Change or remove duplicate matrix elements

1 Ansicht (letzte 30 Tage)
A
A am 26 Aug. 2011
Bearbeitet: Stephen23 am 2 Feb. 2016
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 Kommentare
Sean de Wolski
Sean de Wolski am 26 Aug. 2011
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 am 26 Aug. 2011
That's true. What if on the other hand I want to completely remove any duplicate entries?

Melden Sie sich an, um zu kommentieren.

Akzeptierte Antwort

Sean de Wolski
Sean de Wolski am 26 Aug. 2011
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;
  8 Kommentare
Fangjun Jiang
Fangjun Jiang am 26 Aug. 2011
@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 am 26 Aug. 2011
Thst's true, hence the while-loop I mentioned in the comment above.

Melden Sie sich an, um zu kommentieren.

Weitere Antworten (2)

Fangjun Jiang
Fangjun Jiang am 26 Aug. 2011
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 Kommentare
A
A am 26 Aug. 2011
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 am 26 Aug. 2011
The reason is simple. You added 0.1 to the 2nd duplicated value which will change the comparison of your next loop.

Melden Sie sich an, um zu kommentieren.


MaVo
MaVo am 2 Feb. 2016
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

Kategorien

Mehr zu Structures finden Sie in Help Center und File Exchange

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by