Select all entries in first N-1 dimensions of array

hi, i have a loop in which cell array V keeps expanding:

for it=1:10;
  V{it}= ..... % the number of dimension of V{it} is 'it'
                  % so if it=3: size(V{it})=[2 2 2];
end

later im going to loop through V again, but i want to select only the first N dimensions:

for id=1:9;
  Y{id}=  ... % select first 'id' dimensions of V{id+1}
%so if id=2, W{id}=V{id+1}(:,:,1), but if id=3,  W{id}=V{id+1}(:,:,:,1)

end

I just dont know how to adjust the number of ':' i want to be there. Or equivalently to adjust the numer of times to put in 1:2, ie. W{4}=V{5}(1:2,1:2,1:2,1:2,1).

[Sean de Wolski]

ARGGGHHHH!!

I just went to compliment Oleg for his answer and to up-vote it and it somehow got deleted. My apologies, Oleg and Sargondjani. Hopefully you have the Comma Separated List Expansion somewhere. Oleg, I owe you a beer.

[Sargondjani]

well, i dont care, i have the answer, hahaha

Oleg wrote something like:

v = [repmat({':'},id-1,1);{1}];

Y{id}=V{id+1}(v{:})

[Sargondjani]

o and i thanked oleg, hehe

Answer 1

[Daniel Shub]

Have you thought about using linear indexing?

V{1} = rand([2,2,2]);
x = reshape(V{1}(1:4), repmat(2, 1, 2));
y = V{1}(1:2, 1:2);
isequal(x,y)

EDIT To give a full working example

Create some data:

for it=1:10;
  V{it}= rand([repmat(2, 1, it), 1]);
end

Then get the data you want

for id=1:9;
  Y{id} = reshape(V{id+1}(1:(2^(id))), [repmat(2, 1, id), 1]);
end

[Sargondjani]

the problem is that i dont see how to increase the number of "1:2" after the y=V{1}

but thanks anyway. the problem is solve with the solution that oleg gave...

[Sargondjani]

i see your point after the edit. thanks!! i will need this type of linear indexing later on :-)