CODEWITHSUNDEEP
rperformance
user5497
user5497

Reputation: 243

very slow function in R

I've tried implementing a log likelihood function in R. Here is the function I've used (I'm new to R)

f <- function(t)
{
 s=0
 x=d
 l = dim(x)[1]
 for (i in 1:l)
   {
        vector = d[i,]
        lin_res = t[1] + t[2] * vector[2] + t[3] * vector[3]
        yi = vector[1]
        s = s + yi*lin_res - log(1 + exp(lin_res))
   }
 return (s[1,1])
}

While d is small matrix with the following data:

   y x1         x2    x3         x4
1  0  1 0.29944294   5.0 0.71049142
2  0  2 0.12521669   6.0 0.20554934
3  1  3 0.97254701   3.0 0.43665094
4  0  4 0.79952796   1.0 0.64749898
5  0  5 0.77358425   9.0 0.57564913
6  0  6 0.09983754   5.0 0.32164782
7  1  7 0.46133893  10.0 0.86437213
8  0  8 0.59833493  20.0 0.72545982
9  0  9 0.80005524  80.0 0.35782812
10 0 10 0.02979412 115.0 0.76707371
11 1 11 0.70576655   1.5 0.96908006
12 0 12 0.67138962   2.0 0.37169164
13 0 13 0.33446510   8.0 0.23591223
14 1 14 0.72187427   2.0 0.98578941
15 0 15 0.28193852 200.0 0.87076869
16 1 16 0.11258881   3.0 0.05566943
17 0 17 0.22001868 100.0 0.98197495
18 1 18 0.54681964   4.0 0.53437931
19 0 19 0.03336023   5.0 0.26451825
20 1 20 0.47007378  10.0 0.28463580

For some reason, this function takes a lot of time (running this function 100 time takes ~7 seconds).

d <- structure(list(y = c(0L, 0L, 1L, 0L, 0L, 0L, 1L, 0L, 0L, 0L, 
1L, 0L, 0L, 1L, 0L, 1L, 0L, 1L, 0L, 1L), x1 = 1:20, x2 = c(0.299442944, 
0.125216695, 0.972547007, 0.799527959, 0.773584254, 0.099837539, 
0.461338927, 0.59833493, 0.800055241, 0.029794123, 0.705766552, 
0.671389622, 0.334465098, 0.721874271, 0.281938515, 0.112588815, 
0.220018683, 0.546819639, 0.033360232, 0.470073781), x3 = c(5, 
6, 3, 1, 9, 5, 10, 20, 80, 115, 1.5, 2, 8, 2, 200, 3, 100, 4, 
5, 10), x4 = c(0.710491422, 0.20554934, 0.436650943, 0.647498983, 
0.575649134, 0.321647815, 0.864372135, 0.725459824, 0.357828117, 
0.767073707, 0.969080057, 0.371691641, 0.23591223, 0.985789413, 
0.870768686, 0.055669431, 0.981974949, 0.534379314, 0.26451825, 
0.284635804)), .Names = c("y", "x1", "x2", "x3", "x4"), class = "data.frame", row.names = c(NA, 
-20L))

Can someone please help me in accelerating this function or understand what am I doing wrong?

Thanks!

Upvotes: 1

Views: 1355

Answers (2)

Bhas
Bhas

Reputation: 1834

@Andrie: R uses C (and Fortran in some places) code not C++.

@user5497: the main reason your loop is slow is because you are accessing a dataframe by row. Your d is not a matrix but a dataframe as can be seen from the class argument of structure.

Have a look at this.

f1 is your function

f2 is Julia's function

f2alt is f2 with d replaced by a matrix x as in f4

f3 is the byte compiled version of f1

f4 is the same as f1 but with d converted to a matrix in variable x and with vector set to x[i,]

f5 is the byte compiled version of f4

f1 <- function(t)
{
 s=0
 x=d
 l = dim(x)[1]
 for (i in 1:l)
   {
        vector = d[i,]
        lin_res = t[1] + t[2] * vector[2] + t[3] * vector[3]
        yi = vector[1]
        s = s + yi*lin_res - log(1 + exp(lin_res))
   }
 return (s[1,1])
}

f2 <- function(t)
{
    lin_res = t[1] + t[2] * d[2] + t[3] * d[3]
    return (sum(d[1]*lin_res - log(1 + exp(lin_res))))
}

f2alt <- function(t)
{   
    x <- as.matrix(d)
    lin_res = t[1] + t[2] * x[,2] + t[3] * x[,3]
    return (sum(x[,1]*lin_res - log(1 + exp(lin_res))))
}

library(compiler)
f3 <- cmpfun(f1)

f4 <- function(t)
{
 s <- 0
 x <- as.matrix(d)
 colnames(x) <- NULL
 l <- dim(x)[1]
 for (i in 1:l)
   {
        vector <- x[i,]
        lin_res <- t[1] + t[2] * vector[2] + t[3] * vector[3]
        yi <- vector[1]
        s <- s + yi*lin_res - log(1 + exp(lin_res))
   }
 return (s)
}

f5 <- cmpfun(f4)

tstart <- 1:3

f1(tstart)
f2(tstart)
f2alt(tstart)
f3(tstart)
f4(tstart)
f5(tstart)
all.equal(f1(tstart),f2(tstart))
all.equal(f1(tstart),f2alt(tstart))
all.equal(f1(tstart),f3(tstart))
all.equal(f1(tstart),f4(tstart))
all.equal(f1(tstart),f5(tstart))

library(rbenchmark)

benchmark(f1(tstart),f2(tstart),f2alt(tstart),f3(tstart),f4(tstart),f5(tstart),columns=c("test","elapsed","relative"))

The result is

           test elapsed   relative
1    f1(tstart)   6.912 460.800000
2    f2(tstart)   0.305  20.333333
3 f2alt(tstart)   0.015   1.000000
4    f3(tstart)   6.941 462.733333
5    f4(tstart)   0.032   2.133333
6    f5(tstart)   0.024   1.600000

As you can see byte compiling your function hardly makes a difference. f2 is quick but f2alt, f4 and f5 (byte compiled version of f4) are even quicker and only because they access a matrix and not a dataframe by row.

f2alt is a lot faster than the original f2 because a matrix is accessed and not a dataframe.

Warning: I use R-2.15.1 patched on Mac OS X which does not accept the standard rbenchmark; I have used a slightly modified version.

Upvotes: 6

julia
julia

Reputation: 152

I'm not sure what it does, but when using a loop always think if you could use vectorial operations. This code return the same value as your function f:

f2 <- function(t)
{   
    lin_res = t[1] + t[2] * d[2] + t[3] * d[3]
    return (sum(d[1]*lin_res - log(1 + exp(lin_res))))
}

Random data for t:

tt <- cbind( sample(0:100,100,replace=TRUE), sample(0:100,100,replace=TRUE), sample(0:100,100,replace=TRUE))

The time in my machine:

# original
ptm <- proc.time()
for (t in tt) f(t)
p <- proc.time() - ptm
print(p)
#    user  system elapsed 
#  25.529   0.002  25.533
# new
ptm <- proc.time()
for (t in tt) f2(t)
p <- proc.time() - ptm
print(p)
#    user  system elapsed 
#  1.612   0.001   1.614

Upvotes: 6

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