Non-Sorted nx1 unique values
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Hello, i am attempting to find repeated values in an nx1 matrix, the catch is that the unique function sorts the data. The data should not be sorted. Ideally i would like the output from my mystery function to be 1 for not repeated and 0 for repeated. Any suggestions?!
input: [1;1;1;2;3;4;5;6;6;7] output:[1;0;0;1;1;1;1;1;0;1]
the idea is that i can multiply the output and the original matrix where the data resides so that duplicate records will show as 0 values, but the first stays!
Thanks again! If you would like me to post another question please let me know!
2 Comments
Image Analyst
on 19 Dec 2016
Give a short example of your input and desired output. Why don't you just use unique and histogram(). Are your data integers or floating point values?
Accepted Answer
Roger Stafford
on 19 Dec 2016
Let x be the given n by 1 vector.
[y,p] = sort(x);
d = diff(y)~=0;
b = [d;true] & [true;d];
b(p) = b;
Then b will be your "ideal" desired result.
8 Comments
Roger Stafford
on 21 Dec 2016
@Jan. In William’s original request he states:, “the idea is that i can multiply the output and the original matrix where the data resides so that duplicate records will show as 0 values, but the first stays!” To be able to do such a multiplication (presumably elementwise) the final reordering by “b(p) = b” would be very necessary.
Jan
on 16 Jan 2017
@Roger: I meant "The data should not be sorted." Anyway, both methods are included in your solution.
More Answers (1)
John BG
on 19 Dec 2016
Edited: John BG
on 19 Dec 2016
but this code seems to work with any randomly generated sequence.
If it works for noise it works for any deterministic signal, oder?
N=20;
A=randi([-10 10],1,N);
% nA=[1:1:numel(A)]
A
[B,reloc]=sort(A)
% dB=diff(B)
% B(find(dB~=0))
nv=0;
k=1
while k<=numel(B)-1
k=k+1
u0=B(k-1);u1=B(k);
if u0==u1
nv=[nv k]
k=k+1
while B(k-1)==B(k);
k=k+1
end
end
end
nv(1)=[]
B(nv)
for certain sequences there is an overflow message
Index exceeds matrix dimensions.
Error in find_long_bursts (line 17)
while B(k-1)==B(k);
don't worry, sequence
nv
contains the right indices and therefore
B(bv)
is ok.
Just had to do something else, I am cleaning it if Mr Honjas likes the code :) .
if you John BG's answer useful would you please mark it as Accepted Answer?
To any other reader, please if you find this answer of any help, click on the thumbs-up vote link,
thanks in advance for time and attention
John BG
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