Python 如何缩短列表中对象的代码？

我正在检索要添加到dict的数据库对象，但密钥不能重复，所以我是这样做的：

    carb1 = random.choice(carbs)
    protein1 = random.choice(proteins)
    carb2 = Food.objects.filter(category='Carbs').exclude(name=carb1.name)[0]
    protein2 = Food.objects.filter(category='Protein').exclude(name=protein1.name)[0]
    veg1 = random.choice(vegs)
    veg2 = Food.objects.filter(category='Vegetables').exclude(name=veg1.name)[0]

    meals = [carb1, protein1, carb2, protein2]

    exclude_these = [o.name for o in meals]

    carb3 = Food.objects.filter(category='Carbs').exclude(name__in=exclude_these)[0]
    protein3 = Food.objects.filter(category='Protein').exclude(name__in=exclude_these)[0]
    veg3 = Food.objects.filter(category='Vegetables').exclude(name__in=exclude_these)[0]
    meals.append(carb3)
    meals.append(protein3)
    meals.append(veg3)

    exclude_these = [o.name for o in meals]

    carb4 = Food.objects.filter(category='Carbs').exclude(name__in=exclude_these)[0]
    protein4 = Food.objects.filter(category='Protein').exclude(name__in=exclude_these)[0]
    veg4 = Food.objects.filter(category='Vegetables').exclude(name__in=exclude_these)[0]

    meals.append(carb4)
    meals.append(protein4)
    meals.append(veg4)

    exclude_these = [o.name for o in meals]

    carb5 = Food.objects.filter(category='Carbs').exclude(name__in=exclude_these)[0]
    protein5 = Food.objects.filter(category='Protein').exclude(name__in=exclude_these)[0]
    veg5 = Food.objects.filter(category='Vegetables').exclude(name__in=exclude_these)[0]

    meals.append(carb5)
    meals.append(protein5)
    meals.append(veg5)

    exclude_these = [o.name for o in meals]

    carb6 = Food.objects.filter(category='Carbs').exclude(name__in=exclude_these)[0]
    protein6 = Food.objects.filter(category='Protein').exclude(name__in=exclude_these)[0]
    veg6 = Food.objects.filter(category='Vegetables').exclude(name__in=exclude_these)[0]

但这似乎是一个荒谬的代码量来实现我想要的

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有什么方法可以缩短这个时间吗？

看起来您想从食物表中检索6种碳水化合物、6种蛋白质和6种蔬菜，而不在乎它们是哪种（您在查询中不提供排序，只取第一种）。除了第一个是随机的。也许你不在乎第一个是随机挑选的，或者你希望剩下的5个也是均匀随机挑选的

以下是一些选项：

carb1, carb2, carb3, carb4, carb5, carb6 = tuple(
    Food.objects.filter(category='Carbs')[:6]
)

portein和veg也是如此。坦率地说，不要将变量命名为carbN，只要将它们全部放在一个列表或元组中即可<代码>碳水化合物=食物….[：6]

如果名称列不唯一，并且您确实需要6个名称不同的碳水化合物

carbs = Food.objects.filter(category='Carbs').distinct('name')[:6]

在列上而不是整个记录上使用Distinct在所有数据库后端都不起作用。它在postgres中有效，但我认为它在mysql中不起作用

如果你真的想要6个随机碳水化合物，你可以让数据库为你随机

carbs = Food.objects.filter(category='Carbs').order_by('?')[:6]

如果您的食物表很大，随机订购可能会出现一些性能问题