To change value in cells based on conditions
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I have coding as below.
a=[11 11 33 33 22 44; 33 33 33 11 11 22; 33 33 11 22 22 44; 44 44 33 22 44 11]
I need to change the matrix value based on this conditions and cannot overlap each other:
1. if value 33 occured 3 times continuosly, the end of 33 need to replace with 44
33 33 33 = 33 33 44
2. If value 33 appeared 2 times continously and followed by 22, I need to replace the 22 value with 44 too.
33 33 22 = 33 33 44
2. if value 33 occured then followed by value 11, the value 11 will change into 22
33 11 = 33 22
Therefore, origin matrix will become like this:
c=
11 11 33 33 44 44
33 33 44 11 11 22
33 33 44 22 22 44
44 44 33 22 44 11
I need help to code this problem. Thank you for your help.
采纳的回答
Andrei Bobrov
2012-1-5
n = size(a,1);
K = reshape([a zeros(n,1)].',[],1)';
K(strfind(K,[33 33 33])+2) = 44
K(strfind(K,[33 33 22])+2) = 44
K(strfind(K,[33 11])+1) = 22
out = reshape(K,[],n)';
out = out(:,1:end-1);
ADDED [22:16(UTC+4) 05.01.2012]
n = size(a,1);
A = reshape([a zeros(n,1)].',[],1).';
N = numel(A);
j1 = 1;
while j1 <= N
if j1+2 <= N && (all(A(j1:j1+2) == [33 33 33]) ...
|| all(A(j1:j1+2) == [33 33 22]) ...
|| all(A(j1:j1+2) == [33 33 11]))
A(j1+2) = 44;
j1 = j1+3;
elseif j1+1 <= N && all(A(j1:j1+1) == [33 11])
A(j1+1) = 22;
j1 = j1+2;
else
j1 = j1 + 1;
end
end
out = reshape(A,[],n)';
out = out(:,1:end-1);
3 个评论
Jan
2012-1-5
What does this reply:
strfind([33 33 33 33], [33 33 33])
? If you get [1, 2], your approach will not produce the wanted result.
Andrei Bobrov
2012-1-5
Hi Jan! I am agree with you.
Added variant of code.
yue ishida
2012-1-6
更多回答（3 个）
Walter Roberson
2012-1-5
0 个投票
Use logical vectors for the various conditions. If you think about logical vectors a bit you will come up with a simple way to prevent overlaps between the conditions.
Question:
What should be the output for 33 33 33 33 11
2 个评论
yue ishida
2012-1-5
Walter Roberson
2012-1-5
Okay... it just wasn't clear whether your mention of three 33's was exactly three or was "three or more".
yair suari
2012-1-5
0 个投票
something like (not very efficient)
for i=1:size(a,1) if a(i:i+2)==[33 33 33];a(i:i+2)=[33 33 44] end
1 个评论
Jan
2012-1-5
Or: "a(i+2) = 44" - there is no need to overwrite 33 by 33.
yue ishida
2012-1-5
2 个评论
Jan
2012-1-5
if all(a(i:i+2)==[33 33 33]) % better an explicite ALL
a(i+2)=44; % No need to overwrite the 33 with 33
...
yue ishida
2012-1-6
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