Question

我试图以城市和温度的形式接受用户输入，然后将其保存到字典中。在那之后它应该打印出最低温度是什么以及它是什么城市。虽然我的程序从未到达打印消息，但我试图在没有min函数的情况下迭代地执行此操作。为什么以及如何解决这个问题？谢谢。

编辑：我意识到我可以使用min函数来使这更容易，但这是一个练习练习，我应该通过迭代找到最小值。此外，如果暂时不说，如果2个或更多城市的温度最低，我会忽略。

def main():
city = {}
lowest = 0
while True:
    user = input("Enter city followed by temperature: ")
    if user == "stop":
        print(city)
        for value in city.values():
            if int(value) < lowest:
                lowest = int(value)

        for key in city:
            if city[key] == lowest:
                print("The coldest city is,", city[key])
        break

    name, temperature = user.split()
    city[name] = temperature

main()

Answer 1

您的代码中的问题是您将lowest初始化为0。如果您的所有城市都有正温度，lowest在计算最小值时将永远不会改变，并且任何城市都不会满足if city[key] == lowest:测试。

解决此问题的方法是将lowest初始化为城市的其中一个温度。例如，您可以添加：

lowest = temperature

以下

if user == "stop":

将lowest初始化为用户输入的最后一个温度。

最后，代码中的最后一个问题出现在if city[key] == lowest测试中：构建字典的方式，city[key]是一个字符串，而lowest是一个整数，它们永远不会相等。我建议你在构造字典时将温度转换为整数，这样你就不必在以后处理转换。整体更正后的代码如下所示：

def main():
    city = {}
    while True:
        user = input("Enter city followed by temperature: ")
        if user == "stop":
            print(city)
            lowest = temperature
            for value in city.values():
                if value < lowest:
                    lowest = value

            for key in city:
                if city[key] == lowest:
                    print("The coldest city is,", city)
            break

        name, temperature = user.split()
        temperature = int(temperature)
        city[name] = temperature

main()

Answer 2

找到最低点的更简单方法是使用min

lowest = min(city.values())

“停止”应该break离开你的循环

while True:
    user = input("Enter city followed by temperature: ")
    if user == "stop":
        break
    name, temperature = user.split()
    city[name] = float(temperature)    # convert to float (or perhaps int)


print(city)
lowest = min(city.values())

for key in city:
    if city[key] == lowest:
        print("The coldest city is,", city[key])

当有多个温度最低的城市

时，您可能需要考虑打印不同的消息

Answer 3

您可以使用传递lambda到min来直接获取键/值。

while True:
    user = input("Enter city followed by temperature: ")
    if user == "stop":
        break
    name, temperature = user.split()
    city[name] = float(temperature)    # convert to float (or perhaps int)

lowest_city, lowest_temp = min(city.iteritems(), key=lambda x:x[1])
print("The coldest city is, %s, temperature = %.3f" % (lowest_city, lowest_temp))

这里lambda函数告诉min函数根据选择 iteritems函数返回的元组的第二个值。

在字典中查找最低值

3 个答案: