Looking for a Better Clustering Algorithm

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March 1st, 2013, 11:21 AM

rdkll2k

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Looking for a Better Clustering Algorithm

I have an issue that I've worked out in Python, but to me seems to be more complex than it should be and I'm hoping someone could help me devise a more efficient process.

I have a large set of data in the following format (diff, num1, num2):

SAMPLE:
[[1,5,6]
[1,6,7]
[1,9,10]
[15,15,30]
[16,15,31]
[16,30,46]
[20,42,62]
[20,62,82]
[20,70,90]
[20,82,102]
[32,90,122]
...]

I have those stored in a large python list and need to produce the following results:

{(5,6,7):1,(42,62,82,102):20...}

So I loop through my list of lists, which is sorted by the diff (which is the difference between num1 and num2) and I'm looking for chains where the difference between multiple series of numbers are the same and the first pairs num2 == the second pairs num1 (ie, using the sample above, with a diff of 1, 5,6 is the first pair and 6,7 is the second pair so the result is the tuple 5,6,7 which is the key to a dictionary with a value of 1). The pair 9,10 doesn't work because the chain is broken as there is no 8,9.

The algorithm I've devised is similar to this:

Code:

list1 = [[1,5,6],[1,6,7],[1,9,10],[15,15,30],[16,15,31],[16,30,46],[20,42,62],[20,62,82],[20,70,90],[20,82,102],[32,90,122]]

allCs = {}

for lCount in xrange(list1):
	strC = str(list1[lCount][0]) #used to keep track of the chain being created
	diff = list1[lCount][0]
	numCs = 0

	numCs, chains = processL(diff, lCount, numCs, strC)

	if numCs > 0:
		allCs[chains] = diff


print allCs



def processL(diff, sCount, numCs, strC):
	

	for lCount in xrange(sCount + 1, len(list1)):
		tDiff = list1[lCount][0] #used to keep track of the current lists diff

		if tDiff == diff:
			if list1[lCount][1] == list1[sCount][2]:
				strC += ',' + str(list1[lCount][1])

				numCs += 1

				numCs, chains = processL(diff, lCount, numCs, strC)

				if chains == strC:
					chains += ',' + str(list1[lCount][2])

				break

Any thoughts would be appreciated.

March 1st, 2013, 07:08 PM

b49P23TIvg

Contributing User

Join Date: Aug 2011

Posts: 3,389

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I take it your program doesn't work.

You need to define processL before you use it.
processL doesn't have a return statement, therefor it returns None . You need it to return two values.

Why are you using a string? Something's amiss, you want your output to look like
{(5,6,7):1,(42,62,82,102):20...}

Yet if I print a dictionary that has a key with type str you'll see quotation marks of some sort:

>>> print dict(a=4)
{'a': 4}

I think you really want the key to be a tuple of numbers. But that also displays a little bit differently than you show---see the spaces following comma?
>>> print {(1,2,3):1}
{(1, 2, 3): 1}

---------------
Now, how will we make this work?
Sorting by difference is a great first step.

Code:

def process(L):
    '''
        return a list of all chains
    '''
    keys = []
    while L:
        # I is a list of indexes in L forming a chain
        I = [0]                          # start at beginning of L

        # warning: if you expect len(L) to exceed 6 or so use a binary search
        # see the bisect module
        # The data is sorted and you can calculate the next value to find.
        for i in range(1,len(L)):          # search for the next value
            if L[I[-1]][2] == L[i][1]:
                I.append(i)

        key = L[I[0]][1:2]+[L[i][2] for i in I]

        keys.append(key)

        # remove the used values from L
        for i in reversed(I):
            del L[i]

    return keys

def main(data=[[1,5,6],[1,6,7],[1,9,10],[15,15,30],[16,15,31],[16,30,46],[20,42,62],[20,62,82],[20,70,90],[20,82,102],[32,90,122]]):
    allCs = {}
    i = 0
    while i < len(data):                  # look at all the data
        difference = data[i][0]           # get the next difference
        j = i
        while (j < len(data)) and (data[j][0] == difference): # find span with this same difference
            j += 1
        allCs.update({tuple(key) : difference for key in process(data[i:j])}) # return lists of valid subgroups
        i = j
    return allCs

print(main())

Then if you want to filter against keys of length 2 that's a separate step, or if you want to keep only the longest key per difference that again is another processing step.

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March 4th, 2013, 08:08 AM

rdkll2k

Registered User

Join Date: Feb 2013

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Quote:

Originally Posted by b49P23TIvg

I take it your program doesn't work.

Thanks for the help! This looks so much simpler than what I devised. I'll take a look today and see if I can implement your solution.

You are absolutely correct that what I posted doesn't work. What I'm really doing is a bit more complex than what I posted here. I simply wanted to give a down and dirty sample of the code that I felt was a bit more cumbersome than necessary. I should have made sure that it worked so we weren't focusing on the wrong stuff, but it appears you addressed the sample code bugs and the actual issue at the same time!

Thanks again!

March 8th, 2013, 08:37 AM

rdkll2k

Registered User

Join Date: Feb 2013

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Thanks for the help! The solution you provided worked exactly as I requested.

Unfortunately, I couldn't use it as written because of the complexity of things not mentioned in my initial question, but I was able to use your solution as a basis for what I ultimately needed.

Thanks!

Looking for a Better Clustering Algorithm

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