I have n
data points in some arbitrary space and I cluster them.
The result of my clustering algorithm is a partition represented by an int vector l
If you are going to relabel your partitions, as has been previously suggested, you will potentially need to search through n labels for each of the n items. I.e. the solutions are O(n^2).
Here is my idea: Scan through both lists simultaneously, maintaining a counter for each partition label in each list. You will need to be able to map partition labels to counter numbers. If the counters for each list do not match, then the partitions do not match. This would be O(n).
Here is a proof of concept in Python:
l_1 = [ 1, 1, 1, 0, 0, 2, 6 ]
l_2 = [ 2, 2, 2, 9, 9, 3, 1 ]
l_3 = [ 2, 2, 2, 9, 9, 3, 3 ]
d1 = dict()
d2 = dict()
c1 = []
c2 = []
# assume lists same length
match = True
for i in range(len(l_1)):
if l_1[i] not in d1:
x1 = len(c1)
d1[l_1[i]] = x1
c1.append(1)
else:
x1 = d1[l_1[i]]
c1[x1] += 1
if l_2[i] not in d2:
x2 = len(c2)
d2[l_2[i]] = x2
c2.append(1)
else:
x2 = d2[l_2[i]]
c2[x2] += 1
if x1 != x2 or c1[x1] != c2[x2]:
match = False
print "match = {}".format(match)