Question

我目前能够通过使用大量for循环完成我想要的任务，但我觉得应该有一个更简单的方法。

我希望我的prog返回包含在字符串中找到的单词的列表名称。我现在正在做：

my_car = ["blue","ford"]
dads_car = ["green","audi"]
car = input('Tell me either the make or colour of your car: ').lower()
whos_car = ""
for word in my_car:
    if word in car:
        whos_car = "my_car"
    else:
        whos_car = "unknown"
for word in dads_car:
    if word in car:
        whos_car = "dads_car"
    else:
        whos_car = "unknown"
print(whos_car)

但我想要做的是写一个遍历每个列表的for循环，无论哪个列表包含该单词，都返回列表名称。这不起作用，但希望它会更清楚：

for word in my_car,dads_car:
    if word in car:
        whos_car = the list name it found it in...
    else:
        whos_car = "Unknown"

希望这是有道理的。

非常感谢！

Answer 1

澄清后更新

使用字典和函数：

def get_list(cars, car):
    car_parts = car.split()
    for list_name, car_info in cars.items():
        for item in car_info:
            if item in car_parts:
                return list_name
    return 'unknown'

my_car = ["blue","ford"]
dads_car = ["green","audi"]
cars = {'my_car': my_car, 'dads_car': dads_car}
car = 'the car green'
print(get_list(cars, car))

现在，如果你设置：

car = 'the car green'

打印：

dads_car

但：

car = 'light-green truck'

打印：

unknown

旧尝试

看起来很适合字典：

my_car = ["blue","ford"]
dads_car = ["green","audi"]

cars = {'my_car': my_car, 'dads_car': dads_car}

car = 'green'

for list_name, car_info in cars.items():
    if car in car_info:
        print(list_name)
        break
else:
    print('unknown')

现在，如果你设置：

car = 'green'

打印：

dads_car

但：

car = 'yellow'

打印：

unknown

您可以根据需要向cars添加任意数量的项目，而无需更改代码。所以没有大量的for循环。

如果两个列表中都存在，那么它应该显示它找到的第一个列表。

一旦找到匹配，就会突破循环。仅当else:被执行时才会到达break下的代码。从Python 3.6开始。字典保持它们构建的顺序。因此，首先会搜索my_car。

具有不同逻辑的替代解决方案：

my_car = ["blue","ford"]
dads_car = ["green","audi"]

cars = {'my_car': my_car, 'dads_car': dads_car}

found = False
for list_name, car_info in cars.items():
    for item in car_info:
        if item in car:
            print(list_name)
            found = True
            break
    if found:
        break
else:
    print('unknown')

现在，如果你设置：

car = 'light-green'

打印：

dads_car

或更短的功能：

def get_list(cars, car):
    for list_name, car_info in cars.items():
        for item in car_info:
            if item in car:
                return list_name
    return 'unknown'

print(get_list(cars, car))

Answer 2

您可以使用locals()：

def lookup(query):
    my_car = ["blue","ford"]
    dads_car = ["green","audi"]

    for name, vals in locals().items():
        #  you need a condition to filter the variables you want to search in!
        if name.endswith('_car') and query in vals:
            return name
    return 'unknown'

result = lookup('green')
print(result)

result = lookup('green2')
print(result)

打印：

dads_car
unknown

Answer 3

一种简单的方法是在列表中找不到whos_car时更改my_car = ["blue","ford"] dads_car = ["green","audi"] car = input('Tell me either the make or colour of your car: ').lower() whos_car = None # None is more "Pythonic" for word in my_car: if word in car: whos_car = "my_car" break # no need to continue looping if whos_cas is None: # don't search second list if word has already be found for word in dads_car: if word in car: whos_car = "dads_car" break if whos_car is None: whos_car = "unknow" # if we haven't found anything... print(whos_car)的值，仅在第一个列表中找不到该词时才搜索第二个列表。您的代码可能只是变成：

manage.py

Answer 4

我的解决方案如下，应该很容易理解：

你也可以按照建议使用whos_car结束forloop后的两次休息（如果你把这些列表放大的话）

代码：

my_car = ["blue", "ford"]
dads_car = ["green" , "audi"]

car = str.lower(input('Tell me either the brand or colour of your car: '))

for element in zip(my_car,dads_car):
    if car in element[0]:
        whos_car = "mine car now"
    elif car in element[1]:
        whos_car = "dads car"
    else:
        whos_car = "uknown"

print(whos_car)

输出：

user@user:~$ python3 script.py Tell me either the brand or colour of your car: FORD mine car now

Answer 5

my_car = ["blue","ford"]
dads_car = ["green","audi"]
car = input('Tell me either the make or colour of your car: ').lower()
whos_car = "unknown"
for owner, lst in (('my_car', my_car), ('dads_car', dads_car)):
    for word in lst:
        if word == car:
            whos_car = owner
            break
print(whos_car)

这非常接近您的要求。所以owner和lst是for循环读取的每个元组的共同值。使用break可确保在找到所有者后不会覆盖whos_car。

返回找到字符串的列表名称

5 个答案:

澄清后更新

旧尝试