My list is like
l1 = [ {k1:v1} , {k2:v2}, {v1:k1} ]
Is there any better way to check if any dictionary in the list is having reverse pair?
views:
66answers:
2
+4
A:
I would suggest to transform the dictionaries in tuple and put the tuple in a set. And look in the set if the reverse tuple is in the set. That would have a complexity of O(n) instead of O(n^2).
Xavier Combelle
2010-06-25 08:31:24
Whoops, sorry for my earlier (deleted) comment, I misunderstood the question. +1
Tamás
2010-06-25 08:40:35
If you just want to know whether a reverse pair exists or not, then you can test it with `len(d) != len(set([frozenset(i) for i in d.items()]))`. I think this is what Xavier is suggesting.
Michael Dunn
2010-06-25 08:53:19
+1
A:
This code seems to work without loop:
k1 = 'k1'
k2 = 'k2'
v1 = 'v1'
v2 = 'v2'
l1 = [ {k1:v1} , {k2:v2}, {v1:k1} ]
kv = [e.items()[0] for e in l1]
print(kv)
vk = [(v, k) for (k, v) in kv]
print(vk)
result = [(k, v) for (k, v) in kv if (k, v) in vk]
print(result)
Michał Niklas
2010-06-25 08:47:01
I would suggest vk = set((v, k) for (k, v) in kv) to avoid an O(n^2) complexity
Xavier Combelle
2010-06-25 09:02:45
yess it worked I used like this, result = [(k, v) for (k, v) in kv if (v, k) in kv and k!=v]
Tumbleweed
2010-06-25 09:22:03