'})

#推荐用这个√
df.groupby(['key1']).agg({'data1':'min'})#对data1列，取各组的最小值，名字还是data1

#按key1分组后，aggregate各组data1的最小值和最大值：
df.groupby(['key1'])['data1'].agg({'min','max'})

#推荐用这个√
df.groupby(['key1']).agg({'data1':['min','max']})

# 对data1，把min更名为a，max更名为b
df.groupby(['key1'])['data1'].agg({'a':'min','b':'max'})#这里的'min' 'max'为两个函数名

d:\python27\lib\site-packages\ipykernel_launcher.py:2: FutureWarning: using a dict on a Series for aggregation
is deprecated and will be removed in a future version

df1= df.groupby(['date'])['price'].agg({'sum','count'}).reset_index()