在 Python 中使用范围作为字典键，我有什么选择?

2022-01-24 00:00:00 python dictionary loops range key

问题描述

这是我的第一篇文章，我对编程很陌生，所以我可能无法恰当地传达我的问题，但我会尽力而为！

This is my first post and I'm quite new at programming, so I might not be able to convey my question appropriately, but I'll do my best!

tries_dict = {1:'first', 2:'second', 3:'third', 4:'fourth', ub_tries:'last'}

ub_tries = user input

tries = 1

input ('
Come on make your ' + tries_dict.get(tries) + guess: ')

这 3 个元素是我创建的猜数游戏的一部分，我将它们包含在一个 while 循环中，其中 tries += 1 在每个错误答案之后.

These 3 elements are part of a number guess game I created, and I included them in a while loop where tries += 1 after each wrong answer.

如您所见，在我的字典中，前 4 个答案和游戏结束前最后可能的机会都有自定义值，所以这是我尝试做的:

As you can see, in my dictionary there are custom values for the first 4 answers and the last possible chance before the game is over, so here is what I tried to do:

我想找到一种方法，让第四"和最后"之间的每个答案/键都有NEXT"值.

I wanted to find a way to have the 'NEXT' value for every answer/key between 'fourth' and 'last'.

如:

tries = 5

Come on make your next guess

tries = 6

Come on make your next guess

等等

我确实找到了一些复杂循环的方法，但作为好奇的类型，我想知道更有效/实用的方法来完成这个.

I did find a way with some complex looping, but being the curious type I wanted to know of more efficient/practical ways to accomplish this.

以下是我想过但无法开始工作的一些选项:

Here are some options i thought about but couldn't get to work:

  1. 使用范围作为键
  2. 找到一种方法来生成一个值介于 4 和 ub_tries 之间的列表并将该列表用作键
  1. Using a range as a key
  2. Finding a way to generate a list with values between 4 and ub_tries and using that list as a key

一般来说:如何创建一种方法来为字典中未指定的键提供此一般答案(下一个或其他)?

So generally speaking: how can one create a way to have this general answer (next or whatever) for keys that aren't specified in a dictionary?

任何反馈都将不胜感激，请随时要求澄清，因为我可以告诉自己我的问题有点混乱.

Any feedback would be greatly appreciated, feel free to ask for clarifications since I can tell myself my question is kind of messy.

我希望我在编程和提出相关问题方面变得更加狡猾，到目前为止，我的编程几乎和我的总结技能一样混乱，叹息！

I hope I get more crafty both at programming and asking related questions, so far my programming is nearly as messy as my summary skills, sigh!

解决方案

我不确定这是否是你想要的，但是 dict.get 可能是答案:

I'm not sure whether this is what you want, but dict.get may be the answer:

>>> ub_tries = 20
>>> tries_dict = {1:'first', 2:'second', 3:'third', 4:'fourth', ub_tries:'last'}
>>> tries_dict.get(1, 'next')
'first'
>>> tries_dict.get(4, 'next')
'fourth'
>>> tries_dict.get(5, 'next')
'next'
>>> tries_dict.get(20, 'next')
'last'
>>> tries_dict.get(21, 'next')
'next'

当然，您可以以各种不同的方式将其封装在一个函数中.例如:

Of course you could wrap this up in a function, in various different ways. For example:

def name_try(try_number, ub_tries):
    tries_dict = {1:'first', 2:'second', 3:'third', 4:'fourth', ub_tries:'last'}
    return tries_dict.get(try_number, 'next')

无论如何，dict.get(key, default=None) 类似于 dict[key]，除了如果 key 是不是成员，而不是引发 KeyError，而是返回 default.

At any rate, dict.get(key, default=None) is like dict[key], except that if key is not a member, instead of raising a KeyError, it returns default.

至于你的建议:

使用范围作为键??

当然，您可以这样做(如果您使用的是 Python 2 而不是 3，请使用 xrange 表示 range)，但它有什么帮助呢?

Sure, you can do that (if you're in Python 2 instead of 3, use xrange for range), but how would it help?

d = { range(1, 5): '???', 
      range(5, ub_tries): 'next', 
      range(ub_tries, ub_tries + 1): 'last' }

这是完全合法的——但 d[6] 会引发 KeyError，因为 6 与 <代码>范围(5，ub_tries).

That's perfectly legal—but d[6] is going to raise a KeyError, because 6 isn't the same thing as range(5, ub_tries).

如果你想让它工作，你可以像这样构建一个 RangeDictionary:

If you want this to work, you could build a RangeDictionary like this:

class RangeDictionary(dict):
    def __getitem__(self, key):
        for r in self.keys():
            if key in r:
                return super().__getitem__(r)
        return super().__getitem__(key)

但这远远超出了初学者的 Python"，即使对于这种低效、不完整和不健壮的实现也是如此，所以我不建议这样做.

But that's well beyond "beginners' Python", even for this horribly inefficient, incomplete, and non-robust implementation, so I wouldn't suggest it.

找到一种方法来生成一个值在 4 到 ub_tries 之间的列表，并使用这样的列表作为键

finding a way to generate a list with values between 4 and ub_tries and using such list as a key

你的意思是这样的?

>>> ub_tries = 8
>>> tries_dict = {1:'first', 2:'second', 3:'third', 4:'fourth', ub_tries:'last'}
>>> tries_dict.update({i: 'next' for i in range(5, ub_tries)})
>>> tries_dict
{1: 'first', 2: 'second', 3: 'third', 4: 'fourth', 5: 'next', 6: 'next', 7: 'next', 8: 'last'}
>>> tries_dict[6]
'next'

这可行，但它可能不是一个好的解决方案.

That works, but it's probably not as good a solution.

最后，您可以使用 defaultdict，它可以让您将默认值烘焙到字典中，而不是将其作为每次调用的一部分传递:

Finally, you could use defaultdict, which lets you bake the default value into the dictionary, instead of passing it as part of each call:

>>> from collections import defaultdict
>>> tries_dict = defaultdict(lambda: 'next', 
...                          {1:'first', 2:'second', 3:'third', 4:'fourth', ub_tries:'last'})
>>> tries_dict
defaultdict(<function <lambda> at 0x10272fef0>, {8: 'last', 1: 'first', 2: 'second', 3: 'third', 4: 'fourth'})
>>> tries_dict[5]
'next'
>>> tries_dict
defaultdict(<function <lambda> at 0x10272fef0>, {1: 'first', 2: 'second', 3: 'third', 4: 'fourth', 5: 'next', 8: 'last'})

但是，请注意，这会在您第一次请求时永久创建每个元素，并且您必须创建一个返回默认值的函数.这对于您要更新值并且只希望以默认值作为起点的情况更有用.

However, note that this permanently creates each element the first time you ask for it—and you have to create a function that returns the default value. This makes it more useful for cases where you're going to be updating values, and just want a default as a starting point.

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