How to realize this vector operation ?

I have a vector X and 5 states (each state is a vector like A B C D E as below).
A = [10 11 12];
B = [20 25];
C = [37 37];
D = [49];
E = [57 51 54];
X = [ 10 11 12 20 25 37 37 49 57 51 54 ....];
[10 11 12] corresponds to A and will be represented as 1 in the resultant vector Z, like that 2 for B 3 for C etc.
How can I obtain Z from X. Here Z will be as below : Z = [1 2 3 4 5 .....];

 Accepted Answer

A = [10 11 12];
B = [20 25];
C = [37 37];
D = [49];
E = [57 51 54];
X = [ A C B E D E E A D];
v={A,B,C,D,E};
y=zeros(1,numel(X));
for k=1:numel(v)
id=find(diff([0 ismember(X,v{k})])==1);
y(id)=k;
end
out=y(y~=0)

3 Comments

Thank you sir for the solution, but can I do the same with no loops. :)
I do not see how to do it
sir there was an error......
d4=cos(2*pi*293.665*(0:1/44100:.25));
f4=cos(2*pi*349.228*(0:1/44100:.25));
e4=cos(2*pi*329.628*(0:1/44100:.25));
g4=cos(2*pi*391.995*(0:1/44100:.25));
c5=cos(2*pi*523.251*(0:1/44100:.25));
bb4=cos(2*pi*466.164*(0:1/44100:.25));
a4=cos(2*pi*440*(0:1/44100:.25));
c4=cos(2*pi*261.626*(0:1/44100:.25));
x=[c4 c4 d4 c4 f4 e4 c4 c4 d4 c4 g4 f4 c4 c4 c5 a4 f4 e4 d4 bb4 bb4 a4 f4 g4 f4];
v={a4,bb4,c4,c5,d4,e4,f4,g4};
y=zeros(1,numel(x));
for k=1:numel(v)
id=find(diff([0 ismember(x,v{k})])==1);
y(id)=k;
end
out=y(y~=0)
out =
Columns 1 through 15
8 8 8 8 8 8 1 8 8 8 8 8 8 6 7
Columns 16 through 30
8 8 8 8 8 8 6 8 8 8 8 1 8 8 8
Columns 31 through 41
8 8 8 8 6 8 8 8 8 6 7
But actually the answer must be out=[3 3 5 3 7 6 3 3 5 3 8 7 3 3 4 1 7 6 5 2 2 1 7 8 7] right

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More Answers (1)

A = [10 11 12];
B = [20 25];
C = [37 37];
D = [49];
E = [57 51 54];
v={A,B,C,D,E}
X = [ A C B E D E E A D]
a=cellfun(@(x) find(diff([0 ismember(X,x)])==1),v,'un',0);
b=arrayfun(@(x) x*ones(1,numel(a{x})),1:numel(v),'un',0);
c=cell2mat([a;b])';
d=sortrows(c,1)
out=d(:,2)'

2 Comments

The for loop is much faster then answer 2:
running those answers 1000 times:
Answer 1: Elapsed time is 0.055467 seconds.
Answer 2: Elapsed time is 0.899734 seconds.
thankyou sir for the efforts..... :)

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