Question

我不确定为什么这段代码失败了一些输入（来自CodingBat，链接到问题：Exercise Link）。问题详情如下，我可以用if elif语句来解决这个问题，但我想使用字典。此外，我已经读过，建议不要从字典中获取键值，如下所示。但如果可以指出以下方案中的问题，我将不胜感激。

你驾驶的速度有点太快，一名警察阻止你。编写代码来计算结果，编码为int值：0 =没有票，1 =小票，2 =大票。如果速度为60或更低，则结果为0.如果速度在61和80之间，则结果为1.如果速度为81或更高，则结果为2.除非是您的生日 - 在那一天，您的在所有情况下，速度可以高出5个。

caught_speeding（60，False）→0
caught_speeding（65，False）→1
caught_speeding（65，True）→0

def caught_speeding(speed, is_birthday):
    Bir_dict = {0:speed<=60,1:61<=speed<=80,2:speed>=81}
    NoBir_dict = {0:speed<=65,1:66<=speed<=85,2:speed>=86}
    def getKey(dict,value):
        return [key for key in dict.keys() if (dict[key] == value)]
    if is_birthday:
        out1=getKey(Bir_dict,True)
        return out1[0]
    else:
        out2=getKey(NoBir_dict,True)
        return out2[0]

程序失败

caught_speeding(65, False)
caught_speeding(65, True)

为

工作

caught_speeding(70, False)
caught_speeding(75, False)
caught_speeding(75, True)
caught_speeding(40, False)
caught_speeding(40, True)
caught_speeding(90, False)
caught_speeding(60, False)
caught_speeding(80, False)

Answer 1

看起来像是混合了Bir_dict和NoBir_dict。你能试试下面的代码吗？

def caught_speeding(speed, is_birthday):
        Bir_dict = {0:speed<=65,1:66<=speed<=85,2:speed>=86}
        NoBir_dict = {0:speed<=60,1:61<=speed<=80,2:speed>=81}
        def getKey(dict,value):
            return [key for key in dict.keys() if (dict[key] == value)]
        if is_birthday:
            out1=getKey(Bir_dict,True)
            return out1[0]
        else:
            out2=getKey(NoBir_dict,True)
            return out2[0]

尽管它有效，但我可以建议使用字典的替代方法：票证定义不会相互干扰，换句话说，字典中只能有一个True语句。因此，您可以将代码修改为：

def caught_speeding(speed, is_birthday):
        Bir_dict = {0:speed<=65,1:66<=speed<=85,2:speed>=86}
        NoBir_dict = {0:speed<=60,1:61<=speed<=80,2:speed>=81}
        def getKey(dict):
            return [key for key in dict.keys() if (dict[key] == True)]
        if is_birthday:
            out1=getKey(Bir_dict)
            return out1[0]
        else:
            out2=getKey(NoBir_dict)
            return out2[0]

Answer 2

问题不会认为生日词典具有更高的可用值吗？

如果是我的生日，我要去65岁：我不指望没有票。但是，如果您生成字典并打印它们，您可以看到情况并非如此：

def caught_speeding(speed, is_birthday):
    Bir_dict = {0:speed<=60,1:61<=speed<=80,2:speed>=81}
    print Bir_dict
    NoBir_dict = {0:speed<=65,1:66<=speed<=85,2:speed>=86}
    print NoBir_dict
    def getKey(dict,value):
        return [key for key in dict.keys() if (dict[key] == value)]
    if is_birthday:
        out1=getKey(Bir_dict,True)
        return out1[0]
    else:
        out2=getKey(NoBir_dict,True)
        return out2[0]

输出：

{0: False, 1: True, 2: False}
{0: True, 1: False, 2: False}
0
{0: False, 1: True, 2: False}
{0: True, 1: False, 2: False}
1

您只需在字典对象中交叉值

使用Python字典值返回键作为结果

2 个答案: