Reputation: 807
I have a dictionary that maps 3tuple to 3tuple where key-tuples have some element in common
dict= { (a,b,c):(1,2,3),
(a,b,d):tuple1,
(a,e,b):tuple,
.
(f,g,h):tuple3,
.
.
.
tuple:tuple
}
now how can I find the values that match to (a,b,anyX) in a dictionary ie (1:2:3) and tuple1
this is computer generated and very large thus, it takes effort to determine anyX.
so, any good ways I can do this?
edit:partial matching of (f,g,*),(f, *,g) to tuple3 will also be helpful but not necessary.
Upvotes: 11
Views: 10348
Reputation: 164773
@AshwiniChaudhary's solution can be trivially adapted for an object-oriented solution. You can subclass dict
and add a method:
class tup_dict(dict):
def getitems_fromtup(self, key):
for k, v in self.items():
if all(k1 == k2 or k2 is None for k1, k2 in zip(k, key)):
yield v
d = tup_dict({("foo", 4 , "q"): 9,
("foo", 4 , "r"): 8,
("foo", 8 , "s"): 7,
("bar", 15, "t"): 6,
("bar", 16, "u"): 5,
("baz", 23, "v"): 4})
res = list(d.getitems_fromtup(("foo", 4, None))) # [9, 8]
Upvotes: 1
Reputation: 251061
Lets say if you're passing None
for the missing keys then you can use all
and zip
:
>>> from itertools import permutations
>>> import random
#create a sample dict
>>> dic = {k:random.randint(1, 1000) for k in permutations('abcde', 3)}
def partial_match(key, d):
for k, v in d.iteritems():
if all(k1 == k2 or k2 is None for k1, k2 in zip(k, key)):
yield v
...
>>> list(partial_match(('a', 'b', None), dic))
[541, 470, 734]
>>> list(partial_match(('a', None, 'b'), dic))
[460, 966, 45]
#Answer check
>>> [dic[('a', 'b', x)] for x in 'cde']
[541, 734, 470]
>>> [dic[('a', x, 'b')] for x in 'cde']
[966, 460, 45]
Upvotes: 12
Reputation: 76234
You could reconstruct your dictionary into a triply nested dict.
dict= { ("foo", 4 , "q"): 9,
("foo", 4 , "r"): 8,
("foo", 8 , "s"): 7,
("bar", 15, "t"): 6,
("bar", 16, "u"): 5,
("baz", 23, "v"): 4
}
d = {}
for (a,b,c), value in dict.iteritems():
if a not in d:
d[a] = {}
if b not in d[a]:
d[a][b] = {}
d[a][b][c] = value
Here, d
is equivalent to:
d = {
"foo": {
4:{
"q": 9,
"r": 8
},
8:{
"s": 7
}
},
"bar":{
15:{
"t": 6
}
16:{
"u": 5
}
},
"baz":{
23{
"v": 4
}
}
}
Now you can easily iterate through the possible third keys, given the first and second.
#find all keys whose first two elements are "foo" and 4
a = "foo"
b = 4
for c in d[a][b].iterkeys():
print c
Result:
q
r
This only works for matching the third key. For instance, you wouldn't be able to find all second keys, given the third and the first.
Upvotes: 4
Reputation: 6468
There might be other ways, but assuming you just need to do a single search (in other words there might be ways to build better data structures for repeated searching): (Note that this handles arbitrary lengthed tuple's with the '*' in multiple possible locations)
def match(tup,target):
if len(tup) != len(target):
return False
for i in xrange(len(tup)):
if target[i] != "*" and tup[i] != target[i]:
return False
return True
def get_tuples(mydict,target):
keys = filter(lambda x: match(x,target),mydict.keys())
return [mydict[key] for key in keys]
#example:
dict= { (1,3,5):(1,2,3),
(1,3,6):(1,5,7),
(1,2,5):(1,4,5),
}
print get_tuples(dict,(1,3,'*'))
.
Upvotes: 1