# summing tied (duplicate) numbers in a series

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Vincent Odongo on 8 Apr 2012
Greetings to all, I have a time series data e.g. [4 4 6 7 9 9 9 10 10 13 17 17 17] which is long. I need to identify tied values such that t1=3(three untied values (6, 7, 13)), t2=2 (two ties of extent two (4, 10)), t3=2 (two ties of extent three (9, 17)).
After performing this, I need to sum all the ties in the series using the following equation... ti[(ti-1)(2ti+5)] where ti denotes the magnitude of the tie.
Vincent
Sean de Wolski on 10 Apr 2012
Could you provide the output you expect for the above matrix?
It looks like (histc()||unique)+accumarray() will have a blast with this.

Thomas on 10 Apr 2012
this is a start:
a=[4 4 6 7 9 9 9 10 10 13 17 17 17];
b=unique(a); % this shows unique values
c=find(diff(a)~=0) % finds the differences in the values
d=[0 c length(a)]; % making new matrix to get the values
outputs=sortrows([b' diff(d)'],2)

Andrei Bobrov on 10 Apr 2012
t = [4 4 6 7 9 9 9 10 10 13 17 17 17]
[a n n] = unique(t)
m = histc(n,1:max(n))
tout = accumarray(m',a',[],@(x){x})