How to remove repeating elements but maintain occurrences in an array?
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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
Image Analyst
2015-2-12
I would not use ans as the name of a variable - that has a special meeting in MATLAB.
采纳的回答
Stephen23
2015-2-12
编辑:Stephen23
2015-2-12
>> 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];
更多回答(1 个)
Prashanth
2024-7-19
编辑:Prashanth
2024-7-19
ans=unique(X);
2 个评论
Walter Roberson
2024-7-19
- This has the side effect of sorting the output, which was not desired
- 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
2024-7-19
编辑:DGM
2024-7-19
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)
% sorted output from unique()
usort = unique(X)
% stable output from unique()
usort = unique(X,'stable')
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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