I need to partition an input vector v into smaller series of subvectors each containing the whole array of a periodic pattern given by a local sum of successive elements in the array.

I need to partition an input vector v into smaller series of subvectors each containing the whole array of a periodic pattern given by a local sum of successive elements in the array. I am recording some oscillations in a biological system. The vectors that I analyze represent an array of successive timings between followed events and under some stimulation they follow a certain periodicity . The pattern has two main characteristics: 1- the oscillations are periodic according to period reflected in a constant sum of successive elements of the array that repeats itself. The period varies from one vector to another and cannot be predicted ahead. Hence it is not the same for all vectors 2- many elements in the array are iterated because the system stabilizes at a constant oscillation pattern from which we can easily see the period. However the subvector of the whole pattern is more extended starting with the very early elements whose periodicity are respected by the sum of consecutive elements even if they do not follow exactly the iteration. For example this is a typical vector

v = [14  97  38  41  22   6  27  55  55  22  33  22   6  27  22  33  55   1  35  19  22  33  22  33  22  33  22  33  22  33  55  12  18  23  67  92  12]

in this case the periodicity of oscillation is 55. Note that

22+33 = 22+6+27 = 1+35+19 = 55

I need to extract the subvector with all elements for which the sum of consecutive elements is 55 which is a

a = [22  6  27  55  55  22  33  22   6  27  22  33  55   1  35  19  22  33  22  33  22  33  22  33  22  33  55]

and the starting and ending indices of the subvector from the original vector

i=[5  31]

I have no words to thank you for any suggestion on how to tackle this!

Answer 1

[Stephen23]

v = [14,97,38,41,22,6,27,55,55,22,33,22,6,27,22,33,55,1,35,19,22,33,22,33,22,33,22,33,22,33,55,12,18,23,67,92,12];
N = 55;
idx = v==N;
tmp = v;
for k = 2:numel(v);
    tmp = v + [0,tmp(1:end-1)];
    idy = find(tmp==N);
    idz = bsxfun(@minus,idy(:),k-1:-1:0).';
    idx(idz) = true;
end
out = v(idx);

and check the output:

% test:
a = [22,6,27,55,55,22,33,22,6,27,22,33,55,1,35,19,22,33,22,33,22,33,22,33,22,33,55]
isequal(a,out)

and get the indices:

find(idx,1,'first')
find(idx,1,'last')

[Radu Mihail]

Dear Stephen Thank you for your nice, elegant and very short line of code.I confess it is a lot better than what I was struggling with. Unfortunately the code has to find N (the periodicity) which can not be stated as a known parameter at the beginning of analysis. It varies from one vector to another and it is a daunting task for me (up to this point) how to make my program infer itself a partial sum (N)among very few terms which periodically repeats in one array of a vector but is different from one vector to another.

[Stephen23]

Identifying N will be easy once you specify it for us. What properties does it have? How can we test for it? When you tell us this then we can write a test for it.

Clearly the periodicity is not the mode of v:

>> mode(v)
ans =
    22

I also calculated all sequential sums, and the most common are shown below:

mat = NaN(numel(v));
mat(1,:) = v;
for k = 2:numel(v)
    mat(k,k:end) = mat(k-1,k:end) + v(k:end);
end

>> [N,B] = hist(mat(:),unique(mat(:)));
>> X = find(max(N)==N)
ans =
    38    63   127
>> B(X) % the most common sums (17 occurrences each)
ans =
    66
   132
   330

So how do you want to identify 55 from this data?

[Guillaume]

Stephen, I think you've made a mistake in your mat construction. It should be

mat(k, 1:end-k+1) = mat(k-1, 1:end-k+1) + v(k:end)

You'll find that the mode of that matrix is indeed 55.

[Stephen23]

True. Thank you for the correction, Guillaume. Then this will identify N (for that data set):

v = [14,97,38,41,22,6,27,55,55,22,33,22,6,27,22,33,55,1,35,19,22,33,22,33,22,33,22,33,22,33,55,12,18,23,67,92,12];
mat = NaN(numel(v));
mat(1,:) = v;
for k = 2:numel(v)
    mat(k, 1:end-k+1) = mat(k-1, 1:end-k+1) + v(k:end);
end
N = mode(mat(:));

You might want to check this out too:

hist(mat(:),unique(mat(:)))

which shows how many times each sum occurs in mat.

[Radu Mihail]

This is great indeed! the mode(mat)was the answer. I thank you very much Stephen! I runned a few good recordings through your code and clearly does the job greatly. I also thank Guillaume for his prompt remark!!

[Radu Mihail]

Dear Stephen A quick subsidiary question. As it is now your code writes chunck vectors which as whole sum all the terms of selected subvector to yield multiples of N periodicity(justly found as mode(mat)).However for me only sums of successive adjacent elements in the array have to be N or multiples of it.Now to filter them I use cumsum and select only those results that follow successive cumsum terms multiples of N But I realize I discard a lot of good data For instance your code now selects vectors of the kind for N=7 [1 7 6 7 7 2 7 6 5 7 1] but this is not good for successive elements do not sum to 7 A good one is only if it is of the kind [1 6 7 7 2 5 7 7 3 4 7]All succession of adjacent terms have to sum to N or multiples of it Can you give me some hints how to change the code to filter directly only these from the results Thank you again!

[Stephen23]

Radu Mihail: please give me some exact and complete data to test with. For example:

N = 7;
v = [1 7 6 7 7 2 7 6 5 7 1];
out = ??? % I need to know this too

Please give me three or four examples like this, so that I can test my code and make changes as required.

If you want to make some changes yourself then take a look at the variable idx: this is where you can easily identify if the sums are sequential or are in blocks. Identifying "runs of ones" is a common task that has been solved many times on this forum (and others).

[Stephen23]

@Radu Mihail: sorry, but your one hundred and twenty six line communication via my user page does not help. It includes three copies of my answer code, screeds of unformatted code, and lots of your random comments. Please understand: I am happy to help you, but you really need to consider how you present your information. What you sent me is basically unreadable without me spending two hours trying to decipher it. Please be clear and precise with what you need. 126 lines of text does not help you.

I gave you a very precise request in my last comment: three or four examples like this:

N = 7;
v = [1 7 6 7 7 2 7 6 5 7 1];
out = ??? % I need to know this too

Please make a comment giving these examples. If you also want to explain your requirements, then please be short and precise, something like this:

Return the largest group of consecutive elements meeting the requirements:

• the group must contain at least 10 elements.
• sequential values may add up to the periodicity.
• these sequential values must be ascending in value.
• etc

Also note that you have introduced a new condition which was not in your original question: that multiples of the periodicity are also required. Is there any limit to this? Are any multiples allowed? If so, then 2 is likely to be the most common periodicity.

[Radu Mihail]

Sorry Stephen. I apologize for my overzealous answer. I will give the following 3 examples of soft problems: Ex1 and Ex2 have sum of consecutive terms differ from N

Ex1

N = 14
v = [4,2,1,5,7,1,1,6,14,7,14,7,14,7,7,7,7,1,6,7,1,6,7,7,7,7,7,7,7,1,6,7,1,6,7,1,6,7]
out = [1,5,7,1,1,14,14,14,7,7,7,7,1,6,7,1,6,7,7,7,7,7,7,7,1,6,7,1,6,7,1,6,7]

The first a=[1,5,7,1,1] indices [3,7] Here 1+5+7+1+1=15,which is wrong ,then the next subvector is 14=N

Ex 2

N = 81
v = [1,17,36,6,34,6,35,6,38,12,11,14,13,43,11,14,13,31,12,11,14,56,11,14,13,14,17,23,24,17,70,11,9,61,11,9,72,3,61,6,11]
out [6,34,6,35,6,38,12,11,14,13,43,11,14,13,31,12,11,14,56,11,14,13,14,17,23,24,17,70,11,9,61,11,9,72,3,61,6,11]

The first subvector, a is good for its successive terms all equal with N = 81

a = [ 6,34,6,35,6,38,12,11,14,13,43,11,14,13,31,12,11,14,56,11,14] 
indices[4,24]

for the sum of consecutive adjacent terms equal

N=81 = 6+34+6+35 = 6+38+12+11 +14 = 13+43 +11+ 14 = 56+11+14=81

BUT, THE NEXT subvector b

 b = [13,14,17,23,24,17,70,11,9,61,11,9,72,3,61,6,11]

has problems

13+14+17+23 = 67<81<13+14+17+23+24 = 91

Ex3 there are vectors that the program cannot process

N = 244
v =[2,3,15,15,117,15,15,62,2,3,15,15,117,15,15,62,2,3,15,15,117,15,15,62,2,3,15,15,102,15,15,15,62,2,3]
out Subscript indices must either be real positive integers or logicals.

[Stephen23]

That you, that is much better. I will look at it soon. Can you please also provide (for these three examples) the expected output (i.e. the correct vectors that you need: then I can test and compare).

[Radu Mihail]

Expected vectors

E1

a is bad it should have been

a= [1,5,7,1] indices[3,6]

b=[14], i=9

c=[14], i=11

d=[14 7 7 7 7 1 6 7 1 6 7 7 7 7 7 7 7 1 6 7 1 6 7 1 6 7] indices[13, 38]

Ex 2 subvector a is good

a = [ 6,34,6,35,6,38,12,11,14,13,43,11,14,13,31,12,11,14,56,11,14] indices[4,24]

and the other subvector b should be

b= [17,23,24,17,70,11,9,61,11,9,72,3,61,6,11] indices[27,41]

Because N = 81 = 17 + 23 + 24 +17= 70 + 11 = 9 + 61 + 11= 9 + 72 = 3 + 61 + 6 + 11

Ex 3 a single subvector

a= =[2,3,15,15,117,15,15,62,2,3,15,15,117,15,15,62,2,3,15,15,117,15,15,62,2,3,15,15,102,15,15,15,62] indices [1,33]

[Stephen23]

@Radu Mihail: thank you, that is great. I need to get some sleep, and then I will look at it again. Please be patient :)

[Stephen23]

I have attached to this comment one function (named split) and one test script. I believe that the function does what you want, based on your examples.

The test script runs every example you have given me here, and compare the functions output with the expected output. They all pass :)

[Radu Mihail]

I have no words to thank you Stephen! it works beautifully!!God bless you!