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?
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1What do you mean by 'reverse pair'? – Omnifarious Jun 25 '10 at 07:35
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@Omnifarious it means key as value and value as key @Siperd currently I'm looping over it – shahjapan Jun 25 '10 at 07:43
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http://stackoverflow.com/questions/3116249/how-can-i-check-if-there-exist-any-reverse-element-in-list-of-dict-without-loopin similar question may help you out – shahjapan Aug 05 '10 at 19:14
2 Answers
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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
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Whoops, sorry for my earlier (deleted) comment, I misunderstood the question. +1 – Tamás Jun 25 '10 at 08:40
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1If 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 Jun 25 '10 at 08:53
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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
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2I would suggest vk = set((v, k) for (k, v) in kv) to avoid an O(n^2) complexity – Xavier Combelle Jun 25 '10 at 09:02
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yess it worked I used like this, result = [(k, v) for (k, v) in kv if (v, k) in kv and k!=v] – shahjapan Jun 25 '10 at 09:22
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