通過不同長度的查詢循環

Question

我正在使用zip循環同時處理兩個查詢的結果。但是在某些情況下，結果並不等長。在這種情況下，我希望將首先結束的查詢值設置爲0.始終是由於它是第一個查詢的子集而首先結束的occupancy_agency。具體來說，它是和occupancy_agency_y['available']。我試圖想出一個解決方案，但無法弄清楚如何將它與zip相結合，使我能夠同時循環兩個結果。

def occupancy_data(area_id, description, period, agency_id): 
    occupancy = Occupancy.objects.filter(description=description) \ 
     .values('start_date') \ 
     .annotate(supply_total=Sum('supply')) \ 
     .annotate(available_total=Sum('available')) \ 
     .order_by('start_date') 

    occupancy_agency = Occupancy.objects.filter(description=description, agency_id=agency_id) \ 
     .values('start_date',) \ 
     .annotate(supply=Sum('supply')) \ 
     .annotate(available=Sum('available')) \ 
     .order_by('start_date') 


    x = [] 
    _input = occupancy.values('start_date') 
    for row in _input: 
     x.append("Uge " + str(int(row['start_date'].strftime("%V")))) 

    y = [] 
    for occupancy_y, occupancy_agency_y in zip(occupancy, occupancy_agency): 
     comp_supply = (occupancy_y['supply_total'] - occupancy_agency_y['supply']) 
     comp_available = (occupancy_y['available_total'] - occupancy_agency_y['available']) 

     occupancy_combined = ((comp_supply - comp_available)/comp_supply) 

     y.append(occupancy_combined) 
    return {'x': x, 'y': y} 

Answer 1

itertools提供了zip_longest函數正是這樣做的......

for occupancy_y, occupancy_agency_y in zip_longest(occupancy, occupancy_agency, fillvalue=0): 
    ... 

或者，也許：

for occupancy_y, occupancy_agency_y in zip_longest(occupancy, occupancy_agency, fillvalue={}): 
    comp_supply = (occupancy_y['supply_total'] - occupancy_agency_y.get('supply', 0)) 
    comp_available = (occupancy_y['available_total'] - occupancy_agency_y.get('available', 0)) 

    occupancy_combined = ((comp_supply - comp_available)/comp_supply) 

    y.append(occupancy_combined)