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

2011 3 28
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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

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Matt Fig
Matt Fig 2011-3-28
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
Matt Fig 2011-3-28
And what if the last element, or the last three elements are 33? Do we add an element?
Amir Hamzah UTeM
Amir Hamzah UTeM 2011-3-28
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 2011-3-28
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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Matt Fig
Matt Fig 2011-3-28

1 个投票

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

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Amir Hamzah UTeM
Amir Hamzah UTeM 2011-3-28
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
Amir Hamzah UTeM 2011-3-28
matt,
how about i have Nx7 matrix?how to solve it?
Matt Fig
Matt Fig 2011-3-28
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
Amir Hamzah UTeM 2011-3-28
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 2011-3-28
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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