Brendan Halpin <[email protected]> asked about speeding up the Mata code
to perform
for (i=2; i<=length(s1) + 1; i++) {
for (j=2; j<=length(s2) + 1; j++) {
D[i, j] = min((D[i, j-1] + A[i, j],
D[i-1, j ] + B[i, j],
D[i-1, j-1] + C[i, j] ));
}
}
and he asked if there was a way to time-profile Mata code.
Concerning the latter, see -help mata timer()-.
Concerning the former, I succeeded in reducing the run time by 49%.
I started with the following do-file:
------------------------------------------- time.do --------
clear
local N 100
mata:
void tryit()
{
timer_clear() // clear timers
A = B = C = D = J(`N'+1, `N'+1, 0)
s1 = s2 = J(`N', 1, 0)
timer_on(1) // start timer 1
for (i=2; i<=length(s1)+1; i++) {
for (j=2; j<=length(s2)+1; j++) {
D[i,j] = min((D[i, j-1] + A[i,j],
D[i-1, j ] + B[i,j],
D[i-1, j-1] + C[i,j]))
}
}
timer_off(1) // stop timer 1
timer(1) // report timer 1
}
tryit()
end
------------------------------------------- time.do --------
I then decided one timing did not produce stable enough results and
changed the code to read:
------------------------------------------- time.do --------
clear
local N 100
mata:
void tryit()
{
timer_clear() // clear timers
A = B = C = D = J(`N'+1, `N'+1, 0)
s1 = s2 = J(`N', 1, 0)
for (k=1; k<=50; k++)
timer_on(1) // start timer 1
for (i=2; i<=length(s1)+1; i++) {
for (j=2; j<=length(s2)+1; j++) {
D[i,j] = min((D[i, j-1] + A[i,j],
D[i-1, j ] + B[i,j],
D[i-1, j-1] + C[i,j]))
}
}
timer_off(1) // stop timer 1
}
timer(1) // report timer 1
}
tryit()
end
------------------------------------------- time.do --------
timer(1) reported a run time for 50 iterations of 1.95 seconds and thus
.03906 seconds per iteration.
I then set about playing to speed up the loop. My final version of time.do
read,
------------------------------------------- time.do --------
clear
local N 100
mata:
void tryit()
{
timer_clear() // clear timers
A = B = C = D = J(`N'+1, `N'+1, 0)
s1 = s2 = J(`N', 1, 0)
for (k=1; k<=50; k++)
timer_on(1) // start timer 1
n1 = length(s1) + 1
n2 = length(s2) + 1
for (i=2; i<=length(s1)+1; i++) {
i1 = i-1
for (j=2; j<=length(s2)+1; j++) {
j1 = j-1
a = D[ i, j1] + A[i,j]
b = D[i1, j] + A[i,j]
c = D[i1, j1] + A[i,j]
D[i,j] = (a<b ? (a<c ? a : c) :
(b<c ? b : c))
}
}
timer_off(1) // stop timer 1
}
timer(1) // report timer 1
}
tryit()
end
------------------------------------------- time.do --------
Run time for 50 interations fell from 1.95 to 1.14 seconds.
-- Bill
[email protected]
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