Basic question: how to find range of index values

I want to find the location (range of N index values) over which a variable (mydata) holds a specific value (X)

The variable (mydata) contains many occurrences of X before the occurrence of a plateau - where the value of X is constant for N data points.

ELABORATION:

My problem is a little more complex.

mydata = [1 1 2 3 4 5 6 1 4 4 4 4 4 4 4 1 45 67 8 9 4 4 4 4 4 1 36 2 4 4 4 3 1 1 18 98 99]

I want to find the index values for the 7 (N) consecutive occurrences of '4'. I do not want to find the other occurrences of '4' i.e. one occurrence before the sequence of 7 and later one occurrence of a sequence of 5.

Answer 1

[Etienne O'Brien]

Thank you everyone.

Matt's solution worked for me

S = findstr(mydata,4*ones(1,3));

However, my computer is making heavy weather of this calculation. It's taking several seconds to return the answer as it returns the starting values of sub-occurrences. Unfortunately I realise now that I don't know the precise length of the sequence. For example it could be 1100 or 1500 data points long but I do know that there will be at least 1000. Is it possible to adapt Matt's function to return the first occurrence only?

Thank you.

[Matt Fig]

Several seconds? What size is the data you are looking at? I tested this with data (mydata) which is 1-by-500000 and it took less than a tenth of a second.... For example:

mydata = round((rand(1,500000)>.15)*4);

N = 6; % Look for 6 4s in a row.

tic

M = findstr(mydata,4*ones(1,N));

toc

Elapsed time is 0.005884 seconds.

[Etienne O'Brien]

Thanks Matt,

I'm not quite sure what the reason for this delay was. It has gone now. Thanks for your help with this issue. I have re-posted a somewhat related data question and would be grateful if you have time to look at it also.

[Andrew Newell]

Etienne, I hope that you selected your own answer by mistake. Matt's was clearly the correct choice.

[Etienne O'Brien]

Yes, you are correct. I am sorry.

Answer 2

[Matt Fig]

mydata = [1 1 2 3 4 5 6 1 4 4 4 4 4 4 4 1 45 67 8 9 4 4 4 4 4 1 36 2 4 4 4 3 1 1 18 98 99];
findstr(mydata,4*ones(1,7))

This finds the starting index of a sequence of 7 4s in a row. Change the 4 to a 5 to find a sequence of 7 5s in a row. Change the 7 to a 3 to find a sequence of 3 4s in a row, etc.

Note that the above will return the starting indices of all occurrences of, say, 3 4s in a row - even where those occurrences are sub-occurrences. To limit the search to only those occurrences which are exactly the length desired, simply make one more call to the function:

S = findstr(mydata,4*ones(1,3));
S = S(findstr([0 diff(S)==1 0],[0 0]))

[Andrei Bobrov]

+1

Answer 3

[Laura Proctor]

If you wish to find the corresponding index values where your variable mydata is equal to X, then you can use the find command like this:

iVals = find(mydata==X);

Answer 4

[Chirag Gupta]

I am not exactly clear about the question, but may be this is a starting point:

mydata = [1 1 2 3 4 5 6 1  1 45 67 8 9 1 36 2 3 1 1 18 98 99];
c = find(mydata==1)

You would probably to refine the condition inside the find, especially if your are dealing with doubles

Answer 5

[Laura Proctor]

The following will find the maximum run and return the index values in that run to a variable named reqVals.

I'm sure there's a tidier way to do this, but off the top of my head:

mydata = [1 1 2 3 4 5 6 1 4 4 4 4 4 4 4 1 45 67 8 9 4 4 4 4 4 1 36 2 4 4 4 3 1 1 18 98 99];
iVals = find(mydata==4);
sVals = iVals([true diff(iVals)~=1]);
eVals = iVals([diff(iVals)~=1 true]);
lVals = eVals - sVals + 1;
[num idx] = max(lVals)
reqVals = sVals(idx):eVals(idx)

Answer 6

[Andrei Bobrov]

md = mydata==4;
vf = find([true diff(md)~=0]);
idx = [vf; vf(2:end)-1 length(md)];
indexsevenfour = idx(:,diff(idx)+1== 7)';

MORE variant

idx = bsxfun(@plus,1:7,(1:length(mydata)-7)'-1);
idx(all(bsxfun(@eq,mydata(idx),4*ones(1,7)),2))

Answer 7

[Etienne O'Brien]

Thanks Laura and Andrei,

I've followed up Andrei's as it is the shorter.

I'm sorry, I'm a little lost though - I'm almost a complete beginner.

(1) I get this error message: ??? Error using ==> horzcat CAT arguments dimensions are not consistent.

(2) Could you correct me, if I'm wrong here in my interpretation.

md = mydata==4; % This finds the index values of all occurrences of '4' in the data set

vf = find([true diff(md)~=0]); % What does this do?

idx = [vf; vf(2:end)-1 length(md)]; % What does this do?

indexsevenfour = idx(:,diff(idx)+1== 7)'; % Does this provide the index values for the seven consecutive values of '4'

Thank you.

[Matt Fig]

See my solution above.

[Laura Proctor]

I would go with Matt's response if you want to find sequence of (7) 4's in a row as that is the simplest way to find the index value of the starting point.

The other methods use logical indexing, which assign values of 1 for true and 0 for false.