用Python写一下这个吧

1.自定义函数dict_keys()，实现字典的keys()方法,返回值如下:(['Apple', 'HuWei', 'Vivo', 'OPPO', 'Nokia'])2.自定义函数dict_values()函数，实现字典的values()方法，返回值如下:([4999, 3600, 2999, 3200, 3499])3.自定义函数dict_items()函数，实现字典的items()方法,返回值如下:([('Apple', 4999), ('HuWei', 3600), ('Vivo', 2999), ('OPPO', 3200), ('Nokia', 3499)])1.自定义函数dict_keys()，实现字典的keys()方法,返回值如下:(['Apple', 'HuWei', 'Vivo', 'OPPO', 'Nokia'])2.自定义函数dict_values()函数，实现字典的values()方法，返回值如下:([4999, 3600, 2999, 3200, 3499])3.自定义函数dict_items()函数，实现字典的items()方法,返回值如下:([('Apple', 4999), ('HuWei', 3600), ('Vivo', 2999), ('OPPO', 3200), ('Nokia', 3499)])

d = dict([('Apple', 4999), ('HuWei', 3600), ('Vivo', 2999), ('OPPO', 3200),
          ('Nokia', 3499)])

def dict_keys(dt):
    lt = []
    ii = iter(d)
    try:
        while True:        
            v = next(ii)
            lt.append(v)
    finally:
        return (lt, )
    
res = dict_keys(d)
print(res)

def dict_values(dt):
    res = dict_keys(dt)[0]
    return ([dt[i] for i in res], )
print(dict_values(d))

def dict_items(dt):
    return (list(zip(dict_keys(dt)[0], dict_values(dt)[0])), )
print(dict_items(d))

望采纳！谢谢

def dict_keys():
    return ['Apple', 'HuWei', 'Vivo', 'OPPO', 'Nokia']
def dict_values():
    return [4999, 3600, 2999, 3200, 3499]
def dict_items(n):
    return [dict_keys()[n],dict_values()[n]]

print(dict_keys())
print(dict_values())
print(dict_items(2))