use a range as a dictionary key in Python, what op

问题:

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 ('\nCome on make your ' + tries_dict.get(tries) + guess: ')

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.

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:

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

As in:

tries = 5

Come on make your next guess

tries = 6

Come on make your next guess

and so on

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.

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

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

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?

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

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!

回答1:

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'

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')

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.

As for your suggestions:

using a range as a key??

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' }

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).

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)

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

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

You mean like this?

>>> 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'

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

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'})

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.

回答2:

Have I captured your intent for the game here?

max_tries = 8

tries_verbage = {
    1: 'first',
    2: 'second',
    3: 'third',
    4: 'fourth',
    max_tries: 'last'
    }

for i in xrange(1, max_tries + 1):
    raw_input('Make your %s guess:' % tries_verbage.get(i, 'next'))

returns

Make your first guess:1
Make your second guess:2
Make your third guess:3
Make your fourth guess:4
Make your next guess:5
Make your next guess:6
Make your next guess:7
Make your last guess:8

回答3:

Why don't you just use a list?

MAX_TRIES = 10
tries_list = ["first", "second", "third", "fourth"]

for word in (tries_list[:MAX_TRIES-1] + 
             (["next"] * (MAX_TRIES - len(tries_list) - 1)) + ["last"]):
    result = raw_input("Come on, make your {} guess:".format(word))

Note this code won't work as planned for MAX_TRIES = 0, but I don't think that will be a problem.

回答4:

You can use a dictionary with range as a key:

def switch(x):
    return { 1<x<4: 'several', 5<x<40: 'lots'}[1]


x = input("enter no. of monsters ")
print switch(x)

Hope this helps. :)