Question

public HttpResponseMessage GetAvailableHospitalsByAjax(System.Guid? DirectorateOfHealthID = null, System.Guid? UnitTypeID = null, string DeviceTypeIDs = null)
{
    Context db = new Context();
    var query = db.Hospitals.AsQueryable();
    if (UnitTypeID != null)
    {
        query = query.Where(j => j.HospitalDepartments.Any(www => www.Units.Any(u => u.UnitTypeID == UnitTypeID)));
    }

    if (DirectorateOfHealthID != null)
    {
        query = query.Where(h => h.DirectorateHealthID == DirectorateOfHealthID);
    }


    query = query.Where(j => j.HospitalDepartments.Any(u => u.Units.Any(d => d.Devices.Any(s => s.Status == Enums.DeviceStatus.Free)))
    && j.HospitalDepartments.Any(hd => hd.Units.Any(u => u.Beds.Any(b => b.Status == Enums.BedStatus.Free))));



    var list = query.ToList().Select(w => new HospitalInfo()
    {
        Id = w.ID,
        Name = w.Name 

    }).ToList();


    return Request.CreateResponse(HttpStatusCode.OK, list);
}

我明白了：

pizzas  = ["hawai","salame","vegetable","capriciosa","new york"]
for pizza in pizzas:
    print("I like " + pizza.title() + " pizza!")
print("\n" + "The first three pizzas in the list are: " + str(pizzas[0:3]))
print("\n" + "The last three pizzas in the list are: " + str(pizzas[-1:-3]))

我很困惑。 Isn＆＃39; t

I like Hawai pizza!
I like Salame pizza!
I like Vegetable pizza!
I like Capriciosa pizza!
I like New York pizza!

The first three pizzas in the list are: ['hawai', 'salame', 'vegetable']

The last three pizzas in the list are: []

表示列表中的最后一个元素？我正在为-1编制索引，所以不应该打印我最后3个项目吗？我究竟做错了什么？

Answer 1

您应该使用pizzas[-3:]，这样您的start就是列表末尾的第三个元素，而end就是列表的最后一个。

pizzas  = ["hawai","salame","vegetable","capriciosa","new york"]

for pizza in pizzas:
    print("I like " + pizza.title() + " pizza!")

print("\nThe first three pizzas in the list are: " + str(pizzas[:3]))
print("\nThe last three pizzas in the list are: " + str(pizzas[-3:]))

Answer 2

默认步长为1，因此从-1到-3，步长为1将返回一个空切片。您可以明确地将步骤指示为-1，但这会颠倒项目的顺序：

>>> pizzas[-1:-3:-1] 
['new york', 'capriciosa'] # -3 excluded

但是，要获得最后三项，您需要的是pizza[-3:]

在Python中打印列表的最后三项

2 个答案: