Pandas聚合运算和分组运算的实现示例

import numpy as npimport 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  key20   3   4  a  one1   7   9  b  two2   5   7  c three3   3   4  d  one4   8   7  a  two5   4   7  b three6   8   9  c  one7   4   4  d  two'''#根据key1分组,进行sum()运算df = df.groupby('key1').sum()print(df)'''key1       a    12   10b     8   5c     8   11d    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)) #lambdaprint(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 rangekey1                           a    10 2.828427 5.0  40  12 2.828427 6.0  48b    10 5.656854 5.0  80   8 1.414214 4.0  16c    6 1.414214 3.0  12   9 0.707107 4.5   9d    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 data2key1       a    5.0   12b    5.0   8c    3.0   9d    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  key20   1   5  a  one1   9   3  b  two2   3   6  c three3   6   9  d  one4   8   4  a  two5   5   5  b three6   9   6  c  one7   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_data2key1           a       9     9b      14     8c      12     12d      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  key20   1   5  a  one1   9   3  b  two2   3   6  c three3   6   9  d  one4   8   4  a  two5   5   5  b three6   9   6  c  one7   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_data2key1           a       9     9b      14     8c      12     12d      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  key20   1   5  a  one1   9   3  b  two2   3   6  c three3   6   9  d  one4   8   4  a  two5   5   5  b three6   9   6  c  one7   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_data2key1           a       9     9b      14     8c      12     12d      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_tfprint(df)'''  data1 data2 key1  key2 sum_data1 sum_data2 sum_key20   5   4  a  one     9    12  onetwo1   3   3  b  two     5    12 twothree2   9   2  c three    14     9 threeone3   6   5  d  one    11     9  onetwo4   4   8  a  two     9    12  onetwo5   2   9  b three     5    12 twothree6   5   7  c  one    14     9 threeone7   5   4  d  two    11     9  onetwo'''