Change or remove duplicate matrix elements

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A
A 2011-8-26
编辑： Stephen23 2016-2-2
I have the matrix:
value =
3.1727
5.2495
5.2708
3.3852
5.6222
5.2708
5.1444
4.9834
5.7499
5.7499
3.4728
3.4728
5.3560
5.7499
3.4728
5.7286
6.1225
3.6539
3.0351
4.3020
5.2296
3.8040
4.6747
5.4412
3.6539
and I want to add 0.1 to any duplicate entries, so that all values are unique, and none are removed.
I have tried:
value=sort(value)
for i=1:(length(value)-1)
if value(i+1)==value(i);
value(i+1)=(value(i+1)+0.1);
end
end
But for some reason it has no impact on the matrix..
Your help is greatly appreciated, thanks!
**EDIT: how can i remove all but the first occurance of a duplicate value? unique does not work for me.
  2 个评论
Sean de Wolski
Sean de Wolski 2011-8-26
So what if by adding 0.1 you duplicate a value? You'll have to do a while loop checking to ensure this didn't happen and rerunning the engine again if it did.
A
A 2011-8-26
That's true. What if on the other hand I want to completely remove any duplicate entries?

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采纳的回答

Sean de Wolski
Sean de Wolski 2011-8-26
X = [1 2 3 3 4 5 2]';
X2 = sort(X,1);
idx = [false;diff(X2)<(10^-6)]; %equal to 10^-6th precision
X2(idx) = X2(idx)+.1;
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Fangjun Jiang
Fangjun Jiang 2011-8-26
@Sean, this won't work. The problem is not the floating point comparison. It's the multiple duplication. If the first round has 3 duplications, then after the round, you still have 2 duplications.
X = [1 2 3 3 4 5 2]';
X2 = sort(X,1);
idx = [false;diff(X2)<(10^-6)]; %equal to 10^-6th precision
X2(idx) = X2(idx)+.1
Sean de Wolski
Sean de Wolski 2011-8-26
Thst's true, hence the while-loop I mentioned in the comment above.

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更多回答（2 个）

Fangjun Jiang
Fangjun Jiang 2011-8-26
It works in certain degree but your algorith has a flaw. You'll see it clearly running the following.
value=sort(value);
NewValue=value;
for i=1:(length(NewValue)-1)
if NewValue(i+1)==NewValue(i);
NewValue(i+1)=(NewValue(i+1)+0.1);
end
end
[value NewValue]
  3 个评论
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A
A 2011-8-26
Ah sorry thats a bit hard to read, I didn't realize the numbers woudl be so close together. It is [value NewValue] as you put in your code.
Fangjun Jiang
Fangjun Jiang 2011-8-26
The reason is simple. You added 0.1 to the 2nd duplicated value which will change the comparison of your next loop.

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MaVo
MaVo 2016-2-2
I had a similar issue with a time sequence in 0.005 s steps, but I figured out a solution that worked for me. My problem during debugging was, that it didn't go into the if-condition. I solved this by comparing integers than double values. Maybe this helps someone.
s_length = length(s_time);
ctr = 1;
for ctr = 2:s_length
% Checking values for debugging
new = uint64(s_time(ctr,1)/0.005);
old = uint64(s_time((ctr-1),1)/0.005);
before = (ctr-1);
if new == old
s_time(ctr:end,1) = s_time(ctr:end,1)+0.005;
end
if new > (old + 1)
s_time(ctr:end) = s_time(ctr:end) - 0.005;
end
end

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