Question

我有以下格式的熊猫系列：

23-08-2018  06803fe4-a504-4520-8304-a76a3adcd488    0
23-08-2018  89efbfda-6edc-45a9-a0dd-e520fd8a3e2a    2839
23-08-2018  88f7ff3f-ede7-4dd0-bce9-0d25b598004d    2639
23-08-2018  10f049cb-c165-424a-b2db-99637cc2668c    0
23-08-2018  11a7ec38-a535-4f1c-acc4-c93471401dbd    0
23-08-2018  1292f360-41e5-463e-8547-002858ac0226    0
23-08-2018  145d17c0-9711-4445-8eed-7e7d35f0f896    0
23-08-2018  188d7578-8a3b-4fe5-807a-d098bce1d227    0
23-08-2018  89tfbfda-6edc-45a9-a0dd-e520fd8a3e2a    0
24-08-2018  000a7843-432f-4c67-9d7c-5d3e2ffac439    14000
24-08-2018  000cd8c7-94c7-4cb7-ad70-a60aec275f31    14655
24-08-2018  000dd787-ab81-40a7-a036-a05e4d11fea9    15655
24-08-2018  00115f12-4a50-4412-bc90-940a21a1af65    14655
24-08-2018  0012467d-53c1-4b5e-be8b-fc285d130968    17700
27-08-2018  1e806edd-1c96-4bdb-87b8-b01cb09cdb02    15
27-08-2018  d2c45e73-d5ca-4e28-ba54-4e24b4ee9be3    30

我想要一个具有以下结构的数据框：

Date         AverageInteractionTime

23-08-2018     608.67  // (0+2839+2639+0+0+0+0+0+0)/9
24-08-2018     15333   // (14000+14655+15655+14655+17700)/5
27-08-2018     22.5    // (15+30)/2

基本上，我想根据聊天日期对系列进行分组，并获得第三列的平均值

我该怎么做？

Answer 1

假设您的3列为['Date','Some_ID','AverageInteractionTime']，然后在groupby上使用Date，并在mean上使用AverageInteractionTime为：

df.groupby('Date',as_index=False)['AverageInteractionTime'].mean()

         Date  AverageInteractionTime
0  23-08-2018              608.666667
1  24-08-2018            15333.000000
2  27-08-2018               22.500000

用于将Series转换为所需的Dataframe，然后使用上面的代码：

print(s[:3])
0    23-08-2018 06803fe4-a504-4520-8304-a76a3adcd488 0
1    23-08-2018 89efbfda-6edc-45a9-a0dd-e520fd8a3e2...
2    23-08-2018 88f7ff3f-ede7-4dd0-bce9-0d25b598004...

df = s.str.split(' ',expand=True).rename(columns={0:'Date',1:'Some_ID',2:'AverageInteractionTime'})

Answer 2

如果MultiIndex Series仅将mean与level=0参数一起使用：

print (s.index)
MultiIndex(levels=[['23-08-2018', '24-08-2018', '27-08-2018'], ['000a7843-432f-4c67-9d7c-5d3e2ffac439', '000cd8c7-94c7-4cb7-ad70-a60aec275f31', '000dd787-ab81-40a7-a036-a05e4d11fea9', '00115f12-4a50-4412-bc90-940a21a1af65', '0012467d-53c1-4b5e-be8b-fc285d130968', '06803fe4-a504-4520-8304-a76a3adcd488', '10f049cb-c165-424a-b2db-99637cc2668c', '11a7ec38-a535-4f1c-acc4-c93471401dbd', '1292f360-41e5-463e-8547-002858ac0226', '145d17c0-9711-4445-8eed-7e7d35f0f896', '188d7578-8a3b-4fe5-807a-d098bce1d227', '1e806edd-1c96-4bdb-87b8-b01cb09cdb02', '88f7ff3f-ede7-4dd0-bce9-0d25b598004d', '89efbfda-6edc-45a9-a0dd-e520fd8a3e2a', '89tfbfda-6edc-45a9-a0dd-e520fd8a3e2a', 'd2c45e73-d5ca-4e28-ba54-4e24b4ee9be3']],
           labels=[[0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 2, 2], [5, 13, 12, 6, 7, 8, 9, 10, 14, 0, 1, 2, 3, 4, 11, 15]],
           names=['Date', 'Val'])

df = s.mean(level=0).reset_index(name='AverageInteractionTime')
#alternative solution
#df = s.groupby(level=0).mean().reset_index(name='AverageInteractionTime')
print (df)
         Date  AverageInteractionTime
0  23-08-2018              608.666667
1  24-08-2018            15333.000000
2  27-08-2018               22.500000

如有必要，将索引拆分为MultiIndex：

print (s.index[:3])
Index(['23-08-2018 06803fe4-a504-4520-8304-a76a3adcd488',
       '23-08-2018 89efbfda-6edc-45a9-a0dd-e520fd8a3e2a',
       '23-08-2018 88f7ff3f-ede7-4dd0-bce9-0d25b598004d'],
      dtype='object', name='Date')

s.index = s.index.str.split(expand=True)
df = s.mean(level=0).reset_index(name='AverageInteractionTime')
#alternative solution
#df = s.groupby(level=0).mean().reset_index(name='AverageInteractionTime')

取得熊猫系列对象的平均值

2 个答案: