Pandas 滾動天數並獲得總和

Question

這是我的 dataframe

d= {'dates': ['2020-07-16','2020-07-15','2020-07-14','2020-07-13','2020-07-16','2020-07-15','2020-07-14','2020-07-13'], 
    'location':['Paris','Paris','Paris','Paris','NY','NY','NY','NY'],'T':[100,200,300,400,10,20,30,40]} 
df = pandas.DataFrame(data=d)
df['dates']=pandas.to_datetime(df['dates'])
df

    dates   location    T
0   2020-07-16  Paris   100
1   2020-07-15  Paris   200
2   2020-07-14  Paris   300
3   2020-07-13  Paris   400
4   2020-07-16  NY       10
5   2020-07-15  NY       20
6   2020-07-14  NY       30
7   2020-07-13  NY       40

我想為過去 2 天（包括當前日期）滾動的給定位置提供一些T值。 這是我想要的熊貓：

    dates   location    T     SUM2D
0   2020-07-16  Paris   100     300
1   2020-07-15  Paris   200     500
2   2020-07-14  Paris   300     700
3   2020-07-13  Paris   400     NaN
4   2020-07-16  NY       10      30
5   2020-07-15  NY       20      50
6   2020-07-14  NY       30      70
7   2020-07-13  NY       4      NaN

我試過玩這句話但沒有成功：

df['SUM2D'] = df.set_index('dates').groupby('location').rolling(window=2, freq='D').sum()['T'].values

Answer 1

嘗試在索引之前對 dataframe 進行排序：

df = df.sort_values(['location','dates']).set_index('dates')
df['SUM2D'] = df.groupby('location')['T'].rolling(window=2, freq='D').sum().values

df[::-1]

結果集：

           location    T  SUM2D
dates                          
2020-07-16    Paris  100  300.0
2020-07-15    Paris  200  500.0
2020-07-14    Paris  300  700.0
2020-07-13    Paris  400    NaN
2020-07-16       NY   10   30.0
2020-07-15       NY   20   50.0
2020-07-14       NY   30   70.0
2020-07-13       NY   40    NaN

更緊湊和優雅的解決方案是使用transform ：

df['SUM2D'] = df.sort_values(['dates']).groupby('location')['T'].transform(lambda x: x.rolling(2, 2).sum())

結果現在是：

       dates location    T  SUM2D
0 2020-07-16    Paris  100  300.0
1 2020-07-15    Paris  200  500.0
2 2020-07-14    Paris  300  700.0
3 2020-07-13    Paris  400    NaN
4 2020-07-16       NY   10   30.0
5 2020-07-15       NY   20   50.0
6 2020-07-14       NY   30   70.0
7 2020-07-13       NY   40    NaN