how to find a number in cell and make it NaN?

I have a cell matrix with size 1165x1 cell. In each cell there is one value i.e. number 31 that needs to be found and made NaN.

For example:

A = {[1,2],[3,4],[5 31], [31,6]};

I would like to make 31 = NaN where ever it is seen. Just like matrix B:

B = {[1,2],[3,4],[5 NaN], [NaN,6]};

Any help is greatly appriciated.

Answer 1

[Jan]

Start with a simple loop:

A = {[1,2], [3,4], [5 31], [31,6]};
B = A;
for k = 1:numel(B)
    b = B{k};
    m = (b == 31);
    if any(m)
        b(m) = NaN;
        B{k} = b;
    end
end

This can be condensed into a cellfun method:

p = [1, NaN];
B = cellfun(@(a) a .* p((a == 31) + 1), A, 'UniformOutput', 0)

B = 1×4 cell array
    {[1 2]}    {[3 4]}    {[5 NaN]}    {[NaN 6]}

Answer 2

[Manas Shivakumar]

There are several ways to do this.

1) convert to matrix, replace and convert back to cell array:

tmp = cell2mat(A);
tmp(tmp == 31) == Nan;
A = num2cell(tmp)

2) use cellfun:

A(cellfun(@(elem) elem == 31, A)) = {Nan}

[Jan]

The cellfun approach fails. Replace "Nan" by "NaN", but the nónuniform output will cause troubles also.

[Image Analyst]

Correct Nan to Nan or nan. But still not right. Just to illustrate:

% Method 1
A = {[1,2],[3,4],[5 31], [31,6]}

A = 1×4 cell array
    {[1 2]}    {[3 4]}    {[5 31]}    {[31 6]}

tmp = cell2mat(A);
tmp(tmp == 31) == NaN;
A = num2cell(tmp)

A = 1×8 cell array
    {[1]}    {[2]}    {[3]}    {[4]}    {[5]}    {[31]}    {[31]}    {[6]}

% Method 2:
A = {[1,2],[3,4],[5 31], [31,6]};
A(cellfun(@(elem) elem == 31, A)) = {NaN}

Error using cellfun
Non-scalar in Uniform output, at index 1, output 1.
Set 'UniformOutput' to false.

@Manas Shivakumar Can you try again?

[dpb]

The num2cell conversion also destroys the original cell sizes by placing every element into its own cell. One would have to write something like

 mat2cell(cell2mat(A),[1],[2 2 2 2])
ans =
  1×4 cell array
    {1×2 double}    {1×2 double}    {1×2 double}    {1×2 double}
>> 

to get the original back.

Answer 3

[MP]

Thank you so much to everyone for their valuable time and efforts.

It saved a lot time of mine!