how to replace by comparing in a double array.?
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I have 4 double matrix, one is c(i,j) of size 256*8,having only 0 and 1 as its elements and z=[0 1 1 1 1 1 1 1 ] and y=[1 0 0 0 0 0 0 0] and b=[ 0 1 0 1 0 1 0]
i want to compare each row of c with z,y,b with below condition.
if any row of c(i,j) matches with z or y and if b(k)==0 then that row of c(i,j)=z,
otherwise/else
c(i,j) matches with z or y but b(k)==1 then that row of c(i,j)=y, else c(i,j) values remain same
i have written the code as this
[row]=size(c);
[row1]=size(b)
for i=1:row
if c(i,:)== z | c(i,:)==y
for k=1:row1
if b(k)==0
p(i,:)= z;
else
p(i,:)= y;
end
end
end
end
i could not find where is the wrong why the matrix p does not have the values required..
Thank you in advance.
3 个评论
Jan
2016-11-16
Please format your code using the "{} Code" button to present it in a readable form.
When you claim, that the posted code is "wrong", please add the information, why you think so. Currently the users have to guess or spend time to find this out, while you have this information already. Offering all you have to make an answer as easy as possible encourages the forum users to spend their time to assist you.
KSSV
2016-11-16
How it is different with this?
aditya kumar sahu
2016-11-17
回答（1 个）
A bold guess:
[row]=size(c)
for i=1:row
...
Note that row is a vector and the expression 1:row does most likely not do, what you expect. Try:
row = size(c, 2); % Or: numel(c)
for i = 1:row
Do you know and use the debugger? You can set breakpoint in lines and step through the code line by line. This helps to find such problems and it is much more efficient than asking the forum.
2 个评论
aditya kumar sahu
2016-11-17
Jan
2016-11-21
@aditya: Repeating sentences from the original question does not add any new information.
Did you notice my suggestion:
When you claim, that the posted code is "wrong", please add the information, why you think so.
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2016-11-16
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