Max Series of Consecutive Numbers

Question

I need help finding a way to do this, although it seems as it should be pretty simple. Lets say I have a nx1 array. For example, let X=[1 1 1 .5 .4 .2 -.2 -.3 1 1 0 1]. What I am trying to do is find where the greatest series of consecutive 1's begins and how many 1's are in it. For instance, using X, the greatest series of consecutive series starts at the index 1 and is of length 3. Also, I will be using very large sets of data so I would like to try and find the most efficient and fastest way this can be done.

possible duplicate of [matlab - find index for when values exceeds a threshold n number of times](http://stackoverflow.com/questions/15519695/matlab-find-index-for-when-values-exceeds-a-threshold-n-number-of-times) — Dan, Jul 15 '13 at 07:58

horchler · Accepted Answer · 2013-07-14T20:46:01.480

2

Here's one way that you could accomplish that fairly efficiently:

x = [1 1 1 0.5 0.4 0.2 -0.2 -0.3 1 1 0 1];
x1 = (x==1);
d = diff(x1(:).');
start = find([x1(1) d]==1)
len = find([d -x1(end)]==-1)-start+1

which returns

start =

     1     9    12

len =

     3     2     1

edited Jul 14 '13 at 20:46

answered Jul 14 '13 at 20:15

horchler

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I tweaked the code to make work for both row- and column-vector inputs. – horchler Jul 14 '13 at 20:49
No I am just passing through a column vector. However, the loop is returning empty matrices, even though there are consecutive 1's in the column. – Math244 Jul 14 '13 at 21:02
What vector for example? And are you sure that those values are actually `==1` and not off by some epsilon? If they are just very very close to `1` then you you should be able to adapt the equation for `x1` in my code above to include a small tolerance. – horchler Jul 14 '13 at 21:04
Thank you! That worked by adjusting it to x>=.9. Thank you so much for your help. – Math244 Jul 14 '13 at 21:32
Good to hear. A more general solution would be `x1 = (abs(x-1) – horchler Jul 14 '13 at 22:26

score 0 · Answer 2 · answered Jul 14 '13 at 21:24

A function like this can help

function [start, len] = findLongestRunning(x)
y = find(diff([0;x(:)==1;0]));
[len,J] = max(y(2:2:end)-y(1:2:end));
start = y(2*J(1)-1);
end

Running on your example

>> [start, len] = findLongestRunning(x)
start =
     1
len =
     3

Note that the code returns the first occurence if there are more than one sequence meeting the requirements:

>> [start, len] = findLongestRunning([x 0 x])
start =
     1
len =
     3

Max Series of Consecutive Numbers

2 Answers2

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