I don't know exactly how to describe what I'm looking for, but I will try to make some examples. Let's take three different data series:

  • Series A: 1,2,3,4,5,6,7,8,9,10,9,8,7,6,5,4,3,2,1
  • Series B: 1,2,1,2,1,2,1,2,1,2,1,2,1,2,1
  • Series C: 1,2,3,2,1,2,3,2,1,2,3,2,1

The change from point to point is:

  • Series A: +1,+1,+1,+1,+1,+1,+1,+1,...,-1,-1,-1,-1,-1,-1,-1,-1...
  • Series B: +1,-1,+1,-1,+1,-1,...
  • Series C: +1,+1,-1,-1,+1,+1,...

Or simplified in binary format 1 for +1 and 0 for -1:

  • Series A: 11111111111111...00000000000...
  • Series B: 10101010101010...
  • Series C: 11001100110011...

I'm looking for a function that returns the

  • highest Value for Series A (incrementing data is the same like the previous increment)
  • lowest Value for Series B (data change is always different than the previous)
  • something in between for Series C (data change sometimes same, sometimes different)
  • $\begingroup$ So by "highest value" you mean the longest run of increments in the sequence? Do you have a particular programming language in mind? $\endgroup$ – MånsT May 11 '12 at 11:06
  • $\begingroup$ The returning value should be optimally between 0 and 1, so the highest value would be 1 if there is always the same change in the same direction. My above series A would return something close to 1, say 0.97 since it changes from +1 to -1 sometime in between. Regarding the programming language I am fluent with Java, but also worked with Matlab in the past. For quick tests I run Excel/VBA or Libreoffice Calc. $\endgroup$ – Jens Roth May 11 '12 at 11:32
  • $\begingroup$ A real life example for the data would be temperature during seasons (almost steadily increasing during spring/summer and almost steadily decreasing during fall/winter). The opposite example might be stock market data or just random data. Maybe there is also a standard statistics function available which I don't know of. $\endgroup$ – Jens Roth May 11 '12 at 15:42

I would start with the autocorrelation of the +1/-1 sequence with a lag of 1. It has a range of -1 to 1, but you can convert easily transform it to 0 to 1. Here is a quick example in R:

(note: head(x,-1) drops the last value, tail(x,-1) drops the first)

> x1 <- c(1,1,1,1,1,1,-1,-1,-1,-1,-1)
> x2 <- c(1,-1,1,-1,1,-1,1,-1,1,-1,1)
> x3 <- c(1,1,-1,1,1,-1,1,1,-1,1,1)
> cor(head(x1,-1), tail(x1,-1))
[1] 0.8164966
> cor(head(x2,-1), tail(x2,-1))
[1] -1
> cor(head(x3,-1), tail(x3,-1))
[1] -0.4285714
  • $\begingroup$ Wow, autocorrelation it is :-) The results of -1 to 1 is just perfect. I just installed R and will play with my real data over the weekend and look if the results is what I expected. Thanks Aniko! (Oops, I just saw I need a reputation of 15 to vote up) $\endgroup$ – Jens Roth May 11 '12 at 22:12

Judging from your answer to my comment, it seems that you're looking for a function that gives you the proportion of changes in a sequence. In some sort of pseudocode, with mySequence being a vector of 0's and 1's, that could look like

   if mySequence[index] == mySequence[index-1] then count=count+1
return count/length(mySequence)

The result is 0 if the sequence is monotonely non-increasing and 1 if it is monotonely increasing.

  • $\begingroup$ Your function would give me the same value for all above series, since the starting and ending value are the same. Thus, we have the same number of increases/decreases and that results in something near 0.5. I thought about a real life example and added a comment above. $\endgroup$ – Jens Roth May 11 '12 at 15:42
  • $\begingroup$ You're right! I think I've fixed that with my edit though :) $\endgroup$ – MånsT May 11 '12 at 20:30
  • $\begingroup$ I also love your code and the speed I received the answer - was my first question on this site! It is very simple to rewrite to Java and I will look over the weekend how useful the results are with my data. Thanks MansT! (Oops, I just saw I need a reputation of 15 to vote up) $\endgroup$ – Jens Roth May 11 '12 at 22:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.