1 主要内容
- DataFrame.groupby().sum()
- DataFrame.groupby().agg()
- pandas.concat([DataFrame1,DataFrame2])
- pandas.merge(DataFrame1,DataFrame2,parameters….)
- DataFrame1.join(DataFrame2,lsuffix=’列名 on DataFrame1’,rsuffix=’列名 on DataFrame2’)
- 帮助文档的获取
2 实例
- 构造dataframe如下所示：
food food_id number price user_id weather 0 soup 4 6 1.818250 3 cold 1 soup 8 6 1.834045 4 hot 2 iceream 8 7 3.042422 2 cold 3 chocolate 3 6 5.247564 4 hot 4 iceream 6 3 4.319450 4 cold 5 iceream 5 4 2.912291 1 cold 6 iceream 2 7 6.118529 2 cold 7 soup 8 4 1.394939 2 hot 8 soup 6 8 2.921446 2 hot 9 chocolate 2 1 3.663618 4 hot
实现程序如下所示：
import pandas as pd
from numpy import random
from numpy.random import rand
import numpy as np
random.seed(42)
df = pd.DataFrame({'user_id':random.randint(0,6,10),'food_id':random.randint(1,10,10),
'weather':['cold','hot','cold','hot','cold','cold','cold','hot','hot','hot'],
'food':['soup','soup','iceream','chocolate','iceream','iceream','iceream','soup','soup','chocolate'],
'price':10 * rand(10),'number':random.randint(1,9,10)})
print df
2 groupby函数应用
代码
groupby1 = df.groupby(['user_id']) #按照user_id分组输出所有的值
i = 0
for user_id,group in groupby1:
i = i + 1
print "group", i , user_id
print group
结果
group 1 1
food food_id number price user_id weather
5 iceream 5 4 2.912291 1 cold
group 2 2
food food_id number price user_id weather
2 iceream 8 7 3.042422 2 cold
6 iceream 2 7 6.118529 2 cold
7 soup 8 4 1.394939 2 hot
8 soup 6 8 2.921446 2 hot
group 3 3
food food_id number price user_id weather
0 soup 4 6 1.81825 3 cold
group 4 4
food food_id number price user_id weather
1 soup 8 6 1.834045 4 hot
3 chocolate 3 6 5.247564 4 hot
4 iceream 6 3 4.319450 4 cold
9 chocolate 2 1 3.663618 4 hot
3 groupby和sum等函数结合使用
代码
print(groupby1.sum())#对除了groupby索引以外的每个数值列进行求和 print(groupby1['food_id','number'].sum()) #对除了groupby索引以外的特定数值列进行求和 print(df.groupby(['user_id'],as_index=False).sum())#默认as_index=True,是否将user_id当做索引，默认是 #当然除了sum，还有mean，min，max，median，mode，std,mad等等，操作方法同理 #groupby()中的形参可用help(df.groupby)来查看 #常用的参数axis=0，表示对行进行操作，即指定列中不同值进行分组；axis=1，表示对列进行分组
output[1]:
food_id number price
user_id
1 5 4 2.912291
2 24 26 13.477336
3 4 6 1.818250
4 19 16 15.064678
output[2]:
food_id number
user_id
1 5 4
2 24 26
3 4 6
4 19 16
output[3]:
user_id food_id number price
0 1 5 4 2.912291
1 2 24 26 13.477336
2 3 4 6 1.818250
3 4 19 16 15.064678
df.groupby(['food','weather']).size()
food weather
chocolate hot 2
iceream cold 4
soup cold 1
hot 3
dtype: int64
4 agg函数
代码
print df.groupby(['weather','food']).agg([np.mean,np.median])结果
output[4]:
food_id number price \
user_id
mean median
weather food
cold iceream 2.250000 2
soup 3.000000 3
hot chocolate 4.000000 4
soup 2.666667 2 mean median mean median mean median weather food cold iceream 5.250000 5.5 5.25 5.5 4.098173 3.680936 soup 4.000000 4.0 6.00 6.0 1.818250 1.818250 hot chocolate 2.500000 2.5 3.50 3.5 4.455591 4.455591 soup 7.333333 8.0 6.00 6.0 2.050143 1.834045
5 concat()
代码
print "df :3\n",df[:3] print "df :4\n",df[6:] print pd.concat([df[:3],df[6:]],axis=0)
结果
df :3
food food_id number price user_id weather
0 soup 4 6 1.818250 3 cold
1 soup 8 6 1.834045 4 hot
2 iceream 8 7 3.042422 2 cold
df :4
food food_id number price user_id weather
6 iceream 2 7 6.118529 2 cold
7 soup 8 4 1.394939 2 hot
8 soup 6 8 2.921446 2 hot
9 chocolate 2 1 3.663618 4 hot
df.concat
food food_id number price user_id weather
0 soup 4 6 1.818250 3 cold
1 soup 8 6 1.834045 4 hot
2 iceream 8 7 3.042422 2 cold
6 iceream 2 7 6.118529 2 cold
7 soup 8 4 1.394939 2 hot
8 soup 6 8 2.921446 2 hot
9 chocolate 2 1 3.663618 4 hot
6 merge()和join()
代码
df1=pd.DataFrame({'EmpNr':[5,3,9],'Dest':['The Hague','Amsterdam','Rotterdam']})
df2=pd.DataFrame({'EmpNr':[5,9,7],'Amount':[10,5,2.5]})
print "df1\n",df1
print "df2\n",df2
print "Merge() on Key\n",pd.merge(df1,df2,on='EmpNr')
print "inner join with Merge()\n",pd.merge(df1,df2,how='inner')
print "Dests join tips\n",df1.join(df2,lsuffix='Dest',rsuffix='Tips')
结果
df1
Dest EmpNr
0 The Hague 5
1 Amsterdam 3
2 Rotterdam 9
df2
Amount EmpNr
0 10.0 5
1 5.0 9
2 2.5 7
Merge() on Key
Dest EmpNr Amount
0 The Hague 5 10.0
1 Rotterdam 9 5.0
inner join with Merge()
Dest EmpNr Amount
0 The Hague 5 10.0
1 Rotterdam 9 5.0
Dests join tips
Dest EmpNrDest Amount EmpNrTips
0 The Hague 5 10.0 5
1 Amsterdam 3 5.0 9
2 Rotterdam 9 2.5 7
6帮助文档获取方式
1.help(pd.concat)
2.dir(pd.concat)
3.pd.concat?
...7 参考文献
利用python进行数据分析笔记
python数据分析，Ivan Idris著
本文为转载文章，原文出处：https://blog.csdn.net/ly_ysys629/article/details/72553273