# Why is subscripted assignment so inefficient?

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Why is MATLAB's subscripted assignment so abysmally inefficient with more than 2 subscripts? e.g.,

>> clear, a = []; tic, for j = 1e3:-1:1, for k = 1e3:-1:1, a(j,k,1) = 1; end, end, toc

Elapsed time is 0.193743 seconds.

>> clear, a = []; tic, for j = 1e3:-1:1, for k = 1e3:-1:1, a(j,k) = 1; end, end, toc

Elapsed time is 0.004241 seconds.

##### 1 Comment

Walter Roberson
on 5 Feb 2022

### Accepted Answer

Matt J
on 5 Feb 2022

Edited: Matt J
on 5 Feb 2022

I would guess it's only because you are using 3 subscripts on an array that only has 2 dimensions. This forces Matlab to to do extra work, namely running a check to make sure that in a(j,k,m) the 3rd subscript is m=1 and simplifying the subscripts from (j,k,1) to (j,k). When a actualy does have 3 dimensions, the times are much more comparable:

a=zeros(1e3,1e3,2);

tic,

for j = 1e3:-1:1

for k = 1e3:-1:1

a(j,k,1) = 1;

end

end

toc

a=zeros(1e3,1e3);

tic,

for j = 1e3:-1:1

for k = 1e3:-1:1

a(j,k) = 1;

end

end

toc

##### 7 Comments

Walter Roberson
on 6 Feb 2022

There is a lot of noise in these measurements, and it is difficult to tell what is meaningful or not.

I skip the first few runs in the plot because in nearly all runs there is a bad peak early on.

The second plot here is the more interesting one, showing the log of how the time to allocate increases with amount requested. Noise affects it a lot, so I can only describe trends that I encountered a fair bit:

- time to allocate 1 byte starts off a bit higher
- time to allocate a quite small number of bytes but more than 1, falls a bit, then rises a bit.
- there does seem to be a peak at 256 bytes
- and then it does seem to fall a bit
- until at 2^14 there seems to be a distinct rise. This would correspond to 16384 bytes. I suspect that either 4096 or 16384 bytes is the size of an entry in the "small block pool"
- in some of the plots, there was a leveling from 2^14 to 2^16; the more fine-grained I draw the less I see this, so it might have been a statistical artificat
- from 2^16 onward, the log grows pretty much linearly as log2() of number of bytes increases. To phrase that a different way: from 2^16 onward, the allocation time grows linearly with number of bytes allocated. The actual boundary might possibly be 2^14
- so, below 2^14, there appear to be at least two different allocation strategies, but the boundaries between them are a bit difficult to find. If there were only one allocation strategy for that range, using a free block pool with entries of size 2^14, then you would expect the time to be constant up to that point, but there do seem to be peaks.
- I seem to recall that 256 bytes is the limit below which for some constant allocations and some colon expressions, that MATLAB keeps a copy around as an optimization, to be handed out when encountering another request with the same spelling (spacing even being important!). Perhaps that is why we see a blip at 256??

If someone cares to do a study of how small block allocation is done in MATLAB, I suggest 2^14 as the upper limit to study (and remember to take into account those new hidden copies that MATLAB recently starting taking for small constants.)

N = 50;

G = 2^24;

start = datetime('now');

%timing pushed into a function to allow for potential JIT

[t, gvals] = run_timing(N, G);

stop = datetime('now');

elapsed = seconds(stop - start);

display(elapsed)

figure(1)

%tfilt = filloutliers(t, 'center','movmedian', 3, 1);

startat = 6;

semilogy(startat:N, t(startat:end,:));

xlabel('try #'); ylabel('seconds')

title('absolute time')

%legend("2^{"+string(gvals)+"}", 'NumColumns',4);

figure(2)

tmean = mean(t(startat:end,:),1);

semilogy(gvals, tmean)

xt = 1:ceil(max(gvals));

xticks(xt)

xticklabels("2^{"+string(xt)+"}");

xtickangle(45)

xlabel('bytes')

ylabel('mean seconds')

plin = polyfit(2.^gvals, tmean, 1);

%plog = polyfit(log2(gvals), log2(tmean),1);

fprintf('linear: roughly %g seconds + %g seconds per gigabyte\n', plin(2), plin(1)*2^30);

%fprintf('log: roughly %g seconds + %g seconds per gigabyte\n', 2.^plog(2), plog(1));

function [t,gvals] = run_timing(N,G)

%wb = waitbar(0, '0G');

%cleanMe = onCleanup(@()delete(wb));

gvals = 0:.1:ceil(log2(G));

ng = length(gvals);

t = zeros(N,ng);

for g = 1 : ng

gp = gvals(g);

gv = round(2.^gp);

%waitbar(0, wb, "2^{"+string(gp)+"}");

for n = 1 : N

tic();

initmem(gv);

t(n,g) = toc;

%waitbar(n/N, wb);

end

end

end

%gigabyte 1073741824

%megabyte 1048576

function initmem(M)

ma = ones(1,M,'uint8');

end

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