How to mix integers and chars in matrix?

1. I want to create the following matrix:
A=[1 2 3 A B C]
2. After that, based on the matrix A, I want to generate random matrix of this following matrix:
A1=[1 3 2 A C B]
A2=[2 1 3 B A C]
A2=[2 3 1 B C A]
Does any body can help me?

 Akzeptierte Antwort

Sebastian
Sebastian am 27 Jan. 2011

1 Stimme

As Paulo pointed out, this most likely requires usage of cells, e.g.:
>> A = {1 2 3 'A' 'B' 'C'};
>> A1 = A(randperm(6))
A1 =
[3] 'A' [1] 'C' [2] 'B'
>> A2 = A(randperm(6))
A2 =
[3] [2] 'C' 'A' 'B' [1]
However, if the integers and chars are always 1 digit/characte, then you can maybe also use the following approach:
>> a = '123ABC';
>> a1 = a(randperm(6))
a1 =
13A2CB
>> a2 = a(randperm(6))
a2 =
31BAC2

Weitere Antworten (3)

zakri
zakri am 27 Jan. 2011

0 Stimmen

Thank you very much. I really appreciate it. However, how do we want to make the alphabet change according to the integer? for instance:
Matrix A is :
A=[1 2 3 A B C]
where
A (column 3) is dependant to 1 (column 0)
B(column 4) is dependant to 2 (column1)
C(column 5) is dependant to 3 (column 2)
if the integer 1, 2 and 3 is randomly permute within the columns
0 to 2, the alphabets of A,B and C are also randomly permute within columns 3 to 5.
Sebastian
Sebastian am 27 Jan. 2011

0 Stimmen

Well, you could do something like
>> a = '123ABC';
>> l = size(a,2);
>> rp = randperm(l/2);
>> a1 = a([rp rp+l/2])
a1 =
321CBA
zakri
zakri am 28 Jan. 2011

0 Stimmen

hi again... How to repeatedly do random such: 321CBA 312CAB 213BAC
I type a code as the following:
A='123ABC'
AA=[];
randormsort=[];
l=size(a,2);
rp=randperm(l/2);
for i=1:3
AA(i,:)=A([rp rp+l/2]);
end
randomsort=[AA];
however the result appear as follows:
AA =
49 51 50 65 67 66
49 51 50 65 67 66
49 51 50 65 67 66
the result that i am expected should be something like this:
3 2 1 C B A
2 1 3 B A C
2 3 1 B C A

9 Kommentare

Paulo Silva
Paulo Silva am 28 Jan. 2011
char(AA)
zakri
zakri am 28 Jan. 2011
it returns only
321CBA
Paulo Silva
Paulo Silva am 28 Jan. 2011
find what's wrong with your code, that's the better way to learn.
Paulo Silva
Paulo Silva am 28 Jan. 2011
hint: You got all the code in there but one line of it is in the wrong place :)
zakri
zakri am 29 Jan. 2011
I have change to :
A='123ABC'
AA=[];
randormsort=[];
l=size(A,2);
rp=randperm(l/2);
for i=1:3
AA(i,:)=A([rp rp+l/2]);
end
randomsort=char(AA);
it works. however, the rows in the same value:
213BAC
213BAC
213BAC.
how to make it be in such a way that rows 2 and 3 are different sequence from rows 1?
Paulo Silva
Paulo Silva am 29 Jan. 2011
You are only shuffling the cards once, that's why you get always the same result, shuffle the cards 3 times...
zakri
zakri am 31 Jan. 2011
can you give a hint?
Paulo Silva
Paulo Silva am 31 Jan. 2011
already did lol
c'mon you just need to put the randperm inside the loop, so easy :)
zakri
zakri am 31 Jan. 2011
It works!!thank you.

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