143 views (last 30 days)

The simple case is like this:

2 1 4 6 2

9 4 6 1 2

5 3 2 8 3

7 2 1 9 3

7 1 8 2 4

From the matrix above, i want to insert 3 NaNs in random place. So, my code is like this:

Data = [2,1,4,6,2;9,4,6,1,2;5,3,2,8,3;7,2,1,9,3;7,1,8,2,4];

[rows,cols] = size(Data);

p = 3; %amount of NaN that will we inserted

r = randperm(25); %give the random value from range 1-25

r = r(1:3); %give 3 random number from range 1-25

i = 1;a = 1; b = 1;

while i <= 3 %generate every number in vektor r to be position where NaN is located

n = r(a,b);

b = b+1;

e = 1;

if n <= cols

Data(1,n) = NaN;

else

if n > cols

while n > cols

e = e+1;

k = n - cols;

n = k;

end

Data(e,n) = NaN;

end

end

i = i+1;

end

The output one of the output will be like this:

2 1 4 6 2

9 NaN NaN NaN 2

5 3 2 8 3

7 2 1 9 3

7 1 8 2 4

So, i want to make some constraint such as:

1. every row only can have 2 NaN

2. amount NaN in column 1 have to be less then column 2, and amount NaN in column 2 have to be less then column 3, and so on. eg. output matrix will be like this:

2 1 4 6 2

9 4 6 1 NaN

5 3 2 8 3

7 2 1 NaN 3

7 1 8 2 NaN

for matrix above we can see that:

amount NaN of column 1= 0, column 2=0, column 3=0, column 4=1, column 5= 2.

Somebody can help me to insert those my constraint into my code above? Or there willl be another solution i think.

Thanks before :')

per isakson
on 22 Apr 2012

This is an idea that I have not tested!

jj = 0;

for ii = r

[rr,cc] = ind2sub( size(Data), ii )

if sum(isnan(Data(rr,:))>=2 || sum( isnan(Data(:,cc))>=2

% do nothing

else

Data(rr,cc)=nan;

jj = jj + 1;

if jj = 3, break

end

end

end

--- EDIT ---

The function below will return a result. The constraint is "no more than two NaN in any column or row. However, that was not what you asked for.

function Data = cssm

Data = [2,1,4,6,2;9,4,6,1,2;5,3,2,8,3;7,2,1,9,3;7,1,8,2,4];

p = 3; %amount of NaN that will we inserted

row_vector = randperm(numel(Data));

jj = 0;

for ii = row_vector

[rr,cc] = ind2sub( size(Data), ii );

if sum(isnan(Data(rr,:)))>=2 || sum( isnan(Data(:,cc)))>=2

% do nothing

else

Data(rr,cc)=nan;

jj = jj + 1;

if jj == p, break

end

end

end

end

With the constraint, "amount NaN in column 1 have to be less then column 2, and amount NaN in column 2 have to be less then column 3, and so on.", there is no solution. Do you exclude columns with zero NaN from that constraint?

Thus, (according to my reading) the last column can have two or three NaN and the second last column one or zero NaN. NaN cannot not appear in the other columns.

Richard Brown
on 22 Apr 2012

This is another one of these problems where the simplest way to solve it is to randomly generate candidates until you find one that fits:

A = reshape(randperm(25), 5, 5);

done = false;

while ~done

idx = randperm(25, 3);

[I, J] = ind2sub([5 5], idx);

m = hist(I, unique(I));

n = hist(J, unique(J));

done = all(m <= 2) && all(diff(n) >= 0);

end

A(idx) = NaN;

It's trivial (but a little messier) to make it more general, so I'll leave you to do that if you need to.

EDIT changed code to use randperm instead of randi - only one call to the random number generator is necessary

Richard Brown
on 29 Apr 2012

Here's a much faster method that satisfies both of your constraints. It may be possible to vectorise the loop, but it is, in my opinion, not worth the effort.

First, generate the data

X = rand(1500, 11);

[m,n] = size(X);

nNans = 2000;

We figure out the row and column indices separately. Rows is easy, a single call to randperm does the trick

I = mod(randperm(2*m, nNans), m) + 1;

Then figure out the column positions randomly, going row by row to avoid creating duplicate entries.

J = zeros(1, nNans);

k = 1;

for i = 1:m

idx = (I == i);

J(idx) = randperm(n, nnz(idx));

end

We now need to make sure the columns are ordered correctly. So we construct a logical matrix encoding the position of the NaN entries, and reorder the columns to satisfy your column constraint.

iNan = false(m, n);

iNan(sub2ind([m n], I, J)) = true;

[~, iSorted] = sort(hist(J, 1:n));

iNan = iNan(:, iSorted);

We now have a logical array with the right properties. Last step is to overwrite the entries of X

X(iNan) = nan;

Sign in to comment.

Sign in to answer this question.

Opportunities for recent engineering grads.

Apply Today
## 2 Comments

## Direct link to this comment

https://ww2.mathworks.cn/matlabcentral/answers/36154-how-to-replace-some-of-the-value-in-the-matrix-with-nan#comment_74987

⋮## Direct link to this comment

https://ww2.mathworks.cn/matlabcentral/answers/36154-how-to-replace-some-of-the-value-in-the-matrix-with-nan#comment_74987

## Direct link to this comment

https://ww2.mathworks.cn/matlabcentral/answers/36154-how-to-replace-some-of-the-value-in-the-matrix-with-nan#comment_74996

⋮## Direct link to this comment

https://ww2.mathworks.cn/matlabcentral/answers/36154-how-to-replace-some-of-the-value-in-the-matrix-with-nan#comment_74996

Sign in to comment.