Matthew Dowle
2012-Sep-03 02:04 UTC
[Rd] Possible page inefficiency in do_matrix in array.c
In do_matrix in src/array.c there is a type switch containing : case LGLSXP : for (i = 0; i < nr; i++) for (j = 0; j < nc; j++) LOGICAL(ans)[i + j * NR] = NA_LOGICAL; That seems page inefficient, iiuc. Think it should be : case LGLSXP : for (j = 0; j < nc; j++) for (i = 0; i < nr; i++) LOGICAL(ans)[i + j * NR] = NA_LOGICAL; or more simply : case LGLSXP : for (i = 0; i < nc*nr; i++) LOGICAL(ans)[i] = NA_LOGICAL; ( with some fine tuning required since NR is type R_xlen_t whilst i, nc and nr are type int ). Same goes for all the other types in that switch. This came up on Stack Overflow here : http://stackoverflow.com/questions/12220128/reason-for-faster-matrix-allocation-in-r Matthew
Simon Urbanek
2012-Sep-03 02:32 UTC
[Rd] Possible page inefficiency in do_matrix in array.c
On Sep 2, 2012, at 10:04 PM, Matthew Dowle wrote:> > In do_matrix in src/array.c there is a type switch containing : > > case LGLSXP : > for (i = 0; i < nr; i++) > for (j = 0; j < nc; j++) > LOGICAL(ans)[i + j * NR] = NA_LOGICAL; > > That seems page inefficient, iiuc. Think it should be : > > case LGLSXP : > for (j = 0; j < nc; j++) > for (i = 0; i < nr; i++) > LOGICAL(ans)[i + j * NR] = NA_LOGICAL; > > or more simply : > > case LGLSXP : > for (i = 0; i < nc*nr; i++) > LOGICAL(ans)[i] = NA_LOGICAL; > > ( with some fine tuning required since NR is type R_xlen_t whilst i, nc > and nr are type int ). > > Same goes for all the other types in that switch. > > This came up on Stack Overflow here : > http://stackoverflow.com/questions/12220128/reason-for-faster-matrix-allocation-in-r >That is completely irrelevant - modern compilers will optimize the loops accordingly and there is no difference in speed. If you don't believe it, run benchmarks ;) original> microbenchmark(matrix(nrow=10000, ncol=9999), times=10)Unit: milliseconds expr min lq median uq max 1 matrix(nrow = 10000, ncol = 9999) 940.5519 940.6644 941.136 954.7196 1409.901 swapped> microbenchmark(matrix(nrow=10000, ncol=9999), times=10)Unit: milliseconds expr min lq median uq max 1 matrix(nrow = 10000, ncol = 9999) 949.9638 950.6642 952.7497 961.001 1246.573 Cheers, Simon> Matthew > > ______________________________________________ > R-devel at r-project.org mailing list > https://stat.ethz.ch/mailman/listinfo/r-devel > >