Asked by M.S. Khan
on 26 Jul 2019

M = [0 0 0;... 0 0 3;... 3 3 0;... 0 3 3;... 3 0 0;... 0 0 3;... 3 0 0;... 0 0 0]; I want to use interpolation to fill the gaps between 3s. I tried different methods but no satisfactory answer Is there any other method possible to apply plz Thanks for all cooperation

Answer by Matt J
on 27 Jul 2019

Edited by Matt J
on 27 Jul 2019

One way:

result = sqrt(cummax(M,1).*cummax(M,1,'reverse'))

M.S. Khan
on 27 Jul 2019

Matt J
on 27 Jul 2019

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Answer by dpb
on 27 Jul 2019

for i=1:size(M,2)

ix=find(M(:,i)==3);

if numel(ix)>1

M(ix(1):ix(end),i)=3;

end

end

M.S. Khan
on 27 Jul 2019

Thanks Mr. DP for your support

M = [0 0 0; 2 2 3; 3 3 0; 0 0 0 3 3 0; 2 2 3; 0 0 0; 3 3 2; 0 0 0; 3 3 3] If i have a matrix in this shape. in 1st column, i want only to fill 3 (on 4th position and then on 9th position.) i mean to fill 3s only only 3 and then the next 3, not in all. similary on other columns as well please help me to handle my problems i will be very thankful

dpb
on 27 Jul 2019

LOL! I knew that was coming while writing the above...illustrates that over-simplification gets the right answer to the wrong question.

How large are your arrays and what are actual values in real application? Such pattern matching may well be better suited to casting the values to char() as then can search for string match as patterns...

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Answer by Andrei Bobrov
on 27 Jul 2019

Edited by Andrei Bobrov
on 30 Jul 2019

s = size(M);

[a,b] = regexp(join(string(M)',''),'30+3');

jj = repelem(1:s(2),cellfun(@numel,a));

lo = zeros(s);

lo(sub2ind(s,[a{:}]+1,jj)) = 1;

lo(sub2ind(s,[b{:}],jj)) = -1;

M(cumsum(lo)>0) = 3;

Other variant:

M = [0 0 0; 2 2 3; 3 3 0; 0 0 0; 3 3 0; 2 2 3; 0 0 0; 3 3 2; 0 0 0; 3 3 3];

m = M;

m(m == 0) = nan;

M(fillmissing(m,'previous') == 3 & fillmissing(m,'next') == 3) = 3;

M.S. Khan
on 30 Jul 2019

% Mr. Andrei, can you help me to sort out such problems.

% i have a matrix M.

M =

0 0 3

3 0 0

0 3 0

0 0 0

3 0 3

0 3 0

3 0 0

0 0 0

0 3 3

lets take first column, in the 2nd row is first 3 and next occurence of 3 is in the 4th row. so all should be filled with 3s inside 2nd and 4th row.

it will be like this:

3

3

3

3

then in the same first column, occurence of next 3 should be watached. its 6th row and 8th row.

if there is one zeros among 3's according to rows, it should be filled with 3. if more than one zeros among 3s with respect to rows, that rows of zeros should be filled or replaced with 4.

if its like: 3 % there is only one zeros among rows of 3s

0

3

so it should look like:

3

3

3

i want first column like this:

0

3

3

3

3

3

3

0

0

in 2nd column, first occurence of 3 is in 3rd row and next occurence of 3 is on 6th row.

it will be like this:

3

3

3

3

in the same 2nd column, next occurence of 3 is on 6th rows and next one is 9th rows.

so in next occurence, if 0s (zeros ) between 3's are more than one. it should be filled with 4.

2nd column should look like this:

0

0

3

3

3

3

4

4

3

your cooperation is highely appreciated.

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## dpb (view profile)

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## M.S. Khan (view profile)

## Direct link to this comment

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