Question

共有9个元素，其中包含饮料的名称。我需要做的是删除所有包含“ Americano”的元素

但是，我无法调试我的代码，并且一直显示错误。如何删除包含“ Americano”的list1的元素[0,2,3,6]？

list1 = [ ['3', 'Americano', '7', '14000'], ['4', 'Smoothie_queen', '4', '12000'], ['5', 'Americano', '2', '4000'], ['6', 'Americano', '17', '34000'], ['7', 'Cafe_mocha', '4', '11200'], ['8', 'Cafe_latte', '11', '27500'], ['9', 'Americano', '17', '34000'], ['10', 'Amorparty', '2', '4000'], ['11', 'Plain_yogurt', '13', '45500']]

while True:
        if " Americano" in list1[i]:
            del list1[0]
    if i < x:
        i = i + 1
        continue

    if i >= x:
        break

print(list1)

Answer 1

使用列表推导过滤掉不需要的元素。

@GetMapping("/showFormForAdd")

public String showFormForAdd(Model theModel) {

    //fetch new list(if any) of departments added
    List<Department> theDepartments = departmentService.getDepartments();

    //Create a linkedhash map to hold our department_id-department name information
    final LinkedHashMap<Integer, String> departmentOptions = departmentService.generateDepartmentOptions(theDepartments);



    // create new employee object and attach to our model atrribute.

            //how to add multiple objects?? doing this so i can pre-populate available departments for selection
    theModel.addAttribute("employee", departmentOptions);


    Employee theEmployee = new Employee();
            //how to add multiple objects?? doing this so when program return control to controller it will help me set the attribute of employees so I can save it into the database
    theModel.addAttribute("employee", theEmployee);


    return "customer-form";
}

结果：

list1 = [x for x in list1 if 'Americano' not in x]

Answer 2

使用filter函数，这将创建一个新列表并产生结果

list1 = [ ['3', 'Americano', '7', '14000'], ['4', 'Smoothie_queen', '4', '12000'], ['5', 'Americano', '2', '4000'], ['6', 'Americano', '17', '34000'], ['7', 'Cafe_mocha', '4', '11200'], ['8', 'Cafe_latte', '11', '27500'], ['9', 'Americano', '17', '34000'], ['10', 'Amorparty', '2', '4000'], ['11', 'Plain_yogurt', '13', '45500']]
result =list(filter(lambda x: 'Americano' not in x , list1))
print(result)

输出

[['4', 'Smoothie_queen', '4', '12000'],
 ['6', 'Americano', '17', '34000'],
 ['7', 'Cafe_mocha', '4', '11200'],
 ['8', 'Cafe_latte', '11', '27500'],
 ['10', 'Amorparty', '2', '4000'],
 ['11', 'Plain_yogurt', '13', '45500']]

使用enumerate和list.remove()函数，在现有列表上对其进行修改因此无需创建新列表

list1 = [ ['3', 'Americano', '7', '14000'], ['4', 'Smoothie_queen', '4', '12000'], ['5', 'Americano', '2', '4000'], ['6', 'Americano', '17', '34000'], ['7', 'Cafe_mocha', '4', '11200'], ['8', 'Cafe_latte', '11', '27500'], ['9', 'Americano', '17', '34000'], ['10', 'Amorparty', '2', '4000'], ['11', 'Plain_yogurt', '13', '45500']]
for index, value in enumerate(list1):
    if 'Americano' in value:
        list1.remove(value)

print(list1)

输出

[['4', 'Smoothie_queen', '4', '12000'],
 ['6', 'Americano', '17', '34000'],
 ['7', 'Cafe_mocha', '4', '11200'],
 ['8', 'Cafe_latte', '11', '27500'],
 ['10', 'Amorparty', '2', '4000'],
 ['11', 'Plain_yogurt', '13', '45500']]

Answer 3

如果您不想感到困惑，这是分步代码

list1 = [['3', 'Americano', '7', '14000'], ['4', 'Smoothie_queen', '4', '12000'], ['5', 'Americano', '2', '4000'],
         ['6', 'Americano', '17', '34000'], ['7', 'Cafe_mocha', '4', '11200'], ['8', 'Cafe_latte', '11', '27500'],
         ['9', 'Americano', '17', '34000'], ['10', 'Amorparty', '2', '4000'], ['11', 'Plain_yogurt', '13', '45500']]
lis = []
for row in list1:
    if 'Americano' not in row:
        lis.append(row)
print lis

输出

[['4', 'Smoothie_queen', '4', '12000'], ['7', 'Cafe_mocha', '4', '11200'], ['8', 'Cafe_latte', '11', '27500'], ['10', 'Amorparty', '2', '4000'], ['11', 'Plain_yogurt', '13', '45500']]

如何从列表中删除此元素？

3 个答案: