Need Help Urgent!!!!!!!!!!!!

i have this data in matrix X=[22,33,11,33,33,33,22]

i want to change the data with this form of matrix X=[22,33,22,33,33,33,44]

i have to follow this rules: 1)when find(X==33) next data must change into 22 2)when there have data "33" three times in row,the next data must change to 44

this is my example code for the looping: [a b]=find(X==33)

if X(a,b+1)-X(a,b)==1 if X(a,b+2)-X(a,b+1)==1 X(a,b+3)=44 end end

if X(a,b+1)-X(a,b)==2 X(a,b+2)=22 end

i know im doing this all wrong,can someone give a correct algorithm to get the answer that satisfied the rules.

Amir my email: [e-mail]

[Matt Fig]

Your rules conflict and don't cover all possible cases (at least for a random X). What if there are two 33's in a row? Do we change the second one to a 22? What if there are four 33's in a row? Do we change the last one to a 44, or the element after the four 33's? Please clarify.

[Matt Fig]

And what if the last element, or the last three elements are 33? Do we add an element?

[Amir Hamzah UTeM]

Thank Matt for replying. this is my rules

1.if only have one 33 in row,

if X=[22,33,11,33,11,11,22] change -> X=[22,33,22,33,22,11,22]

2. if there have two 33's in a row,only change the next data to 11.

if X=[22,33,22,33,33,22,22] only change next data to 11 -> X=[22,33,11,33,33,11,22]

3. if have three 33's in a row, the next data must be change to 44

if X=[22,33,22,33,33,33,22] must change into X=[22,33,22,33,33,33,44] -> X=[22,33,11,33,33,33,44]

[Amir Hamzah UTeM]

4. if there have four 33's element in row,the 4th element change to 44

5. if the last three elements are 33 in row,we must add an element to 44

Answer 1

[Matt Fig]

You did not completely clarify, but here is a non-vectorized code that will do part of the job. Note that for this problem, it might be quicker to use a loop anyway.

cnt = 1;
Y = X==33;
while cnt <= length(X)-2
    if Y(cnt)
        if Y(cnt+1:cnt+2)  % Three in a row.
            X(cnt+3) = 44;
            cnt = cnt + 4;
        elseif Y(cnt+1) % Two in a row.
            X(cnt+2) = 11;
            cnt = cnt + 3;
        else  % Just one 33.
            X(cnt+1) = 22;
            cnt = cnt + 2;
        end
    else
        cnt = cnt + 1;
    end
end

[Amir Hamzah UTeM]

thanks Matt.. i think it can works with my project and i need to proceed to the next stage. catch you later,thanks a lot.. :D

[Amir Hamzah UTeM]

matt,

how about i have Nx7 matrix?how to solve it?

[Matt Fig]

Put everything in a FOR loop over the rows of X. Change only the X assignments to X(ii,cnt+3) = 44 (for example), where ii is the loop index.

[Amir Hamzah UTeM]

i have tried to do this,but it all wrong. did i miss something?

X=[33 11 33 33 33 33 22

22 11 33 33 11 33 22

33 33 33 11 33 33 11];

cnt = 1;

Y = X==33;

while cnt <= length(X)-2

if Y(cnt)

if Y(cnt+1:cnt+2) % Three in a row.

for i=1:3

X(i,cnt+3) = 44;

cnt = cnt + 4;

end

elseif Y(cnt+1) % Two in a row.

for i=1:3

X(i,cnt+2) = 11;

cnt = cnt + 3;

end

else % Just one 33.

for i=1:3

X(i,cnt+1) = 22;

cnt = cnt + 2;

end

end

else

cnt = cnt + 1;

end

end

[Matt Fig]

I said, "Put everything in a FOR loop over the rows of X." It looks like you just put one clause of the IF statement in a FOR loop. Everything means EVERYTHING (except or course your X variable creation.)

for ii = 1:size(X,1)

cnt = 1;

Y = X(ii,:)==33;

.....

end