使用同一词典中的另一个值声明词典？

Question

I'm using python trying to basically do this:

myDict = {"key1" : 1, "key2" : myDict["key1"]+1}

...if you catch my drift. Possible without using multiple statements?

EDIT: Also, if anyone could tell me a better way to state this question more clearly that would be cool. I don't really know how to word what I'm asking.

EDIT2: Seems to be some confusion - yes, it's more complex than just "key2":1+1, and what I'm doing is mostly for code readability as it will get messy if I have to 2-line it.

Here's a bit more accurate code sample of what I'm trying to do...though it's still not nearly as complex as it gets :P

lvls={easy:  {mapsize:(10,10), winPos:(mapsize[0]-1,mapsize[1]-1)},
      medium:{mapsize:(15,15), winPos:(mapsize[0]-RANDOMINT,mapsize[1]-1)},
      hard:  {mapsize:(20,20), winPos:(mapsize[0]-RANDOMINT,mapsize[1]-RANDOMINT)}
     }

Answer 1

No, this isn't possible in general without using multiple statements.

In this particular case, you could get around it in a hacky way. For example:

myDict = dict(zip(("key1", "key2"), itertools.count(1))

However, that will only work when you want to specify a single start value and everything else will be sequential, and presumably that's not general enough for what you want.

If you're doing this kind of thing a lot, you could wrap those multiple statements up in some suitably-general function, so that each particular instance is just a single expression. For example:

def make_funky_dict(*args):
    myDict = {}
    for key, value in zip(*[iter(a)]*2):
        if value in myDict:
            value = myDict[value] + 1
        myDict[key] = value
    return myDict

myDict = make_funky_dict("key1", 1, "key2", "key1")

But really, there's no good reason not to use multiple statements here, and it will probably be a lot clearer, so… I'd just do it that way.

Answer 2

It's not possible without using multiple statements, at least not using some of the methods from your problem statement. But here's something, using dict comprehension:

>>> myDict = {"key" + str(key): value for (key, value) in enumerate(range(7))}
>>> myDict
{'key0': 0,
 'key1': 1,
 'key2': 2,
 'key3': 3,
 'key4': 4,
 'key5': 5,
 'key6': 6}

Of course those aren't in order, but they're all there.

Answer 3

The only variable you are trying to use is an integer. How about a nice function:

def makelevel(size,jitterfunc=lambda:0):
    return {'mapsize':(size,size), 'winPos':(size-1+jitterfunc(),size-1+jitterfunc())}

lvls = {hardness,makelevel(size) for hardness, size in [('easy',10),('medium',15), ('hard',20)]}

Of course, this function looks a bit like a constructor. Maybe you should be using objects?

Answer 4

If you want a dict that will allow you to to have values that are evaluated on demand you can do something like this:

class myDictType(dict):                                                                
  def __getitem__(self, key):                                                      
    retval = dict.__getitem__(self,key)                                            
    if type(retval) == type(lambda: 1):                                            
      return retval()                                                              
    return retval                                                                  

myDict = myDictType()                                                                     

myDict['bar'] = lambda: foo['foo'] + 1                                                

myDict['foo'] = 1                                                                     
print myDict['bar']   #This'll print a 2                                                       
myDict['foo'] = 2                                                                   
print myDict['bar']   #This'll print a 3                                                     

This overrides __getitem__ in the dictionary to return whatever is stored in it (like a normal dictionary,) unless what is stored there is a lambda. If the value is a lambda, it instead evaluates it and returns the result.