Pandas聚合运算和分组运算的实现示例

import numpy as np
import pandas as pd
#创建df对象
dict_data = {
  'key1':['a','b','c','d','a','b','c','d'],
  'key2':['one','two','three','one','two','three','one','two'],
  'data1':np.random.randint(1,10,8),
  'data2':np.random.randint(1,10,8)
}
 
df = pd.DataFrame(dict_data)
print(df)
'''
  data1 data2 key1  key2
0   3   4  a  one
1   7   9  b  two
2   5   7  c three
3   3   4  d  one
4   8   7  a  two
5   4   7  b three
6   8   9  c  one
7   4   4  d  two
'''
#根据key1分组,进行sum()运算
df = df.groupby('key1').sum()
print(df)
'''
key1       
a    12   10
b     8   5
c     8   11
d    16   13
'''
#内置的聚合函数
print(df.groupby('key1').sum())
print('*'*50)
print(df.groupby('key1').max())
print('*'*50)
print(df.groupby('key1').min())
print('*'*50)
print(df.groupby('key1').mean())
print('*'*50)
print(df.groupby('key1').size())
print('*'*50)
#分组中非Nan数据的数量
print(df.groupby('key1').count())
print('*'*50)
print(df.groupby('key1').describe())

#自定义聚合函数
#最大值-最小值
def peak_range(df):
  #返回数据范围差值
  return df.max()**2 - df.min()**2
 
#agg() 可以将聚合计算的结果祖闯成一个dataframe对象返回
 
print(df.groupby('key1').agg(peak_range))
 
#lambda
print(df.groupby('key1').agg(lambda df:df.max()-df.min()))

#应用多个聚合函数,默认列索引为函数名
#通过元素重新命名列索引('列索引',函数)
print(df.groupby('key1').agg(['sum','std','mean',('range',peak_range)]))
'''
   data1           data2           
    sum    std mean range  sum    std mean range
key1                           
a    10 2.828427 5.0  40  12 2.828427 6.0  48
b    10 5.656854 5.0  80   8 1.414214 4.0  16
c    6 1.414214 3.0  12   9 0.707107 4.5   9
d    15 0.707107 7.5  15   8 2.828427 4.0  32
'''

#指定每一列使用某个聚合运算函数
print(df.groupby('key1').agg({'data1':'mean','data2':'sum'}))
'''
   data1 data2
key1       
a    5.0   12
b    5.0   8
c    3.0   9
d    7.5   8
'''

#创建df对象
dict_data = {
  'key1':['a','b','c','d','a','b','c','d'],
  'key2':['one','two','three','one','two','three','one','two'],
  'data1':np.random.randint(1,10,8),
  'data2':np.random.randint(1,10,8)
}
 
df = pd.DataFrame(dict_data)
print(df)
'''
  data1 data2 key1  key2
0   1   5  a  one
1   9   3  b  two
2   3   6  c three
3   6   9  d  one
4   8   4  a  two
5   5   5  b three
6   9   6  c  one
7   4   1  d  two
'''
#按照key1分组,进行sum()运算
#在运算结果的列索引前添加前缀
k1_sum = df.groupby('key1').sum().add_prefix('sum_')
print(k1_sum)
'''
   sum_data1 sum_data2
key1           
a       9     9
b      14     8
c      12     12
d      10     10
'''

#创建df对象
dict_data = {
  'key1':['a','b','c','d','a','b','c','d'],
  'key2':['one','two','three','one','two','three','one','two'],
  'data1':np.random.randint(1,10,8),
  'data2':np.random.randint(1,10,8)
}
 
df = pd.DataFrame(dict_data)
print(df)
'''
  data1 data2 key1  key2
0   1   5  a  one
1   9   3  b  two
2   3   6  c three
3   6   9  d  one
4   8   4  a  two
5   5   5  b three
6   9   6  c  one
7   4   1  d  two
'''
#按照key1分组,进行sum()运算
#在运算结果的列索引前添加前缀
k1_sum = df.groupby('key1').sum().add_prefix('sum_')
print(k1_sum)
'''
   sum_data1 sum_data2
key1           
a       9     9
b      14     8
c      12     12
d      10     10
'''
#将运算结果和原始数据拼接到一起
#参数1:原始数据
#参数2:运算结果数据
pd.merge(df,k1_sum,left_on='key1',right_index=True)

#创建df对象
dict_data = {
  'key1':['a','b','c','d','a','b','c','d'],
  'key2':['one','two','three','one','two','three','one','two'],
  'data1':np.random.randint(1,10,8),
  'data2':np.random.randint(1,10,8)
}
 
df = pd.DataFrame(dict_data)
print(df)
'''
  data1 data2 key1  key2
0   1   5  a  one
1   9   3  b  two
2   3   6  c three
3   6   9  d  one
4   8   4  a  two
5   5   5  b three
6   9   6  c  one
7   4   1  d  two
'''
#按照key1分组,进行sum()运算
#在运算结果的列索引前添加前缀
k1_sum = df.groupby('key1').sum().add_prefix('sum_')
print(k1_sum)
'''
   sum_data1 sum_data2
key1           
a       9     9
b      14     8
c      12     12
d      10     10
'''
#transform() 计算 会将计算的结果按照原始数据的排序组装成一个dataframe对象
k1_sum_tf = df.groupby('key1').transform(np.sum).add_prefix('sum_')
# print(k1_sum_tf.columns)
 
#把运算结果数据拼接到原始数据后
df[k1_sum_tf.columns] = k1_sum_tf
print(df)
'''
  data1 data2 key1  key2 sum_data1 sum_data2 sum_key2
0   5   4  a  one     9    12  onetwo
1   3   3  b  two     5    12 twothree
2   9   2  c three    14     9 threeone
3   6   5  d  one    11     9  onetwo
4   4   8  a  two     9    12  onetwo
5   2   9  b three     5    12 twothree
6   5   7  c  one    14     9 threeone
7   5   4  d  two    11     9  onetwo
'''