How to remove repeating elements but maintain occurrences in an array?

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Vignesh
Vignesh 2015-2-12
编辑： DGM 2024-7-19
Is there a way that I can get the first occurrence of consecutively repeating values, even if the same value occurs at few different places in the matrix?
Say I have a matrix
X=[2 2 2 2 -1 -1 -1 6 6 6 6 6 5 5 5 2 2 2 2 7 7 6 6]
I want the result to be
ans=[2 -1 6 5 2 7 6].
I've checked a similar question How to remove repeating elements from an array but using unique and keeping the same order as original matrix gives
ans=[2 -1 6 5 7].
I know I can use a loop like below. But I was wondering if there's a function for this.
for i=1:(length(X)-1)
if (X(i+1,6)-X(i,6))~=0
ans(i,1)=X(i+1,6);
else
ans(i,1)=NaN;
end
end
A function to do this will be great.
  3 个评论
显示 1更早的评论
Image Analyst
Image Analyst 2015-2-12
I would not use ans as the name of a variable - that has a special meeting in MATLAB.
Vignesh
Vignesh 2015-2-15
Thanks Stephen! That helped. Image Analyst, it was just an example for the question. Thanks for pointing out anyway.

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采纳的回答

Stephen23
Stephen23 2015-2-12
编辑：Stephen23 2015-2-12
You could do this with diff:
>> X=[2 2 2 2 -1 -1 -1 6 6 6 6 6 5 5 5 2 2 2 2 7 7 6 6];
>> Y = [true,diff(X)~=0];
>> X(Y)
ans = [2,-1,6,5,2,7,6]
This works well with integers. If you are working with floating point numbers, then you might need to include some kind of tolerance in the diff:
>> Y = [true,abs(diff(X))>tol];
  1 个评论
Stephen23
Stephen23 2015-2-12
编辑：Stephen23 2015-2-12
If you want this as a function, then you can easily define your own:
>> start = @(v)v([true,diff(v)~=0]);
>> start(X)
ans = 2 -1 6 5 2 7 6

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更多回答（1 个）

Prashanth
Prashanth 2024-7-19
编辑：Prashanth 2024-7-19
ans=unique(X);
  2 个评论
Walter Roberson
Walter Roberson 2024-7-19
  1. This has the side effect of sorting the output, which was not desired
  2. This as the side effect of only outputing any one value once. We see from the example input and output that 2 is expected as the output twice.
DGM
DGM 2024-7-19
编辑：DGM 2024-7-19
Ran in:
For example:
% our input vector
X = [2 2 2 2 -1 -1 -1 6 6 6 6 6 5 5 5 2 2 2 2 7 7 6 6];
% Stephen's answer
start = @(v)v([true,diff(v)~=0]);
start(X)
ans = 1x7
2 -1 6 5 2 7 6
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% sorted output from unique()
usort = unique(X)
usort = 1x5
-1 2 5 6 7
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% stable output from unique()
usort = unique(X,'stable')
usort = 1x5
2 -1 6 5 7
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While the ordering can be resolved by using the 'stable' option, we're not actually after unique values at all. The goal is to reduce each run of a given value to a single instance of that value. Since there are multiple runs of 2 and 6, the output from unique(...,'stable') still results in a loss of information.
As an aside, using 'ans' as a variable name is a problem waiting to happen.

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