Matlab: Convert elements larger (smaller) than 1 (-1) into a sequence of 1 (-1) - matlab

Question

UPDATE: I've done some testing, and the solution of Jonas is the fastest for a range of different size input vectors. In particular, as angainor points out, the solution scales up to large sizes incredibly well - an important test as it is usually the large size problems that prompt us to pose these kind of questions on SO. Thanks to both Jonas and tmpearce for your solutions - based on the efficiency of the solution for large size problems I'm giving the answer tick to Jonas.
My Question: I have this column vector:
Vec = [0; 1; 2; -1; -3; 0; 0; 2; 1; -1];
I would like to convert every element greater than one into a sequence of ones that has length equal to the value of the element. Similarly, I want to convert every element less than minus one into a sequence of minus ones. Thus my output vector should look like this:
VecLong = [0; 1; 1; 1; -1; -1; -1; -1; 0; 0; 1; 1; 1; -1];
Note that each 2 has been changed into two 1's, while the -3 has been changed into three -1's. Currently, I solve the problem like this:
VecTemp = Vec;
VecTemp(VecTemp == 0) = 1;
VecLong = NaN(sum(abs(VecTemp)), 1);
c = 1;
for n = 1:length(Vec)
if abs(Vec(n)) <= 1
VecLong(c) = Vec(n);
c = c + 1;
else
VecLong(c:c + abs(Vec(n))) = sign(Vec(n));
c = c + abs(Vec(n));
end
end
This doesn't feel very elegant. Can anyone suggest a better method? Note: You can assume that Vec will contain only integer values. Thanks in advance for all suggestions.

Answer 1

You can use the good old cumsum-approach to repeating the entries properly. Note that I'm assigning a few temporary variables that you can get rid of, if you want to put everything into one line.
%# create a list of values to repeat
signVec = sign(Vec);
%# create a list of corresponding indices that repeat
%# as often as the value in signVec has to be repeated
tmp = max(abs(Vec),1); %# max: zeros have to be repeated once
index = zeros(sum(tmp),1);
index([1;cumsum(tmp(1:end-1))+1])=1; %# assign ones a pivots for cumsum
index = cumsum(index); %# create repeating indices
%# repeat
out = signVec(index);
out'
out =
0 1 1 1 -1 -1 -1 -1 0 0 1 1 1 -1

Answer 2

Edit: I thought of another (slightly obscure) but shorter way to do this, and it is faster than the loop you've got.
for rep=1:100000
#% original loop-based solution
end
toc
Elapsed time is 2.768822 seconds.
#% bsxfun-based indexing alternative
tic;
for rep=1:100000
TempVec=abs(Vec);TempVec(Vec==0)=1;
LongVec = sign(Vec(sum(bsxfun(#gt,1:sum(TempVec),cumsum(TempVec)))+1))
end
toc
Elapsed time is 1.798339 seconds.
This answer scales pretty well too, compared to the original - at least, to a point. There's a performance sweet spot.
Vec = repmat(OrigVec,10,1);
#% test with 100,000 loops
#% loop-based solution:
Elapsed time is 19.005226 seconds.
#% bsxfun-based solution:
Elapsed time is 4.411316 seconds.
Vec = repmat(OrigVer,1000,1);
#% test with 1,000 loops - 100,000 would be horribly slow
#% loop-based solution:
Elapsed time is 18.105728 seconds.
#% bsxfun-based solution:
Elapsed time is 98.699396 seconds.
bsxfun is expanding the vector into a matrix, then collapsing it with sum. With very large vectors this is needlessly memory heavy compared to the loop, so it ends up losing. Before then though, it does quite well.
Original, slow answer:
Here's a one-liner:
out=cell2mat(arrayfun(#(x) repmat(((x>0)*2)-1+(x==0),max(1,abs(x)),1),Vec,'uni',0));
out' =
0 1 1 1 -1 -1 -1 -1 0 0 1 1 1 -1
What's going on:
((x>0)*2)-1 + (x==0) #% if an integer is >0, make it a 1, <0 becomes -1, 0 stays 0
max(1,abs(x)) #% figure out how many times to replicate the value
arrayfun(#(x) (the above stuff), Vec, 'uni', 0) #% apply the function
#% to each element in the array, generating a cell array output
cell2mat( (the above stuff) ) #% convert back to a matrix