Effacer les filtres
Effacer les filtres

How would I create my own 'unique' function without using the built in unique?

5 vues (au cours des 30 derniers jours)
I understand that the built in Unique function is stating that whatever function you define as unique is the same data set but without repetitions.
should I create a function file such as
function[unique,c] = myuni_1(A)
rows(A)= size(A,1);
columns(A) = size(A,2);
and then proceed to define the rows and columns as itself? I'm a bit confused here.
  3 commentaires
Alexander
Alexander le 29 Avr 2014
My TA mentioned my professor may ask a question like this on my final exam and I would like to be prepared for anything. Also, my own curiosity is driving me to ask it.
I've looked at the function file on Matlab and it is a very long code. I'm curious if this is even possible with an alternative method that uses less code.
Jan
Jan le 30 Avr 2014
Modifié(e) : Jan le 30 Avr 2014
What exactly should your function do? Which types of input are expected? What is the 2nd output c? Please do not let us guess the details.
Notice that "rows(A)= size(A,1);" is no valid Matlab syntax.

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Jan
Jan le 30 Avr 2014
A short version:
function U = myUnique(A)
[As, SV] = sort(A(:));
UV(SV) = ([1; diff(As)] ~= 0); % [1, As(2:nA) - As(1:nA-1)];
U = A(UV);
  1 commentaire
Alexander
Alexander le 30 Avr 2014
works like a charm, i was using 3 different for loops and trying to sort each one by the size of A but got nothing.
Thank you friend!

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Plus de réponses (2)

Walter Roberson
Walter Roberson le 29 Avr 2014
Suppose you have two "bags", A and B. Start with A empty.
If B is empty, you are done and A holds the unique values. Otherwise, take one one item, C, out of B (removing it from B). Is there already a copy of C in A? If there is, then throw away C. Otherwise, add C to A. Now restart this paragraph.
  1 commentaire
Alexander
Alexander le 30 Avr 2014
im just confused about how to pull out a value and place it into the new matrix

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Sean de Wolski
Sean de Wolski le 30 Avr 2014
Modifié(e) : Sean de Wolski le 30 Avr 2014
How about this?
x = [1 2 3 3 3 3 2 4 5];
[uv,ia] = intersect(x,x)

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