Question

Thanks for the answers, I have not used StackOverflow before so I was suprised by the number of answers and the speed of them – its fantastic.

I have not been through the answers properly yet, but thought I should add some information to the problem specification. See the image below.

I can’t post an image in this because i don’t have enough points but you can see an image
at http://journal.acquitane.com/2010-01-20/image003.jpg

This image may describe more closely what I’m trying to achieve. So you can see on the horizontal lines across the page are price points on the chart. Now where you get a clustering of lines within 0.5% of each, this is considered to be a good thing and why I want to identify those clusters automatically. You can see on the chart that there is a cluster at S2 & MR1, R2 & WPP1.

So everyday I produce these price points and then I can identify manually those that are within 0.5%. – but the purpose of this question is how to do it with a python routine.

I have reproduced the list again (see below) with labels. Just be aware that the list price points don’t match the price points in the image because they are from two different days.

[YR3,175.24,8]
[SR3,147.85,6]
[YR2,144.13,8]
[SR2,130.44,6]
[YR1,127.79,8]
[QR3,127.42,5]
[SR1,120.94,6]
[QR2,120.22,5]
[MR3,118.10,3]
[WR3,116.73,2]
[DR3,116.23,1]
[WR2,115.93,2]
[QR1,115.83,5]
[MR2,115.56,3]
[DR2,115.53,1]
[WR1,114.79,2]
[DR1,114.59,1]
[WPP,113.99,2]
[DPP,113.89,1]
[MR1,113.50,3]
[DS1,112.95,1]
[WS1,112.85,2]
[DS2,112.25,1]
[WS2,112.05,2]
[DS3,111.31,1]
[MPP,110.97,3]
[WS3,110.91,2]
[50MA,110.87,4]
[MS1,108.91,3]
[QPP,108.64,5]
[MS2,106.37,3]
[MS3,104.31,3]
[QS1,104.25,5]
[SPP,103.53,6]
[200MA,99.42,7]
[QS2,97.05,5]
[YPP,96.68,8]
[SS1,94.03,6]
[QS3,92.66,5]
[YS1,80.34,8]
[SS2,76.62,6]
[SS3,67.12,6]
[YS2,49.23,8]
[YS3,32.89,8]

I did make a mistake with the original list in that Group C is wrong and should not be included. Thanks for pointing that out.

Also the 0.5% is not fixed this value will change from day to day, but I have just used 0.5% as an example for spec’ing the problem.

Thanks Again.
Mark

PS. I will get cracking on checking the answers now now.

Hi:

I need to do some manipulation of stock prices. I have just started using Python, (but I think I would have trouble implementing this in any language). I’m looking for some ideas on how to implement this nicely in python.

Thanks
Mark

Problem:
I have a list of lists (FloorLevels (see below)) where the sublist has two items (stockprice, weight). I want to put the stockprices into groups when they are within 0.5% of each other. A groups strength will be determined by its total weight. For example:

Group-A
115.93,2
115.83,5
115.56,3
115.53,1
-------------
TotalWeight:12
-------------
Group-B
113.50,3
112.95,1
112.85,2
-------------
TotalWeight:6
-------------    

FloorLevels[   
[175.24,8]
[147.85,6]
[144.13,8]
[130.44,6]
[127.79,8]
[127.42,5]
[120.94,6]
[120.22,5]
[118.10,3]
[116.73,2]
[116.23,1]
[115.93,2]
[115.83,5]
[115.56,3]
[115.53,1]
[114.79,2]
[114.59,1]
[113.99,2]
[113.89,1]
[113.50,3]
[112.95,1]
[112.85,2]
[112.25,1]
[112.05,2]
[111.31,1]
[110.97,3]
[110.91,2]
[110.87,4]
[108.91,3]
[108.64,5]
[106.37,3]
[104.31,3]
[104.25,5]
[103.53,6]
[99.42,7]
[97.05,5]
[96.68,8]
[94.03,6]
[92.66,5]
[80.34,8]
[76.62,6]
[67.12,6]
[49.23,8]
[32.89,8]
]

Answer 1

I suggest a repeated use of k-means clustering — let’s call it KMC for short. KMC is a simple and powerful clustering algorithm… but it needs to “be told” how many clusters, k, you’re aiming for. You don’t know that in advance (if I understand you correctly) — you just want the smallest k such that no two items “clustered together” are more than X% apart from each other. So, start with k equal 1 — everything bunched together, no clustering pass needed;-) — and check the diameter of the cluster (a cluster’s “diameter”, from the use of the term in geometry, is the largest distance between any two members of a cluster).

If the diameter is > X%, set k += 1, perform KMC with k as the number of clusters, and repeat the check, iteratively.

In pseudo-code:

def markCluster(items, threshold):
    k = 1
    clusters = [items]
    maxdist = diameter(items)
    while maxdist > threshold:
        k += 1
        clusters = Kmc(items, k)
        maxdist = max(diameter(c) for c in clusters)
    return clusters

assuming of course we have suitable diameter and Kmc Python functions.

Does this sound like the kind of thing you want? If so, then we can move on to show you how to write diameter and Kmc (in pure Python if you have a relatively limited number of items to deal with, otherwise maybe by exploiting powerful third-party add-on frameworks such as numpy) — but it’s not worthwhile to go to such trouble if you actually want something pretty different, whence this check!-)