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From: Walter Roberson on 18 Jun 2010 22:43 wahyoe Unggul wrote: > I do not use Matlab profiler, I calculate the time to execute a matrix > with the manual, I think there is an error function "assembly", when > entering data (matrix) small size can be executed quickly but instead, > jiak data (matrix) large size is very slow (time old) to be executed, > can you help to correct the above functions so that data (matrix) which > can be executed with large quick times It appears I have to repeat myself as you are having trouble finding my previous posting: === begin what I already posted === Is it correct that you do not care what the values of B and C are after the loop? If so then your loop becomes for i = 1:length(tA) A(tA(i,2),tA(i,3)) = tA(i,1) + A(tA(i,2),tA(i,3)); end Which can be optimized to a plain vector calculation as T = sub2ind(size(A),tA(i,2),tA(i,3)); A(T) = A(T) + tA(:,1); If, though, the tA(i,2),tA(i,3) pairs might be duplicated, then T = sub2ind(size(A),tA(i,2),tA(i,3)); T1 = accumarray(T,tA(:,1)); T2 = unique(T); A(T2) = A(T2) + T1(T2); === end what I already posted === Your existing code does a lot of needless copying. Your existing code does not pre-allocate memory properly for the case where max(tA(:,3)) is greater than max(tA(:,2)) *Are* duplicate pairs of tA(i,2), tA(i,3) possible, or is every pair unique?
From: wahyoe Unggul on 19 Jun 2010 07:02 Walter Roberson <roberson(a)hushmail.com> wrote in message <G9WSn.1$Yo5.0(a)newsfe01.iad>... > wahyoe Unggul wrote: > > I do not use Matlab profiler, I calculate the time to execute a matrix > > with the manual, I think there is an error function "assembly", when > > entering data (matrix) small size can be executed quickly but instead, > > jiak data (matrix) large size is very slow (time old) to be executed, > > can you help to correct the above functions so that data (matrix) which > > can be executed with large quick times > > It appears I have to repeat myself as you are having trouble finding my > previous posting: > > === begin what I already posted === > > Is it correct that you do not care what the values of B and C are after > the loop? If so then your loop becomes > > for i = 1:length(tA) > A(tA(i,2),tA(i,3)) = tA(i,1) + A(tA(i,2),tA(i,3)); > end > > Which can be optimized to a plain vector calculation as > > T = sub2ind(size(A),tA(i,2),tA(i,3)); > A(T) = A(T) + tA(:,1); > > > If, though, the tA(i,2),tA(i,3) pairs might be duplicated, then > > T = sub2ind(size(A),tA(i,2),tA(i,3)); > T1 = accumarray(T,tA(:,1)); > T2 = unique(T); > A(T2) = A(T2) + T1(T2); > > > === end what I already posted === > > Your existing code does a lot of needless copying. > > Your existing code does not pre-allocate memory properly for the case > where max(tA(:,3)) is greater than max(tA(:,2)) > > *Are* duplicate pairs of tA(i,2), tA(i,3) possible, or is every pair unique? +++++++++++++++++++++++++++++++++++++++++++++++++ very-very fantastic thank ypu very^2 much of B and C do not affect the function or loop after replacing my script for i=1:length(tA); > C=A; > B(tA(i,2),tA(i,3))=tA(i,1); > A(tA(i,2),tA(i,3))=B(tA(i,2),tA(i,3))+C(tA(i,2),tA(i,3)); > end; into this script for i = 1:length(tA) > A(tA(i,2),tA(i,3)) = tA(i,1) + A(tA(i,2),tA(i,3)); > end very^2 fantastic I once had to wait hours to complete the data (matrix) are large, now less than a minute I got some results whether additional below inserted after the "end" or between "for" and "end"? T = sub2ind(size(A),tA(i,2),tA(i,3)); > A(T) = A(T) + tA(:,1); > > or > T = sub2ind(size(A),tA(i,2),tA(i,3)); > T1 = accumarray(T,tA(:,1)); > T2 = unique(T); > A(T2) = A(T2) + T1(T2); thanks before
From: Walter Roberson on 19 Jun 2010 09:44 wahyoe Unggul wrote: > whether additional below inserted after the "end" or between "for" and > "end"? > T = sub2ind(size(A),tA(i,2),tA(i,3)); >> A(T) = A(T) + tA(:,1); >> >> or >> T = sub2ind(size(A),tA(i,2),tA(i,3)); >> T1 = accumarray(T,tA(:,1)); >> T2 = unique(T); >> A(T2) = A(T2) + T1(T2); Either of those *replace* the loop. The first of them can be used if there are no cases in which [tA(i,2),tA(i,2)] == [tA(k,2),tA(k,2)] for some i and k. The second of them should work (I think) in the case where there are repeated pairs.
From: wahyoe Unggul on 19 Jun 2010 13:49 Walter Roberson <roberson(a)hushmail.com> wrote in message <aR3Tn.8779$1Q5.2332(a)newsfe08.iad>... > wahyoe Unggul wrote: > > > whether additional below inserted after the "end" or between "for" and > > "end"? > > T = sub2ind(size(A),tA(i,2),tA(i,3)); > >> A(T) = A(T) + tA(:,1); > >> > >> or > >> T = sub2ind(size(A),tA(i,2),tA(i,3)); > >> T1 = accumarray(T,tA(:,1)); > >> T2 = unique(T); > >> A(T2) = A(T2) + T1(T2); > > Either of those *replace* the loop. The first of them can be used if > there are no cases in which [tA(i,2),tA(i,2)] == [tA(k,2),tA(k,2)] for > some i and k. The second of them should work (I think) in the case where > there are repeated pairs. ================================================== case back to the beginning of the function if the C and B influence on the loop, how to modify the loop in order to complete the matrix that is big enough?
From: Walter Roberson on 19 Jun 2010 14:19
wahyoe Unggul wrote: > Walter Roberson <roberson(a)hushmail.com> wrote in message > <aR3Tn.8779$1Q5.2332(a)newsfe08.iad>... >> wahyoe Unggul wrote: >> >> > whether additional below inserted after the "end" or between "for" >> and > "end"? >> > T = sub2ind(size(A),tA(i,2),tA(i,3)); >> >> A(T) = A(T) + tA(:,1); >> >> >> >> or >> >> T = sub2ind(size(A),tA(i,2),tA(i,3)); >> >> T1 = accumarray(T,tA(:,1)); >> >> T2 = unique(T); >> >> A(T2) = A(T2) + T1(T2); >> >> Either of those *replace* the loop. The first of them can be used if >> there are no cases in which [tA(i,2),tA(i,2)] == [tA(k,2),tA(k,2)] for >> some i and k. The second of them should work (I think) in the case >> where there are repeated pairs. > ================================================== > case back to the beginning of the function if the C and B influence on > the loop, how to modify the loop in order to complete the matrix that is > big enough? I do not understand your question, sorry. |