To generate alternate 0's and 1's

Say I have A=[2 4 3 4 100 253 77 10 45]. I want to generate alternate 0's and 1' such that we have 2 zeros (i.e, zeros equal to size of 1st element), then 4 ones (ones equal to the size of 2nd element) and so on

Result should be : [0 0 1 1 1 1 0 0 0 1 1 1 1 . . . .]. How to do this?

[Rik]

I'm guessing RunLength should help out.

Answer 1

[Andrei Bobrov]

out = repelem(mod(0:numel(A)-1,2),A);

[brijesh soni]

Thanks for the answer !

Answer 2

[dpb]

One brute force way--

 B=[];
 for i=1:numel(A)
   B=[B (1-mod(i,2))*ones(1,A(i))];
 end

Answer 3

[Jos (10584)]

bitget(repelem(0:numel(A)-1, A), 1)

[update] I modified my original but erroneous answer bitget(repelem(1:numel(A), A), 2). In his comments below, Andrei hit a serious flaw in this one. One does need to look at the last bit!

Thanks Andrei, for pointing this out.

[Andrei Bobrov]

maybe so:

bitget(repelem(2:numel(A)+1, A), 1)

[Jos (10584)]

Sure, but why add an extra + operation?

[Andrei Bobrov]

Hi Jos!

What am I doing wrong?

My laptop:

>> A=[2 4 3 4]
A =
     2     4     3     4
>> bitget(repelem(1:numel(A), A), 2) % your variant
ans =
     0     0     1     1     1     1     1     1     1     0     0     0     0
>> bitget(repelem(2:numel(A)+1, A), 1) % with change
ans =
     0     0     1     1     1     1     0     0     0     1     1     1     1
>> bitget(repelem(0:numel(A)-1, A), 1) % or
ans =
     0     0     1     1     1     1     0     0     0     1     1     1     1
>> 

[Jos (10584)]

Sorry! You're complete right to look only at the last bit!