# Mutual array between matrixes ( complete )

3 vues (au cours des 30 derniers jours)
alexaa1989 le 11 Août 2014
Commenté : alexaa1989 le 15 Août 2014
Thank you all for your previous answers however it seems that I could not resolve the problem so I decided to explain it in more details hope someone can help
I have 3 matrixes: x1=[8,4,2, 1,7,3 ,5,6] x2=[4,3,5, 7,6,2 ,1,8] o=[8,1,3,5,2,6,4,7]
I generate two other matrixes as follow using x1 and x2 and o y1=[5,2,8, 1,7,3 ,6,4] y2=[8,1,3, 7,6,2 ,5,4]
here is how y1 and y2 are produced: for the middle section of y1 and y2 I used exactly the same part in x1 and x2 that are pointed with Bold after that for the parts in the right side of Bold I use parts of (o)that are not appeared in y1 and y2 so far
for the left side of Bold I use all arrays from x2 that are not still used in y1 and same for x1 and y2 I have coded some of it as follow
x1=[8,4,2,1,7,3,5,6]
x2=[4,3,5,7,6,2,1,8]
o=[8,1,3,5,2,6,4,7]
c1=3;
c2=6;
y1=[ V1 x1(c1+1:c2) V2 ]
y2=[ V3 x2(c1+1:c2) V4 ]
I need to define V1 V2 V3 V4
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Joseph Cheng le 11 Août 2014
Modifié(e) : Joseph Cheng le 11 Août 2014
in your direct message you wrote
v1 v2 v3 v4 are the codes to define what I have been wanting
of course if you can code the program in a different way I would be happy to use it
my comment is can you supply what they should be manually so i can better understand your conditions.
alexaa1989 le 11 Août 2014
My apologies, I misundrestood.
V1=[8,4,2]
x1(c1+1:c2)=[1,7,3]
V2=[5,6]
V3=[4,3,5]
x2(c1+1:c2)=[7,6,2]
V4=[1,8]

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### Réponses (2)

Christopher Berry le 14 Août 2014
I think that your algorithm description and examples values are contradictory, so its hard to answer this exactly. But, I will suggest you look at the function setdiff. This will let you find the elements of one set that are not present in another set. See the documentation below
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alexaa1989 le 15 Août 2014
thank you for your response but I have written it successfully

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Roger Stafford le 14 Août 2014
Christopher is right. The function you need is 'setdiff' using the 'stable' option.
n = length(x1);
y1 = x1((c1+1):c2);
v2 = setdiff(o,y1,'stable');
v2 = v2(end-n+c2+1:end);
y2 = x2((c1+1):c2);
v4 = setdiff(o,y2,'stable');
v4 = v4(end-n+c2+1:end);
v1 = setdiff(x2,[y1,v2],'stable');
v3 = setdiff(x1,[y2,v4],'stable');
y1 = [v1,y1,v2];
y2 = [v3,y2,v4];
As Christopher pointed out, you contradicted yourself in your comment, Alexaa1989. The v's there are different from those in the original problem statement. I have assumed that the original statement is the correct one.
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alexaa1989 le 15 Août 2014
thank you for your response but I have written it successfully

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