來自以下Pandas數據框。df = pd.DataFrame({'Id': [102,102,102,303,303,944,944,944,944],'A':[1.2,1.2,1.2,0.8,0.8,2.0,2.0,2.0,2.0],'B':[1.8,1.8,1.8,1.0,1.0,2.2,2.2,2.2,2.2], 'A_scored_time':[10,25,0,33,0,40,0,90,0],'B_scored_time':[0,0,30,0,41,0,75,0,95]})我試圖創建源自的組合的['A_scored_time','B_scored_time']列表，以獲得以下與unique對應的列表Id：Id(102) = A_Time = [10,25], B_Time = [30]Id(303) = A_Time = [33], B_Time = [41]Id(944) = A_Time = [40,90], B_Time = [75,95]該列表將在下面的功能中應用。x1 = [1,0,0] x2 = [0,1,0] x3 = [0,0,1]k = 100 # constanttotal_timeslot = 100 # same as kA_Time = [] B_Time = [] 對于范圍內的i（區別ID），df在此處具有3個不同的ID。對于每個i，概率陣列y。y = np.array([1-(A + B)/k, A/k, B/k]) def sum_squared_diff(x1, x2, x3, y): ssd = [] for k in range(total_timeslot): if k in A_Time: ssd.append(sum((x2 - y) ** 2)) elif k in B_Time: ssd.append(sum((x3 - y) ** 2)) else: ssd.append(sum((x1 - y) ** 2)) return ssd輸出將是len k的數組。一旦獲得此值，我將對所有n（n個不同的Id）數組求和。這是我所追求的。結果為df：Id(102) = sum(sum_squared_diff(x1, x2, x3, y)) =5.872800000000018Id(303) = sum(sum_squared_diff(x1, x2, x3, y)) = 3.9407999999999896Id(944) = sum(sum_squared_diff(x1, x2, x3, y)) =7.760800000000006給予 toatl sum = 17.574400000000015.
1 回答
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TA貢獻1829條經驗 獲得超9個贊
要回答標題中的問題，請使用：
df.groupby('Id')[['A_scored_time','B_scored_time']]\
.agg(lambda x: x[x != 0].tolist())\
.reset_index()
輸出：
Id A_scored_time B_scored_time
0 102 [10, 25] [30]
1 303 [33] [41]
2 944 [40, 90] [75, 95]
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