Insert zeros in random positions following another zero

Hi All,

I have a large array such as this one: X = [2 1 0 4 5 0 778 90 0 3 88 0 77 66 0 12 23 0 45 80 0 89 67 0 34 67 0 76 32 0]; (just much larger)

I'd like to insert 20% of additional zeros at random positions following another 0 such as:

new_X = [2 1 0 0 4 5 0 778 90 0 3 88 0 77 66 0 12 23 0 0 45 80 0 89 67 0 34 67 0 76 32 0];

I need to find the indeces of the zeros in X but then I can't get the "random" 20% of zeros correct...

Does anybody know how to do that?

Kind regards

Phil

[Phillip]

% This does what I want...but do you know of a quicker way?

X = [2 1 0 4 5 0 778 90 0 3 88 0 77 66 0 12 23 0 45 80 0 89 67 0 34 67 0 76 32 0];

% Find indices of zeros in X

zero_indices = find(X == 0);

% Calculate additional 0s to insert (20% of the total number of 0 in X)

num_existing_zeros = length(zero_indices);

num_additional_zeros = round(0.2 * num_existing_zeros);

% Randomly select positions following existing 0s

additional_zero_positions = randsample(num_existing_zeros, num_additional_zeros);

% Sort 'em

additional_zero_positions = sort(additional_zero_positions);

% Insert 0s

for i = 1:num_additional_zeros

insert_index = zero_indices(additional_zero_positions(i)) + i;

X = [X(1:insert_index+1) 0 X(insert_index+2:end)];

zero_indices = zero_indices + 1;

end

Answer 1

[Steven Lord]

You can use repelem to replicate the elements.

X = [2 1 0 4 5 0 778 90 0 3 88 0 77 66 0 12 23 0 45 80 0 89 67 0 34 67 0 76 32 0];

Start off by "replicating" every element once.

replicationFactor = ones(size(X));

Identify where the zero elements are located.

iszero = X == 0;
zeroLocations = find(iszero)

zeroLocations = 1×10

3 6 9 12 15 18 21 24 27 30

numZeros = numel(zeroLocations)

numZeros = 10

Let's select 20% of the 0 locations (rounded up) to replicate twice. To do this, update the selected element in replicationFactor from 1 to 2.

elementsInZeroLocationsToReplicate = randperm(numZeros, ceil(0.2*numZeros));
randomSelection = zeroLocations(elementsInZeroLocationsToReplicate)

randomSelection = 1×2

6 15

replicationFactor(randomSelection) = 2

replicationFactor = 1×30

1 1 1 1 1 2 1 1 1 1 1 1 1 1 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

Now replicate with repelem.

Y = repelem(X, replicationFactor)

Y = 1×32

2 1 0 4 5 0 0 778 90 0 3 88 0 77 66 0 0 12 23 0 45 80 0 89 67 0 34 67 0 76

[Phillip]

Thank you!!!

Answer 2

[Davide Masiello]

This should work without the need for a for loop.

X = [2 1 0 4 5 0 778 90 0 3 88 0 77 66 0 12 23 0 45 80 0 89 67 0 34 67 0 76 32 0];

Select two random locations and increase the index by one (i.e. the new zero goes after the old random zero).

idx = sort(randsample(find(X==0),2));
idx = idx+[1:length(idx)]

idx = 1×2

4 32

Create a new vector of NaNs of the length required for the new vector.

newX = nan(1,length(X)+length(idx));

First place the new additional zeros.

newX(idx) = 0;

Then add the old vector where the NaNs are left.

newX(isnan(newX)) = X;

See the result below

disp(newX)

     2     1     0     0     4     5     0   778    90     0     3    88     0    77    66     0    12    23     0    45    80     0    89    67     0    34    67     0    76    32     0     0

[Phillip]

Thank you!!!