Run length of consecutive integers in array

I am working with an array and need to determine the run length of consecutive integers in the 4th column of the array. Here is a sample from my data:

In the 4th column, where the numbers increase by 1, I want to determine the amount of times they consecutively increase by 1 and output to another variable. So for the first two numbers I want an output run length of 2 since 58 and 59 are consecutive. The next 3 numbers (275,276,277) would produce a run length of 3. It needs to iterate all the way through the 4th column and produce a run length for all numbers that are consecutive. It would also be nice to take the date of the last number in the run length to be an index for each run length. Example for first 5 rows:

• Date: 1948 , 12 , 27 , Run Length: 2
• Date: 1950 , 3 , 4 , Run Length: 3

[Azzi Abdelmalek]

How can we test your data? post your data as a text instead of an image.

[Tyler Smith]

attached is a .mat file of the data.

Answer 1

[Azzi Abdelmalek]

v=[1949 4 5 57;1949 12 27 58;1950 7 8 275;1950 7 6 276 ;1950 7 3 277 ]
a=v(:,end)
id=[0;diff(a)]==1
ii1=strfind(id',[0 1])
ii2=strfind([id' 0],[1 0])
ii=ii2-ii1
for k=1:numel(ii)
  ix=ii2(k);
  out{k,1}=sprintf('date %d  %d  %d  Run Length: %d',[v(ix,1:3) ii(k)])
end

[Tyler Smith]

Thanks so much!

Answer 2

[dpb]

ix=find(diff(v)~=1);  % indices
rl=diff([0; ix]);     % runlength

[Tyler Smith]

Thank you!

[dpb]

No problem but I'm curious why you'd choose the more complex of the two solutions? Because I left it to you to write the output expression and you didn't follow how to use the index array, mayhaps?

>> a=[y m d v];
>> ix=find(diff(a(:,end))~=1);
>> rl=diff([0; ix])
>> fprintf('Date: %d, %3d, %3d, Run Length: %2d\n',[a(ix,1:3) rl].')
Date: 1948,  12,  27, Run Length:  2
Date: 1950,   3,   4, Run Length:  3
Date: 1950,  11,  26, Run Length:  5
Date: 1950,  12,  11, Run Length:  4
>>