finding consecutive numbers in an array

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Hi,

I'am trying to find minimum 320 consecutive numbers in an array. Is there any simple way to find it?

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Jan 2016-1-31

编辑：Jan 2016-1-31

Please explain the term "consecutive numbers" in an example. Do you mean, that the difference between two neigboring numbers is +1? Or is -1 accepted also? Do you mean integer values in a double array? Which result is wanted? The first index? What exactly is "minimum 320 numbers"? I assume it is a block of 320 or more elements?

Huseyin 2016-1-31

Assume that I have an array like;

a= [8 9 10 11 15 16 17 180 181 182 183 184 185 190]

In that 1x14 double, I want to find the beginning and end of the consecutive 6 numbers. Answer should be like it begins at 180 and ends with 185

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Answer 1

Jan 2016-1-31

编辑：Jan 2016-1-31

在 MATLAB Online 中打开

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With FEX: RunLength:

Data  = randi([1, 10], 1, 1e6);  % Any test data
Len   = 5;                       % 320 in your case
[B, N, Index] = RunLength(diff(Data));
Match = find(N >= Len & B == 1);
% Output to control results:
for k = 1:numel(Match)
  iResult = Index(Match(k));
  fResult = iResult + N(Match(k)) - 1;
  disp(Data(iResult:fResult));
end

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Huseyin 2016-1-31

编辑：Huseyin 2016-1-31

I set your RunLength package to the path but Matlab gives a message like 'too many output arguments'. It gives the same message for your example also.

Jan 2016-1-31

编辑：Jan 2016-1-31

Please post a complete copy of the error message.

Start RunLength once without input arguments for a compilation of the C-code. If you do not have a C-compiler, use the slower RunLength_M.

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finding consecutive numbers in an array

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