On Sun, 28 May 2017, Riccardo (Jack) Lucchetti wrote: anyway: something like this? <hansl> n = 100000 a = mrandgen(i,1,1000,n,2) target = 365 column = 1 sel = a[,column] .= target c = selifr(seq(1,n)'~a, sel) print c </hansl> ------------------------------------------------------- Riccardo (Jack) Lucchetti Dipartimento di Scienze Economiche e Sociali (DiSES) Università Politecnica delle Marche (formerly known as Università di Ancona) r.lucchetti(a)univpm.it http://www2.econ.univpm.it/servizi/hpp/lucchetti -------------------------------------------------------

On Sun, 28 May 2017, Marcin Błażejowski wrote: Thanks. However, it seems that a loop-based version is actually faster (much to my surprise): <hansl> set verbose off function scalar find_0(matrix a, scalar column, scalar target) n = rows(a) sel = a[,column] .= target c = selifr(seq(1,n)'~a, sel) return c[1,1] end function function scalar find_1(matrix a, scalar column, scalar target) n = rows(a) chk = 0 i = 1 loop while !chk --quiet chk = a[i++,column] == target endloop return i-1 end function n = 100000 a = mrandgen(i,1,1000,n,2) target = 365 column = 1 set stopwatch loop 1000 find_0(a, column, target) endloop t0 = $stopwatch loop 1000 find_1(a, column, target) endloop t1 = $stopwatch print t0 t1 </hansl> ------------------------------------------------------- Riccardo (Jack) Lucchetti Dipartimento di Scienze Economiche e Sociali (DiSES) Università Politecnica delle Marche (formerly known as Università di Ancona) r.lucchetti(a)univpm.it http://www2.econ.univpm.it/servizi/hpp/lucchetti -------------------------------------------------------

W dniu 28.05.2017 o 12:34, Riccardo (Jack) Lucchetti pisze: Jack, I don't get it: simpler (I think) version based on loop for/break is even faster: <hansl> set verbose off function scalar find_0(matrix a, scalar column, scalar target) n = rows(a) sel = a[,column] .= target c = selifr(seq(1,n)'~a, sel) return c[1,1] end function function scalar find_1(matrix a, scalar column, scalar target) #n = rows(a) chk = 0 i = 1 loop while !chk --quiet chk = a[i++,column] == target endloop return i-1 end function function scalar find_2(matrix a, scalar column, scalar target) n = rows(a) loop for i=1..n --quiet if (a[i,column] == target) index = i break endif endloop return index end function n = 100000 a = mrandgen(i,1,1000,n,2) target = 365 column = 1 set stopwatch loop 1000 find_0(a, column, target) endloop t0 = $stopwatch loop 1000 find_1(a, column, target) endloop t1 = $stopwatch loop 1000 find_2(a, column, target) endloop t2 = $stopwatch print t0 t1 t2 </hansl>

W dniu 28.05.2017 o 20:57, Riccardo (Jack) Lucchetti pisze: Ok, so do we known the theshold for staying at find_0 approach and switching to loop-based approach? Marcin

On Sun, 28 May 2017, Riccardo (Jack) Lucchetti wrote: Ok, two more things on this: (1) it turn out that if you combine the matrix-based algorithm with a divide-and-conquer strategy you do get some improvement. (2) The timings of the loop-based versions depend heavily on how many different values are contaned in the column to be serched: this is the scalar called H in the script below. Have fun. <hansl> set verbose off function scalar find_0(const matrix a, scalar column, scalar target) n = rows(a) c = a[,column] .= target if maxc(c) == 0 return NA endif c = selifr(seq(1,n)', c) return c[1,1] end function function scalar find_1(const matrix a, scalar column, scalar target) #n = rows(a) chk = 0 i = 1 loop while !chk --quiet chk = a[i++,column] == target endloop return i-1 end function function scalar find_2(const matrix a, scalar column, scalar target) n = rows(a) loop for i=1..n --quiet if (a[i,column] == target) index = i break endif endloop return index end function function scalar find_3(const matrix a, scalar column, scalar target) n = rows(a) h = floor(n/2) c = find_0(a[1:h,column], 1, target) if ok(c) return c else c = find_3(a[h+1:,column], column, target) endif return c end function n = 50000 H = 450 # try different values here a = mrandgen(i,1,H,n,2) column = 1 rep = 2000 u = mrandgen(i,1,H,n,1) set stopwatch loop i=1..rep find_0(a, column, u[i]) endloop t0 = $stopwatch loop i=1..rep find_1(a, column, u[i]) endloop t1 = $stopwatch loop i=1..rep find_2(a, column, u[i]) endloop t2 = $stopwatch loop i=1..rep find_3(a, column, u[i]) endloop t3 = $stopwatch print t0 t1 t2 t3 </hansl> ------------------------------------------------------- Riccardo (Jack) Lucchetti Dipartimento di Scienze Economiche e Sociali (DiSES) Università Politecnica delle Marche (formerly known as Università di Ancona) r.lucchetti(a)univpm.it http://www2.econ.univpm.it/servizi/hpp/lucchetti -------------------------------------------------------

On Mon, 29 May 2017, Riccardo (Jack) Lucchetti wrote: Oooof. c = find_3(a[h+1:,1], 1, target) should be c = find_3(a[h+1:,column], 1, target) I really should get into the habit of reading what I wrote before I send :) ------------------------------------------------------- Riccardo (Jack) Lucchetti Dipartimento di Scienze Economiche e Sociali (DiSES) Università Politecnica delle Marche (formerly known as Università di Ancona) r.lucchetti(a)univpm.it http://www2.econ.univpm.it/servizi/hpp/lucchetti -------------------------------------------------------

W dniu 28.05.2017 o 11:32, Riccardo (Jack) Lucchetti pisze: I take care of it and all values are uniqe for sure. -- Marcin Błażejowski