# Replacing the column of array elements with NaN.

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##### 0 Comments

### Accepted Answer

Geoff Hayes
on 29 May 2014

Edited: Geoff Hayes
on 29 May 2014

It seems that the rows of x indicate which ranges of values in A should be preserved, with the rest of the entries in the second column of matrix A set to NaN. A looping solution is as follows:

% pre-allocate array of indices indicating values in A to keep/preserve

keepInA = zeros(size(A(:,2)));

% loop over all rows of x

for i=1:size(x,1)

% find where in the first column of A are the two values for the ith row of x

mems = ismember(A(:,1),x(i,:));

% set that range in keepInA to be all ones indicating all values in that range

% are to be kept (there's is probably a better way to do this)

keepInA(find(mems,1,'first'):find(mems,1,'last')) = 1;

% now just set all those elements in the second row of A to be NaN if they

% are NOT to be kept (i.e. zero)

A(keepInA==0,2) = NaN;

The above assumes that there is no overlap of ranges in x and that those ranges can be found in A. If the assumptions are not true, then the above code would have to be modified.

##### 2 Comments

Geoff Hayes
on 29 May 2014

I think that there will always be a loop, whether it is implicit or explicit like the above. Here is a crazy different approach that has no explicit loop:

y = x'; % transpose x so that ranges are column-wise

z = ismember(A(:,1),y(:)); % note that the second input is a column

k = or(z,mod(cumsum(z),2));

A(k==0,2) = NaN;

So what is going on in the above? We convert x to a column vector (via the assignment to y and y(:)) so that the ismember returns a combination of the outputs from ismember in the previous code but as one vector:

z =

0

0

1

0

1

0

1

0

0

1

0

Which is almost okay but we need to fill in all the zeroes in between two neighbouring ones so that we get the correct ranges. If we do a cumulative sum via cumsum then we see

0

0

1

1

2

2

3

3

3

4

4

which is not quite what we want. In fact, all we really want are the odd numbers and the first even number that follows the set of consecutive odd numbers. We can remove all even numbers via mod(cumsum(z),2)) and then "add" back in the missing ones (corresponding to 0.4 and 0.9) via or, so that

or(z,mod(cumsum(z),2));

0

0

1

1

1

0

1

1

1

1

0

is the list of indices that we wish to preserve/keep. So the 5 lines (or so) of the above for loop could be replaced by the 3-4 lines from above. Is it any better? Probably not as this logic is more confusing to follow than the straight-forward for loop.

### More Answers (3)

Andrei Bobrov
on 29 May 2014

Out_Arr = A;

Out_Arr(all(bsxfun(@lt,A(:,1),x(:,1)')|bsxfun(@gt,A(:,1),x(:,2)'),2),2) = nan;

##### 0 Comments

Udit Gupta
on 29 May 2014

This should do the trick -

index1 = A(:,1)<x(1,1) | A(:,1)>x(1,2);

index2 = A(:,1)<x(2,1) | A(:,1)>x(2,2);

A(index1 & index2, 2) = NaN

##### 2 Comments

Udit Gupta
on 29 May 2014

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