CODEWITHSUNDEEP
pythondictionary
sholsapp
sholsapp

Reputation: 16080

What do I do when I need a self referential dictionary?

I'm new to Python, and am sort of surprised I cannot do this. 

dictionary = {
    'a' : '123',
    'b' : dictionary['a'] + '456'
}

I'm wondering what the Pythonic way to correctly do this in my script, because I feel like I'm not the only one that has tried to do this.

EDIT: Enough people were wondering what I'm doing with this, so here are more details for my use cases. Lets say I want to keep dictionary objects to hold file system paths. The paths are relative to other values in the dictionary. For example, this is what one of my dictionaries may look like.

dictionary = {
    'user': 'sholsapp',
    'home': '/home/' + dictionary['user']
}

It is important that at any point in time I may change dictionary['user'] and have all of the dictionaries values reflect the change. Again, this is an example of what I'm using it for, so I hope that it conveys my goal.

From my own research I think I will need to implement a class to do this.

Upvotes: 71

Views: 29227

Answers (11)

Davis Herring
Davis Herring

Reputation: 40023

Class statements can be used to define dictionaries with string keys in multiple steps like this:

def data(name,bases,values): return values
class dictionary(metaclass=data):
  a='123'
  b=a+'456'

The “metaclass” just uses the locals of the class statement as the value of the class definition (which is assigned here to dictionary). This approach avoids relying on dict.__getitem__ in a way that will be inconsistent with other access via dict.get, dict.popitem, etc.

Like several other answers here, though, it does not support the “spreadsheet” use case of redefining values implicitly when other values change. (That sort of object isn’t a dictionary of values at all, but rather of nullary functions that can be evaluated when desired.)

Upvotes: 0

Rivers
Rivers

Reputation: 1923

The jsbueno answer in Python 3 :

class MyDict(dict):
    def __getitem__(self, item):
        return dict.__getitem__(self, item).format(self)

dictionary = MyDict({
    'user' : 'gnucom',
    'home' : '/home/{0[user]}',
    'bin' : '{0[home]}/bin' 
})

print(dictionary["home"])
print(dictionary["bin"])

Here we use the python 3 string formatting with curly braces {} and the .format() method.

Documentation : https://docs.python.org/3/library/string.html

Upvotes: 2

CMCDragonkai
CMCDragonkai

Reputation: 6382

This is very easy in a lazily evaluated language (haskell).

Since Python is strictly evaluated, we can do a little trick to turn things lazy:

Y = lambda f: (lambda x: x(x))(lambda y: f(lambda *args: y(y)(*args)))

d1 = lambda self: lambda: {
  'a': lambda: 3,
  'b': lambda: self()['a']()
}

# fix the d1, and evaluate it
d2 = Y(d1)()

# to get a
d2['a']() # 3

# to get b
d2['b']() # 3

Syntax wise this is not very nice. That's because of us needing to explicitly construct lazy expressions with lambda: ... and explicitly evaluate lazy expression with ...(). It's the opposite problem in lazy languages needing strictness annotations, here in Python we end up needing lazy annotations.

I think with some more meta-programmming and some more tricks, the above could be made more easy to use.

Note that this is basically how let-rec works in some functional languages.

Upvotes: 1

om-nom-nom
om-nom-nom

Reputation: 62855

If you, just like me wandering how to make @jsbueno snippet work with {} style substitutions, below is the example code (which is probably not much efficient though): 

import string

class MyDict(dict):
    def __init__(self, *args, **kw):
        super(MyDict,self).__init__(*args, **kw)
        self.itemlist = super(MyDict,self).keys()
        self.fmt = string.Formatter() 

    def __getitem__(self, item):
        return self.fmt.vformat(dict.__getitem__(self, item), {}, self)


xs = MyDict({
    'user' : 'gnucom',
    'home' : '/home/{user}',
    'bin' : '{home}/bin'
})


>>> xs["home"]
'/home/gnucom'
>>> xs["bin"]
'/home/gnucom/bin'

I tried to make it work with the simple replacement of % self with .format(**self) but it turns out it wouldn't work for nested expressions (like 'bin' in above listing, which references 'home', which has it's own reference to 'user') because of the evaluation order (** expansion is done before actual format call and it's not delayed like in original % version).

Upvotes: 4

intuited
intuited

Reputation: 24084

As sort of an extended version of @Tony's answer, you could build a dictionary subclass that calls its values if they are callables:

class CallingDict(dict):
    """Returns the result rather than the value of referenced callables.

    >>> cd = CallingDict({1: "One", 2: "Two", 'fsh': "Fish",
    ...                   "rhyme": lambda d: ' '.join((d[1], d['fsh'],
    ...                                                d[2], d['fsh']))})
    >>> cd["rhyme"]
    'One Fish Two Fish'
    >>> cd[1] = 'Red'
    >>> cd[2] = 'Blue'
    >>> cd["rhyme"]
    'Red Fish Blue Fish'
    """
    def __getitem__(self, item):
        it = super(CallingDict, self).__getitem__(item)
        if callable(it):
            return it(self)
        else:
            return it

Of course this would only be usable if you're not actually going to store callables as values. If you need to be able to do that, you could wrap the lambda declaration in a function that adds some attribute to the resulting lambda, and check for it in CallingDict.__getitem__, but at that point it's getting complex, and long-winded, enough that it might just be easier to use a class for your data in the first place.

Upvotes: 2

aaronasterling
aaronasterling

Reputation: 71064

This is an interesting problem. It seems like Greg has a good solution. But that's no fun ;)

jsbueno as a very elegant solution but that only applies to strings (as you requested).

The trick to a 'general' self referential dictionary is to use a surrogate object. It takes a few (understatement) lines of code to pull off, but the usage is about what you want:

S = SurrogateDict(AdditionSurrogateDictEntry)
d = S.resolve({'user': 'gnucom',
               'home': '/home/' + S['user'],
               'config': [S['home'] + '/.emacs', S['home'] + '/.bashrc']})

The code to make that happen is not nearly so short. It lives in three classes:

import abc

class SurrogateDictEntry(object):
    __metaclass__ = abc.ABCMeta
    def __init__(self, key):
        """record the key on the real dictionary that this will resolve to a 
           value for
        """
        self.key = key

    def resolve(self, d):
        """ return the actual value"""
        if hasattr(self, 'op'):
            # any operation done on self will store it's name in self.op. 
            # if this is set, resolve it by calling the appropriate method 
            # now that we can get self.value out of d
            self.value = d[self.key]
            return getattr(self, self.op + 'resolve__')()
        else:
            return d[self.key]

    @staticmethod
    def make_op(opname):
        """A convience class. This will be the form of all op hooks for subclasses
           The actual logic for the op is in __op__resolve__ (e.g. __add__resolve__)
        """
        def op(self, other):
            self.stored_value = other
            self.op = opname
            return self
        op.__name__ = opname
        return op

Next, comes the concrete class. simple enough.

class AdditionSurrogateDictEntry(SurrogateDictEntry):

    __add__ = SurrogateDictEntry.make_op('__add__')
    __radd__ = SurrogateDictEntry.make_op('__radd__')

    def __add__resolve__(self):
        return self.value + self.stored_value 

    def __radd__resolve__(self):
        return self.stored_value + self.value

Here's the final class

class SurrogateDict(object):
    def __init__(self, EntryClass):
        self.EntryClass = EntryClass

    def __getitem__(self, key):
        """record the key and return""" 
        return self.EntryClass(key)

    @staticmethod
    def resolve(d):
        """I eat generators resolve self references"""
        stack = [d]
        while stack:
            cur = stack.pop()
            # This just tries to set it to an appropriate iterable
            it = xrange(len(cur)) if not hasattr(cur, 'keys') else cur.keys()
            for key in it:
                # sorry for being a duche. Just register your class with
                # SurrogateDictEntry and you can pass whatever.
                while isinstance(cur[key], SurrogateDictEntry):
                    cur[key] = cur[key].resolve(d)
                # I'm just going to check for iter but you can add other
                # checks here for items that we should loop over. 
                if hasattr(cur[key], '__iter__'):
                    stack.append(cur[key])
        return d

In response to gnucoms's question about why I named the classes the way that I did.

The word surrogate is generally associated with standing in for something else so it seemed appropriate because that's what the SurrogateDict class does: an instance replaces the 'self' references in a dictionary literal. That being said, (other than just being straight up stupid sometimes) naming is probably one of the hardest things for me about coding. If you (or anyone else) can suggest a better name, I'm all ears.

I'll provide a brief explanation. Throughout S refers to an instance of SurrogateDict and d is the real dictionary.

  1. A reference S[key] triggers S.__getitem__ and SurrogateDictEntry(key) to be placed in the d.

  2. When S[key] = SurrogateDictEntry(key) is constructed, it stores key. This will be the key into d for the value that this entry of SurrogateDictEntry is acting as a surrogate for.

  3. After S[key] is returned, it is either entered into the d, or has some operation(s) performed on it. If an operation is performed on it, it triggers the relative __op__ method which simple stores the value that the operation is performed on and the name of the operation and then returns itself. We can't actually resolve the operation because d hasn't been constructed yet.

  4. After d is constructed, it is passed to S.resolve. This method loops through d finding any instances of SurrogateDictEntry and replacing them with the result of calling the resolve method on the instance.

  5. The SurrogateDictEntry.resolve method receives the now constructed d as an argument and can use the value of key that it stored at construction time to get the value that it is acting as a surrogate for. If an operation was performed on it after creation, the op attribute will have been set with the name of the operation that was performed. If the class has a __op__ method, then it has a __op__resolve__ method with the actual logic that would normally be in the __op__ method. So now we have the logic (self.op__resolve) and all necessary values (self.value, self.stored_value) to finally get the real value of d[key]. So we return that which step 4 places in the dictionary.

  6. finally the SurrogateDict.resolve method returns d with all references resolved.

That'a a rough sketch. If you have any more questions, feel free to ask.

Upvotes: 6

jsbueno
jsbueno

Reputation: 110666

No fear of creating new classes - You can take advantage of Python's string formating capabilities and simply do:

class MyDict(dict):
   def __getitem__(self, item):
       return dict.__getitem__(self, item) % self

dictionary = MyDict({

    'user' : 'gnucom',
    'home' : '/home/%(user)s',
    'bin' : '%(home)s/bin' 
})


print dictionary["home"]
print dictionary["bin"]

Upvotes: 62

Greg
Greg

Reputation: 47164

What you're describing in your edit is how an INI config file works. Python does have a built in library called ConfigParser which should work for what you're describing.

Upvotes: 7

Jochen Ritzel
Jochen Ritzel

Reputation: 107736

Write a class, maybe something with properties:

class PathInfo(object):
    def __init__(self, user):
        self.user = user

    @property
    def home(self):
        return '/home/' + self.user

p = PathInfo('thc')
print p.home # /home/thc

Upvotes: 3

Tony Veijalainen
Tony Veijalainen

Reputation: 5565

Nearest I came up without doing object:

dictionary = {
    'user' : 'gnucom',
    'home' : lambda:'/home/'+dictionary['user'] 
}

print dictionary['home']()
dictionary['user']='tony'
print dictionary['home']()

Upvotes: 24

Daniel DiPaolo
Daniel DiPaolo

Reputation: 56418

>>> dictionary = {
... 'a':'123'
... }
>>> dictionary['b'] = dictionary['a'] + '456'
>>> dictionary
{'a': '123', 'b': '123456'}

It works fine but when you're trying to use dictionary it hasn't been defined yet (because it has to evaluate that literal dictionary first).

But be careful because this assigns to the key of 'b' the value referenced by the key of 'a' at the time of assignment and is not going to do the lookup every time. If that is what you are looking for, it's possible but with more work.

Upvotes: 8

Related Questions