Element-wise array replication according to a coun

Question

This question already has an answer here:

• Repeat copies of array elements: Run-length decoding in MATLAB 5 answers

My question is similar to this one, but I would like to replicate each element according to a count specified in a second array of the same size.

An example of this, say I had an array v = [3 1 9 4], I want to use rep = [2 3 1 5] to replicate the first element 2 times, the second three times, and so on to get [3 3 1 1 1 9 4 4 4 4 4].

So far I'm using a simple loop to get the job done. This is what I started with:

vv = [];
for i=1:numel(v)
    vv = [vv repmat(v(i),1,rep(i))];
end

I managed to improve by preallocating space:

vv = zeros(1,sum(rep));
c = cumsum([1 rep]);
for i=1:numel(v)
    vv(c(i):c(i)+rep(i)-1) = repmat(v(i),1,rep(i));
end

However I still feel there has to be a more clever way to do this... Thanks

Answer 1

accumarray function can be used to make the code work if zeros exit in rep array

function vv = repeatElements(v, rep)
index = accumarray(cumsum(rep)'+1, 1);
vv = v(cumsum(index(1:end-1))+1);
end

This works similar to solution of gnovice, except that indices are accumulated instead being assigned to 1. This allows to skip some indices (3 and 6 in the example below) and remove corresponding elements from the output.

>> v = [3 1 42 9 4 42];
>> rep = [2 3 0 1 5 0];
>> index = accumarray(cumsum(rep)'+1, 1)'

index =

     0     0     1     0     0     2     1     0     0     0     0     2

>> cumsum(index(1:end-1))+1

ans =

     1     1     2     2     2     4     5     5     5     5     5

>> vv = v(cumsum(index(1:end-1))+1)

vv =

     3     3     1     1     1     9     4     4     4     4     4

Answer 2

What you are trying to do is to run-length decode. A high level reliable/vectorized utility is the FEX submission rude():

% example inputs
counts = [2, 3, 1];
values = [24,3,30];

the result

rude(counts, values)
ans =
    24    24     3     3     3    30

Note that this function performs the opposite operation as well, i.e. run-length encodes a vector or in other words returns values and the corresponding counts.

Answer 3

Here's one way I like to accomplish this:

>> index = zeros(1,sum(rep));
>> index(cumsum([1 rep(1:end-1)])) = 1;

index =

     1     0     1     0     0     1     1     0     0     0     0

>> index = cumsum(index)

index =

     1     1     2     2     2     3     4     4     4     4     4

>> vv = v(index)

vv =

     3     3     1     1     1     9     4     4     4     4     4

This works by first creating an index vector of zeroes the same length as the final count of all the values. By performing a cumulative sum of the rep vector with the last element removed and a 1 placed at the start, I get a vector of indices into index showing where the groups of replicated values will begin. These points are marked with ones. When a cumulative sum is performed on index, I get a final index vector that I can use to index into v to create the vector of heterogeneously-replicated values.

Answer 4

To add to the list of possible solutions, consider this one:

vv = cellfun(@(a,b)repmat(a,1,b), num2cell(v), num2cell(rep), 'UniformOutput',0);
vv = [vv{:}];

This is much slower than the one by gnovice..