Fast way to store different length Matrix rows into Cell - arrays

Question

I have three matrices:
Values = [200x7] doubles
numOfStrings = [82 78 75 73 72 71 70] %(for example)
numOfColumns = [1 4]
The numOfColumns may contain any set of distinct values from 1 to 7. For example [1 2 3] or [4 7] or [1 5 6 7]. Obviously, biggest numOfColumns that can be is [1 2 3 4 5 6 7].
numOfColumns show the columns I want to get. numOfStrings shows the rows of that columns I need. I.e. in my example I want to get columns 1 and 4. So from the 1 column I want to get the 82 first rows and from the 4th get the 73 first rows.
i.e. if numOfColumns = [1 4] then
myCell{1} = Values( 1:82,1); % Values(numOfStrings(numOfColumn(1)), numOfColumn(1))
myCell{2} = Values( 1:73,4); % Values(numOfStrings(numOfColumn(2)), numOfColumn(2))
P.S. That's not necessary to save it into cell array. If you can offer any another solution Ill be grateful to you.
I'm looking for the fastest way to do this, which is likely to be by avoiding for loops and using vectorization.
I think a lot about sub2ind function. But I can't figure out how to return arrays of the different size! Because myCell{1} - [82x1] and myCell{2} - [73x1]. I suppose I can't use bsxfun or arrayfun.
RESULTS:
Using for loops alike #Will 's answer:
for jj = 1:numel(numOfColumns)
myCell{rowNumber(numOfColumns(jj)),numOfColumns(jj)} = Values( 1:numOfStrings(numOfColumns(jj)),numOfColumns(jj));
end
Elapsed time is 157 seconds
Using arrayfun alike #Yishai E 's answer:
myCell(sub2ind(size(myCell),rowNumber(numOfColumns),numOfColumns)) = arrayfun( #(nOC) Values( 1:numOfStrings(nOC),nOC), numOfColumns, 'UniformOutput', false);
Elapsed time is 179 seconds
Using bsxfun alike #rahnema1 's answer:
idx = bsxfun(#ge,numOfStrings , (1:200).');
extracted_values = Values (idx);
tempCell = mat2cell(extracted_values,numOfStrings);
myCell(sub2ind(size(myCell),rowNumber(numOfColumns),numOfColumns)) = myCell(numOfColumns)';
Elapsed time is 204 seconds
SO, I got a lot of working answers, and some of them are vectorized as I asked, but for loops still fastest!

Answer 1

This should solve your problem, using arrayfun, which "vectorizes" the application of the indexing function. Not really, but it doesn't call the interpreter for each entry from numOfColumns. Interestingly enough, this is slower than the non-vectorized code in the other answer! (for 1e5 entries, 0.95 seconds vs. 0.23 seconds...)
arrayfun(#(nOC)Values(1:numOfStrings(nOC), nOC), numOfColumns, 'UniformOutput', false)

Answer 2

% Get number of elements in NUMOFCOLUMNS
n = numel(numOfColumns);
% Set up output
myCell = cell(1,n);
% Loop through all NUMOFCOLUMNS values, storing to cell
for i = 1:n
myCell{i} = Values(1:numOfStrings(numOfColumns(i)), numOfColumns(i));
end
Which for your example gives output
myCell =
[82x1 double] [73x1 double]

Answer 3

You can create logical indices for extraction of the desired elements :
idx = bsxfun(#ge,numOfStrings , (1:200).');
that in MATLAB R2016b or Octave (thanks to broadcasting/expansion) can be written as:
idx = numOfStrings >= (1:200).';
extract values:
extracted_values = Values (idx);
then using mat2cell convert data to cell :
myCell = mat2cell(extracted_values,numOfStrings);
all in one line :
myCell = mat2cell(Values (numOfStrings >= (1:200).'), numOfStrings);
If you want to use different numOfColumns with different sizes to extract elements of the cell you can each time do this:
result = myCell(numOfColumns);
If both numOfStrings and numOfColumns change and you need to compute the result once do this:
%convert numOfColumns to logical index:
numcols_logical = false(1,7);
numcols_logical(numOfColumns) = true;
extracted_values = Values ((numOfStrings .* numcols_logical) >= (1:200).');
if you need cell array
result= mat2cell(extracted_values,numOfStrings(numcols_logical));