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Amir Hamzah UTeM
Amir Hamzah UTeM on 28 Mar 2011
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: noksworld@yahoo.com
  4 Comments
Amir Hamzah UTeM
Amir Hamzah UTeM on 28 Mar 2011
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
Amir Hamzah UTeM on 28 Mar 2011
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

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Accepted Answer

Matt Fig
Matt Fig on 28 Mar 2011
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
  5 Comments
Amir Hamzah UTeM
Amir Hamzah UTeM on 28 Mar 2011
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
Matt Fig on 28 Mar 2011
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

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