To change value in cells based on conditions

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.

Answer 1

[Andrei Bobrov]

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);

[Jan]

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]

Hi Jan! I am agree with you.

Added variant of code.

[yue ishida]

thank you very much.

Answer 2

[Walter Roberson]

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

[yue ishida]

the output is 33 33 44 33 22

[Walter Roberson]

Okay... it just wasn't clear whether your mention of three 33's was exactly three or was "three or more".

Answer 3

[yair suari]

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

[Jan]

Or: "a(i+2) = 44" - there is no need to overwrite 33 by 33.

Answer 4

[yue ishida]

i still get error for my answer, maybe we need to change a bit of this coding? I was trying this code based on your code, but still don't get the answer...

for i=1:size(a,1) 
    if a(i:i+2)==[33 33 33]
        a(i:i+2)=[33 33 44]; 
    elseif a(i:i+2)==[33 33 22]
        a(i:i+2)=[33 33 44];
    elseif a(i:i+1)==[33 11]
        a(i:i+1)=[33 22];    
    end
end

[Jan]

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]

thanks for guide...