更新时间:2023-12-01 11:40:04
好吧,这有点复杂.通常,我会尝试使用transform进行一些操作,但是如果有人有以下方面的功能,我会感到很高兴:
Ok this got a bit more complicated. Normally I'd try something with transform but I'd be glad if someone had something better than the following:
使用groupby
并将df发送到func,其中 df.loc
,最后使用pd.concat
将数据帧重新粘合在一起:
Use groupby
and send df to func where df.loc
is used, lastly use pd.concat
to glue the dataframe together again:
import pandas as pd
data = {'matchID': {0: 1, 1: 1, 2: 1, 3: 1, 4: 1, 5: 1, 6: 1, 7: 1, 8: 2, 9: 2, 10: 2,
11: 2, 12: 2, 13: 2, 14: 2, 15: 2},
'court': {0: 'A', 1: 'D', 2: 'D', 3: 'A', 4: 'A', 5: 'A', 6: 'D', 7: 'D', 8: 'A',
9: 'D', 10: 'D', 11: 'A', 12: 'A', 13: 'A', 14: 'D', 15: 'D'},
'speed': {0: 100, 1: 200, 2: 300, 3: 100, 4: 120, 5: 250, 6: 110, 7: 100, 8: 100,
9: 200, 10: 300, 11: 100, 12: 120, 13: 250, 14: 110, 15: 100},
'server': {0: 1, 1: 2, 2: 3, 3: 4, 4: 1, 5: 2, 6: 3, 7: 4, 8: 1, 9: 2, 10: 3,
11: 4, 12: 1, 13: 2, 14: 3, 15: 4}}
df = pd.DataFrame(data)
def func(dfx):
dfx['meanSpeedCourtA13'],dfx['meanSpeedCourtD13'] = \
(dfx.loc[(dfx.server.isin((1,3))) & (dfx.court == 'A'),'speed'].mean(),
dfx.loc[(dfx.server.isin((1,3))) & (dfx.court == 'D'),'speed'].mean())
return dfx
newdf = pd.concat(func(dfx) for _, dfx in df.groupby('matchID'))
print(newdf)
返回
court matchID server speed meanSpeedCourtA13 meanSpeedCourtD13
0 A 1 1 100 110.00 205.00
1 D 1 2 200 110.00 205.00
2 D 1 3 300 110.00 205.00
3 A 1 4 100 110.00 205.00
4 A 1 1 120 110.00 205.00
5 A 1 2 250 110.00 205.00
6 D 1 3 110 110.00 205.00
7 D 1 4 100 110.00 205.00
8 A 2 1 100 110.00 205.00
9 D 2 2 200 110.00 205.00
10 D 2 3 300 110.00 205.00
11 A 2 4 100 110.00 205.00
12 A 2 1 120 110.00 205.00
13 A 2 2 250 110.00 205.00
14 D 2 3 110 110.00 205.00
15 D 2 4 100 110.00 205.00