Re: CALCULATION TIME

*To*: mathgroup at smc.vnet.net*Subject*: [mg21431] Re: CALCULATION TIME*From*: "Allan Hayes" <hay at haystack.demon.co.uk>*Date*: Fri, 7 Jan 2000 00:20:41 -0500 (EST)*References*: <851ek8$4b5@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

Jerry, Re the summation: Why not use Dot? Example: y = Table[Random[], {30000}]; Sum[y[[i]]^2, {i, 20000}] // Timing {2.42 Second, 6672.59} #.# &[Take[y, 20000]] // Timing {0.39 Second, 6672.59} Allan --------------------- Allan Hayes Mathematica Training and Consulting Leicester UK www.haystack.demon.co.uk hay at haystack.demon.co.uk Voice: +44 (0)116 271 4198 Fax: +44 (0)870 164 0565 "Blimbaum Jerry DLPC" <BlimbaumJE at ncsc.navy.mil> wrote in message news:851ek8$4b5 at smc.vnet.net... > I have some code performing (roughly) the following Do Loops: > > Do[......... > Do[x[[m,n+1]] = x[[m,n]] + B* y[[n+m-1]]*e[[n]] > > ----------------------------------------------, > > eps+Sum[y[[n-q]]^2,{q,0,M-1}] > {m,M} ], > > {n,M,Length[y]}] > > where y, e are lists (y is data that contains about 30000 elements), > etc. Running a data set takes about 10.5 secs. I wanted to shorten the > time but couldnt figure out how to write it either in Nest or Compile form > so I went thru it piece meal to see if I could find a culprit and from this > I noticed if I replace the Sum term in the denominator by a number, lets say > 1, and then perform the calculations it only takes about 2.5 seconds! I > did a sample Sum[expr^2] with a similar Do loop separately and it takes no > time at all. > > What is wrong? > > thanks. Jerry Blimbaum NSWC Panama City, Fl >