pandas数据聚合与分组运算的实现

# 通过列名进行分组
import pandas as pd
df = pd.DataFrame({"key":['c','b','c','a','b','b','a','c','a'],
                "data":[2,4,6,8,10,1,14,16,19]
})
print(df)
'''
  key  data
0   c     2
1   b     4
2   c     6
3   a     8
4   b    10
5   b     1
6   a    14
7   c    16
8   a    19
'''
# 按照key列进行分组
print(df.groupby(by='key'))
'''<pandas.core.groupby.generic.DataFrameGroupBy object at 0x0000000008216688>
'''
group_obj = df.groupby('key')
for i in group_obj:
    print(i)
'''
('a',   key  data
3   a     8
6   a    14
8   a    19)
('b',   key  data
1   b     4
4   b    10
5   b     1)
('c',   key  data
0   c     2
2   c     6
7   c    16)
'''

import numpy as np
import pandas as pd
df = pd.DataFrame({
    'key1':['A','A','B','B','A'],
    'key2':['one','two','one','two','one'],
    'data1':['2','3','4','6','8'],
    'data2':['3','5','6','3','7']
})
print(df)
'''
  key1 key2 data1 data2
0    A  one     2     3
1    A  two     3     5
2    B  one     4     6
3    B  two     6     3
4    A  one     8     7
'''
se = pd.Series(['a','b','c','a','b'])
print(se)
'''
0    a
1    b
2    c
3    a
4    b
dtype: object
'''
group_obj = df.groupby(se) # 定义series对象进行分组
for i in group_obj:
    print(i)
'''
('a',   key  data
0   c     2
3   a     8)
('b',   key  data
1   b     4
4   b    10)
('c',   key  data
2   c     6)
'''

# 通过字典进行分组
from pandas import DataFrame,Series
num_df = DataFrame({'a':[1,2,3,4,5],
                   'b':[6,7,8,9,10],
                   'c':[11,12,13,14,15],
                   'd':[5,4,3,2,1],
                   'e':[10,9,8,7,6]})
print(num_df)
'''
   a   b   c  d   e
0  1   6  11  5  10
1  2   7  12  4   9
2  3   8  13  3   8
3  4   9  14  2   7
4  5  10  15  1   6
'''
# 定义分组规则
mapping = {'a':'第一组','b':'第二组','c':'第一组','d':'第三组','e':'第二组'}
by_column = num_df.groupby(mapping, axis=1)
for i in by_column:
    print(i)
'''
('第一组',    a   c
0  1  11
1  2  12
2  3  13
3  4  14
4  5  15)
('第三组',    d
0  5
1  4
2  3
3  2
4  1)
('第二组',     b   e
0   6  10
1   7   9
2   8   8
3   9   7
4  10   6)
'''

key_list = ['a', 'a', 'a', 'b', 'b']
num_df.groupby([len, key_list]).min()