标签：poi imp itertools ide type das frame index 列表
任何分组(groupby)操作都涉及原始对象的以下操作之一:
在许多情况下，我们将数据分成多个集合，并在每个子集上应用一些函数。在应用函数中，可以执行以下操作:
下面来看看创建一个DataFrame对象并对其执行所有操作 -
import pandas as pd ipl_data = {‘Team‘: [‘Riders‘, ‘Riders‘, ‘Devils‘, ‘Devils‘, ‘Kings‘, ‘kings‘, ‘Kings‘, ‘Kings‘, ‘Riders‘, ‘Royals‘, ‘Royals‘, ‘Riders‘], ‘Rank‘: [1, 2, 2, 3, 3,4 ,1 ,1,2 , 4,1,2], ‘Year‘: [2014,2015,2014,2015,2014,2015,2016,2017,2016,2014,2015,2017], ‘Points‘:[876,789,863,673,741,812,756,788,694,701,804,690]} df = pd.DataFrame(ipl_data) print (df)
输出结果：
Points Rank Team Year
0 876 1 Riders 2014
1 789 2 Riders 2015
2 863 2 Devils 2014
3 673 3 Devils 2015
4 741 3 Kings 2014
5 812 4 kings 2015
6 756 1 Kings 2016
7 788 1 Kings 2017
8 694 2 Riders 2016
9 701 4 Royals 2014
10 804 1 Royals 2015
11 690 2 Riders 2017
Pandas对象可以分成任何对象。有多种方式来拆分对象，如 -
现在来看看如何将分组对象应用于DataFrame对象
示例
import pandas as pd ipl_data = {‘Team‘: [‘Riders‘, ‘Riders‘, ‘Devils‘, ‘Devils‘, ‘Kings‘, ‘kings‘, ‘Kings‘, ‘Kings‘, ‘Riders‘, ‘Royals‘, ‘Royals‘, ‘Riders‘], ‘Rank‘: [1, 2, 2, 3, 3,4 ,1 ,1,2 , 4,1,2], ‘Year‘: [2014,2015,2014,2015,2014,2015,2016,2017,2016,2014,2015,2017], ‘Points‘:[876,789,863,673,741,812,756,788,694,701,804,690]} df = pd.DataFrame(ipl_data) print (df.groupby(‘Team‘))
输出结果：
<pandas.core.groupby.DataFrameGroupBy object at 0x00000245D60AD518>
import pandas as pd ipl_data = {‘Team‘: [‘Riders‘, ‘Riders‘, ‘Devils‘, ‘Devils‘, ‘Kings‘, ‘kings‘, ‘Kings‘, ‘Kings‘, ‘Riders‘, ‘Royals‘, ‘Royals‘, ‘Riders‘], ‘Rank‘: [1, 2, 2, 3, 3,4 ,1 ,1,2 , 4,1,2], ‘Year‘: [2014,2015,2014,2015,2014,2015,2016,2017,2016,2014,2015,2017], ‘Points‘:[876,789,863,673,741,812,756,788,694,701,804,690]} df = pd.DataFrame(ipl_data) print (df.groupby(‘Team‘).groups)
输出结果：
{
‘Devils‘: Int64Index([2, 3], dtype=‘int64‘),
‘Kings‘: Int64Index([4, 6, 7], dtype=‘int64‘),
‘Riders‘: Int64Index([0, 1, 8, 11], dtype=‘int64‘),
‘Royals‘: Int64Index([9, 10], dtype=‘int64‘),
‘kings‘: Int64Index([5], dtype=‘int64‘)
}
按多列分组
import pandas as pd ipl_data = {‘Team‘: [‘Riders‘, ‘Riders‘, ‘Devils‘, ‘Devils‘, ‘Kings‘, ‘kings‘, ‘Kings‘, ‘Kings‘, ‘Riders‘, ‘Royals‘, ‘Royals‘, ‘Riders‘], ‘Rank‘: [1, 2, 2, 3, 3,4 ,1 ,1,2 , 4,1,2], ‘Year‘: [2014,2015,2014,2015,2014,2015,2016,2017,2016,2014,2015,2017], ‘Points‘:[876,789,863,673,741,812,756,788,694,701,804,690]} df = pd.DataFrame(ipl_data) print (df.groupby([‘Team‘,‘Year‘]).groups)
输出结果：
{
(‘Devils‘, 2014): Int64Index([2], dtype=‘int64‘),
(‘Devils‘, 2015): Int64Index([3], dtype=‘int64‘),
(‘Kings‘, 2014): Int64Index([4], dtype=‘int64‘),
(‘Kings‘, 2016): Int64Index([6], dtype=‘int64‘),
(‘Kings‘, 2017): Int64Index([7], dtype=‘int64‘),
(‘Riders‘, 2014): Int64Index([0], dtype=‘int64‘),
(‘Riders‘, 2015): Int64Index([1], dtype=‘int64‘),
(‘Riders‘, 2016): Int64Index([8], dtype=‘int64‘),
(‘Riders‘, 2017): Int64Index([11], dtype=‘int64‘),
(‘Royals‘, 2014): Int64Index([9], dtype=‘int64‘),
(‘Royals‘, 2015): Int64Index([10], dtype=‘int64‘),
(‘kings‘, 2015): Int64Index([5], dtype=‘int64‘)
}
使用groupby对象，可以遍历类似itertools.obj的对象。
import pandas as pd ipl_data = {‘Team‘: [‘Riders‘, ‘Riders‘, ‘Devils‘, ‘Devils‘, ‘Kings‘, ‘kings‘, ‘Kings‘, ‘Kings‘, ‘Riders‘, ‘Royals‘, ‘Royals‘, ‘Riders‘], ‘Rank‘: [1, 2, 2, 3, 3,4 ,1 ,1,2 , 4,1,2], ‘Year‘: [2014,2015,2014,2015,2014,2015,2016,2017,2016,2014,2015,2017], ‘Points‘:[876,789,863,673,741,812,756,788,694,701,804,690]} df = pd.DataFrame(ipl_data) grouped = df.groupby(‘Year‘) for name,group in grouped: print (name) print (group) print (‘\n‘)
输出结果：
2014
Points Rank Team Year
0 876 1 Riders 2014
2 863 2 Devils 2014
4 741 3 Kings 2014
9 701 4 Royals 2014
2015
Points Rank Team Year
1 789 2 Riders 2015
3 673 3 Devils 2015
5 812 4 kings 2015
10 804 1 Royals 2015
2016
Points Rank Team Year
6 756 1 Kings 2016
8 694 2 Riders 2016
2017
Points Rank Team Year
7 788 1 Kings 2017
11 690 2 Riders 2017
默认情况下，groupby对象具有与分组名相同的标签名称。
使用get_group()方法，可以选择一个组。参考以下示例代码 -
import pandas as pd ipl_data = {‘Team‘: [‘Riders‘, ‘Riders‘, ‘Devils‘, ‘Devils‘, ‘Kings‘, ‘kings‘, ‘Kings‘, ‘Kings‘, ‘Riders‘, ‘Royals‘, ‘Royals‘, ‘Riders‘], ‘Rank‘: [1, 2, 2, 3, 3,4 ,1 ,1,2 , 4,1,2], ‘Year‘: [2014,2015,2014,2015,2014,2015,2016,2017,2016,2014,2015,2017], ‘Points‘:[876,789,863,673,741,812,756,788,694,701,804,690]} df = pd.DataFrame(ipl_data) grouped = df.groupby(‘Year‘) print (grouped.get_group(2014))
输出结果：
Points Rank Team Year
0 876 1 Riders 2014
2 863 2 Devils 2014
4 741 3 Kings 2014
9 701 4 Royals 2014
聚合函数为每个组返回单个聚合值。当创建了分组(group by)对象，就可以对分组数据执行多个聚合操作。
import pandas as pd import numpy as np ipl_data = {‘Team‘: [‘Riders‘, ‘Riders‘, ‘Devils‘, ‘Devils‘, ‘Kings‘, ‘kings‘, ‘Kings‘, ‘Kings‘, ‘Riders‘, ‘Royals‘, ‘Royals‘, ‘Riders‘], ‘Rank‘: [1, 2, 2, 3, 3,4 ,1 ,1,2 , 4,1,2], ‘Year‘: [2014,2015,2014,2015,2014,2015,2016,2017,2016,2014,2015,2017], ‘Points‘:[876,789,863,673,741,812,756,788,694,701,804,690]} df = pd.DataFrame(ipl_data) grouped = df.groupby(‘Year‘) print (grouped[‘Points‘].agg(np.mean))
输出结果：
Year
2014 795.25
2015 769.50
2016 725.00
2017 739.00
Name: Points, dtype: float64
另一种查看每个分组的大小的方法是应用size()函数 -
import pandas as pd import numpy as np ipl_data = {‘Team‘: [‘Riders‘, ‘Riders‘, ‘Devils‘, ‘Devils‘, ‘Kings‘, ‘kings‘, ‘Kings‘, ‘Kings‘, ‘Riders‘, ‘Royals‘, ‘Royals‘, ‘Riders‘], ‘Rank‘: [1, 2, 2, 3, 3,4 ,1 ,1,2 , 4,1,2], ‘Year‘: [2014,2015,2014,2015,2014,2015,2016,2017,2016,2014,2015,2017], ‘Points‘:[876,789,863,673,741,812,756,788,694,701,804,690]} df = pd.DataFrame(ipl_data) grouped = df.groupby(‘Team‘) print (grouped.agg(np.size))
输出结果：
Team
Devils 2 2 2
Kings 3 3 3
Riders 4 4 4
Royals 2 2 2
kings 1 1 1
通过分组系列，还可以传递函数的列表或字典来进行聚合，并生成DataFrame作为输出
import pandas as pd import numpy as np ipl_data = {‘Team‘: [‘Riders‘, ‘Riders‘, ‘Devils‘, ‘Devils‘, ‘Kings‘, ‘kings‘, ‘Kings‘, ‘Kings‘, ‘Riders‘, ‘Royals‘, ‘Royals‘, ‘Riders‘], ‘Rank‘: [1, 2, 2, 3, 3,4 ,1 ,1,2 , 4,1,2], ‘Year‘: [2014,2015,2014,2015,2014,2015,2016,2017,2016,2014,2015,2017], ‘Points‘:[876,789,863,673,741,812,756,788,694,701,804,690]} df = pd.DataFrame(ipl_data) print(df) print(‘\n‘) grouped = df.groupby(‘Team‘) agg = grouped[‘Points‘].agg([np.sum, np.mean, np.std]) print (agg)
输出结果：
Team Rank Year Points
0 Riders 1 2014 876
1 Riders 2 2015 789
2 Devils 2 2014 863
3 Devils 3 2015 673
4 Kings 3 2014 741
5 kings 4 2015 812
6 Kings 1 2016 756
7 Kings 1 2017 788
8 Riders 2 2016 694
9 Royals 4 2014 701
10 Royals 1 2015 804
11 Riders 2 2017 690
sum mean std
Team
Devils 1536 768.000000 134.350288
Kings 2285 761.666667 24.006943
Riders 3049 762.250000 88.567771
Royals 1505 752.500000 72.831998
kings 812 812.000000 NaN
分组或列上的转换返回索引大小与被分组的索引相同的对象。因此，转换应该返回与组块大小相同的结果。
import pandas as pd import numpy as np ipl_data = {‘Team‘: [‘Riders‘, ‘Riders‘, ‘Devils‘, ‘Devils‘, ‘Kings‘, ‘kings‘, ‘Kings‘, ‘Kings‘, ‘Riders‘, ‘Royals‘, ‘Royals‘, ‘Riders‘], ‘Rank‘: [1, 2, 2, 3, 3,4 ,1 ,1,2 , 4,1,2], ‘Year‘: [2014,2015,2014,2015,2014,2015,2016,2017,2016,2014,2015,2017], ‘Points‘:[876,789,863,673,741,812,756,788,694,701,804,690]} df = pd.DataFrame(ipl_data) grouped = df.groupby(‘Team‘) score = lambda x: (x - x.mean()) / x.std()*10 print (grouped.transform(score))
输出结果：
Points Rank Year
0 12.843272 -15.000000 -11.618950
1 3.020286 5.000000 -3.872983
2 7.071068 -7.071068 -7.071068
3 -7.071068 7.071068 7.071068
4 -8.608621 11.547005 -10.910895
5 NaN NaN NaN
6 -2.360428 -5.773503 2.182179
7 10.969049 -5.773503 8.728716
8 -7.705963 5.000000 3.872983
9 -7.071068 7.071068 -7.071068
10 7.071068 -7.071068 7.071068
11 -8.157595 5.000000 11.618950
过滤根据定义的标准过滤数据并返回数据的子集。filter()函数用于过滤数据。
import pandas as pd import numpy as np
ipl_data = {‘Team‘: [‘Riders‘, ‘Riders‘, ‘Devils‘, ‘Devils‘, ‘Kings‘, ‘kings‘, ‘Kings‘, ‘Kings‘, ‘Riders‘, ‘Royals‘, ‘Royals‘, ‘Riders‘], ‘Rank‘: [1, 2, 2, 3, 3,4 ,1 ,1,2 , 4,1,2], ‘Year‘: [2014,2015,2014,2015,2014,2015,2016,2017,2016,2014,2015,2017], ‘Points‘:[876,789,863,673,741,812,756,788,694,701,804,690]} df = pd.DataFrame(ipl_data) filter = df.groupby(‘Team‘).filter(lambda x: len(x) >= 3) print (filter)
输出结果：
Points Rank Team Year
0 876 1 Riders 2014
1 789 2 Riders 2015
4 741 3 Kings 2014
6 756 1 Kings 2016
7 788 1 Kings 2017
8 694 2 Riders 2016
11 690 2 Riders 2017
在上述过滤条件下，要求返回三次以上参加IPL的队伍。
标签：poi imp itertools ide type das frame index 列表
原文地址：https://www.cnblogs.com/Summer-skr--blog/p/11705892.html
