Most efficient way to vertically concatenate numeric data?

In the profiled output picture below, mergedDataPerRank is a cell array storing 53 double matrices with size (6300, 33). Using vertcat, it takes approx. 0.5 seconds (20 calls) to vertically concatenate the data. Is there a more efficient way to do this or is 0.5 seconds already fairly good?

[Walter Roberson]

0.5 seconds is already good for 20 calls each vertically concatenating 53 of 6300 x 33;

C = arrayfun(@(idx) rand(6300,33), (1:53).', 'uniform', 0);

tic; for K = 1 : 20; result1 = cell2mat(C); end; t1 = toc;
tic; for K = 1 : 20; result2 = vertcat(C{:}); end; t2 = toc;
format long g
[t1, t2]

ans = 1×2

1.811011 0.86993

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On my desktop, the vertcat takes roughly 0.286 seconds

[Steven Lord]

nrows = 6300;
ncols = 33;
ncells = 53;
C = arrayfun(@(idx) rand(nrows,ncols), (1:ncells).', 'uniform', 0);

tic; for K = 1 : 20; result1 = cell2mat(C); end; t1 = toc;
tic; for K = 1 : 20; result2 = vertcat(C{:}); end; t2 = toc;

Let's look at the "preallocate and fill" approach as an alternative. Of course, this code assumes that each cell contains a matrix of the same size and all of them are double matrices.

tic;
for k = 1:20
    result3 = zeros(nrows*ncells, ncols);
    beforefirst = 0;
    for whichcell = 1:ncells
        result3(beforefirst+(1:nrows), :) = C{whichcell};
        beforefirst = beforefirst + nrows;
    end
end
t3 = toc;

format long g
[t1, t2, t3]

ans = 1×3

1.878186 0.88884 0.856936

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Its time is pretty close to the vertcat approach. Do the three approaches give the same answer?

check = isequal(result1, result2, result3)

check = logical
   1

[Konstantin]

But what is "Merge_filesAcrossRanks"? It looks like it is some your own custom function to read a lot of files in a given directory. If it is so (and if the size of all tables is quaranteed a priori), then I would recommend to combine reading and merging: just preallocate the whole final matrix (53*6300 tall, 33 wide), create a "current position (line)" variable, and then read files into this table while advancing the "current position".

[Walter Roberson]

@Konstantin

That approach turns out to be slower.

nmat = 53;
nrow = 6300;
ncol = 33;

rng(12345)
tic
C = cell(nmat, 1);
for K = 1 : nmat; C{K} = rand(nrow, ncol); end
R1 = vertcat(C{:});
t1 = toc;

rng(12345);
R2 = zeros(nmat*nrow,ncol);
counter = 1;
for K = 1 : nmat; R2(counter:counter+nrow-1,:) = rand(nrow, ncol); counter = counter + nrow; end
t2 = toc;

format long g
[t1, t2]

ans = 1×2

0.159601 0.273057

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