There is no direct route. It's pretty easy with list comprehensions, though;
[k for k, v in d.iteritems() if v == desired_value]
If you need to do this occasionally and don't think it's worth while indexing it the other way as well, you could do something like:
class bidict(dict):
def key_with_value(self, value, default=None):
for k, v in self.iteritems():
if v == value:
return v
return default
def keys_with_value(self, value, default=None):
return [v for k, v in self.iteritems() if v == value]
Then d.key_with_value
would behave rather like d.get
, except the other way round.
You could also make a class which indexed it both ways automatically. Key and value would both need to be hashable, then. Here are three ways it could be implemented:
In two separate dicts, with the exposing some dict-like methods; you could perhaps do foo.by_key[key]
or foo.by_value[value]
. (No code given as it's more complicated and I'm lazy and I think this is suboptimal anyway.)
In a different structure, so that you could do d[key]
and d.inverse[value]
:
class bidict(dict):
def __init__(self, *args, **kwargs):
self.inverse = {}
super(bidict, self).__init__(key, value)
def __setitem__(self, key, value):
super(bidict, self).__setitem__(key, value)
self.inverse[value] = key
def __delitem__(self, key):
del self.inverse[self[key]]
super(bidict, self).__delitem__(key)
In the same structure, so that you could do d[key]
and d[value]
:
class bidict(dict):
def __setitem__(self, key, value):
super(bidict, self).__setitem__(key, value)
super(bidict, self).__setitem__(value, key)
def __delitem__(self, key):
super(bidict, self).__delitem__(self[key])
super(bidict, self).__delitem__(key)
(Notably absent from these implementations of a bidict
is the update
method which will be slightly more complex (but help(dict.update)
will indicate what you'd need to cover). Without update
, bidict({1:2})
wouldn't do what it was intended to, nor would d.update({1:2})
.)
Also consider whether some other data structure would be more appropriate.