Efficient colon operator for multiple start and end points

Question

Suppose I have the following two variables:

start_idx = [1 4 7];
end_idx   = [2 6 15];

I want to efficiently (no for loop if possible) generate a single row which consists of the colon operator being applied between corresponding elements of start_idx and end_idx. For this example, this would result in:

result = [1:2 4:6 7:15];

Therefore:

results = [1 2 4 5 6 7 8 9 10 11 12 13 14 15];

The method to do this should be usable inside Simulink's MATLAB Function block. Thank you very much!

Answer 1

Here's a vectorized approach based on cumulative summation -

% Get lengths of each group
lens = end_idx - start_idx + 1;

% Determine positions in o/p array where groups would shift
shift_idx = cumsum(lens(1:end-1))+1

% Initialize ID array and at shifting positions place strategically created
% numbers, such that when ID array is cumulatively summed would result in
% desired "ramped" array
id_arr = ones(1,sum(lens));
id_arr([1 shift_idx]) = [start_idx(1) start_idx(2:end) - end_idx(1:end-1)];
out = cumsum(id_arr)

Sample run -

start_idx =
     6     8    13
end_idx =
    11    11    15
out =
     6     7     8     9    10    11     8     9    10    11    13    14    15

Answer 2

As I indicated in the comments, a one liner to solve this would be :

out=cell2mat(arrayfun(@(x,y)[x:y],start_idx,end_idx,'uniformoutput',false));

---

The arrayfun call will create a cell array whose each cell is a part of your output :

ans =

     1     2


ans =

     4     5     6


ans =

  Columns 1 through 8

     7     8     9    10    11    12    13    14

  Column 9

    15

By wrapping it inside a cell2mat call you get the expected output :

out =

  Columns 1 through 8

     1     2     4     5     6     7     8     9

  Columns 9 through 14

    10    11    12    13    14    15

Answer 3

This is cumbersome, but perhaps faster:

x = min(start_idx):max(end_idx);
m = sum(bsxfun(@ge,x,start_idx(:)),1)==numel(end_idx)-sum(bsxfun(@le,x,end_idx(:)),1)+1;
result = x(m);

It correctly handles empty ranges, i.e.

start_idx = [1 4 16]
end_idx   = [2 6 15];

gives

result =
     1     2     4     5     6

Answer 4

If you measure the elapsed time for this piece of code, after clearing the workspace, you will see that it takes an average of 0.004 sec while Divakar's code also takes roughly the same amount i.e. 0.007 sec..

start_idx=[2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 34 36 38 40 42 44];
end_idx=[100 200 300 400 500 600 700 800 900 1000 1100 1200 1300 ...
1400 1500 1600 1700 1800 1900 2000 2100 2200];
tic
final_arr=[];
for i=1:length(start_idx)
final_arr=[final_arr,start_idx(i):end_idx(i)];
end
toc
final_arr

As you can see, I have used start and end idx arrays of longer length and made sure that the end array elements are very far away from their respective start array elements.

The elapsed time which comes after the command 'toc' always changes according to the load on the CPU.. When I measure the time, I had only 1-2 apps open other than MATLAB and cleared the workspace before executing this code. The final_arr has a count of ~24k elements yet the time it took to process the output is not very much.

Hope it helps.