由于 Petr 已经为您提供了(理想的)Django 风格的答案,这里有一个关于如何使用列表理解将任何可迭代对象放入列表的 Pythonic 答案。
# Leave out .all(), if you already have .filter()
dates_queryset = Profile.objects.filter(user=request.user)
dates = [qr.date for qr in dates_queryset]
weights = [qr.weight for qr in dates_queryset]
num = len(dates_queryset)
您可以使用 zip() 进一步缩短它:List comprehension with tuple assignment
dates_queryset = Profile.objects.filter(user=request.user)
dates, weights = zip(*[(qr.date, qr.weight) for qr in dates_queryset])
num = len(dates_queryset)