Le 26/11/10 21:13, Romain Francois a écrit :
Hello,
Can we really make the assumption that the data is sorted. The original
example was not:
I am working on a function to make a duplicated value unique. For
example,
the original vector would be like : a = c(2,1,1,3,3,3,4)
If we can make the assumption, here is a C++ based version:
nodup_cpp_assumingsorted <- cxxfunction( signature( x_ = "numeric" ), '
// since we modify x, we need to make a copy
NumericVector x = clone<NumericVector>(x_);
int n = x.size() ;
double current, previous = x[0] ;
int index ;
for( int i=1; i<n; i++){
current = x[i] ;
if( current == previous ){
x[i] = current + (++index) / 100.0 ;
} else {
index = 0 ;
}
previous = current ;
}
return x ;
', plugin = "Rcpp" )
with these results:
> x <- sort( sample( 1:100000, size = 300000, replace = TRUE ) )
> system.time( nodup3( x ) )
utilisateur système écoulé
0.090 0.004 0.094
> system.time( nodup3a( x ) )
utilisateur système écoulé
0.036 0.005 0.040
> system.time( nodup4( x ) )
utilisateur système écoulé
0.025 0.004 0.029
> system.time( nodup_cpp_assumingsorted( x) )
utilisateur système écoulé
0.003 0.001 0.004
Now, if we don't make the assumption that the data is sorted, here is
another C++ based version:
require( inline )
require( Rcpp )
nodup_cpp <- cxxfunction( signature( x_ = "numeric" ), '
// since we modify x, we need to make a copy
NumericVector x = clone<NumericVector>(x_);
typedef std::map<double,int> imap ;
typedef imap::value_type pair ;
imap index ;
int n = x.size() ;
double current, previous = x[0] ;
index.insert( pair( previous, 0 ) );
imap::iterator it = index.begin() ;
for( int i=1; i<n; i++){
current = x[i] ;
if( current == previous ){
x[i] = current + ( ++(it->second) / 100.0 ) ;
} else {
it = index.find(current) ;
if( it == index.end() ){
it = index.insert(
current > previous ? it : index.begin(),
pair( current, 0 )
) ;
} else {
x[i] = current + ( ++(it->second) / 100.0 ) ;
}
previous = current ;
}
}
return x ;
', plugin = "Rcpp" )
which gives me this :
> x <- sample( 1:100000, size = 300000, replace = TRUE )
>
> system.time( nodup_cpp( x ) )
utilisateur système écoulé
0.111 0.002 0.113
> system.time( nodup3( sort( x ) ) )
utilisateur système écoulé
0.162 0.011 0.172
> system.time( nodup3a( sort( x ) ) )
utilisateur système écoulé
0.099 0.009 0.109
> system.time( nodup4( sort( x ) ) )
utilisateur système écoulé
0.089 0.004 0.094
so nodup4 is still faster, but the values are not in the right order:
> x <- c( 2, 1, 1, 2 )
> nodup4( sort( x ) )
[1] 1.00 1.01 2.00 2.01
> nodup_cpp( x )
[1] 2.00 1.00 1.01 2.01
Romain
I think this gives a more fair comparison :
> system.time( nodup_cpp( x ) )
utilisateur système écoulé
0.113 0.002 0.114
> system.time( { oo <- order(order(x)) ; nodup3( sort( x ) )[oo] } )
utilisateur système écoulé
0.336 0.012 0.347
> system.time( { oo <- order(order(x)) ; nodup3a( sort( x ) )[oo] } )
utilisateur système écoulé
0.251 0.011 0.262
> system.time( { oo <- order(order(x)) ; nodup4( sort( x ) )[oo] } )
utilisateur système écoulé
0.287 0.006 0.294
Romain
Le 26/11/10 20:01, William Dunlap a écrit :
-----Original Message-----
From: William Dunlap
Sent: Thursday, November 25, 2010 9:31 AM
To: 'randomcz'; r-help@r-project.org
Subject: RE: [R] help: program efficiency
If the input vector t is known to be ordered
(or if you only care about runs of duplicated
values, not all duplicated values) the following
is pretty quick
nodup3<- function (t) {
t + (sequence(rle(t)$lengths) - 1)/100
}
If you don't know if the the input will be ordered
then ave() will do it a bit faster than your
code
nodup2<- function (t) {
ave(t, t, FUN = function(x) x + (seq_along(x) - 1)/100)
}
E.g., for a sorted sequence of 300,000 numbers drawn with
replacement from 1:100,000 I get:
a2<- sort(sample(1:1e5, size=3e5, replace=TRUE))
system.time(v<- nodup(a2))
user system elapsed
2.78 0.05 3.97
system.time(v2<- nodup2(a2))
user system elapsed
1.83 0.02 2.66
system.time(v3<- nodup3(a2))
user system elapsed
0.18 0.00 0.14
identical(v,v2)&& identical(v,v3)
[1] TRUE
If speed is truly an issue, the built-in sequence may
be replaced by a faster one that does the same thing:
nodup3a<- function (t) {
faster.sequence<- function(nvec) {
seq_len(sum(nvec)) - rep(cumsum(c(0L, nvec[-length(nvec)])),
nvec)
}
t + (faster.sequence(rle(t)$lengths) - 1)/100
}
That took 0.05 seconds on the a2 dataset and produced
identical results.
rle() computes a sort of second difference and
nodup3a computes a cumsum on that second diffence,
to get back to a first difference. The following
avoids that wasted operation (along with rle's
computation of the values component of its output).
nodup4<- function(t) {
n<- length(t)
p<- c(0L, which(t[-1L] != t[-n]), n)
t + ( seq_len(n) - rep.int(p[-length(p)] + 1L, diff(p)) ) /100
}
That reduced nodup3a's time by about 30% on that dataset.
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
-----Original Message-----
From: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org] On Behalf Of randomcz
Sent: Thursday, November 25, 2010 6:49 AM
To: r-help@r-project.org
Subject: [R] help: program efficiency
hey guys,
I am working on a function to make a duplicated value unique.
For example,
the original vector would be like : a = c(2,1,1,3,3,3,4)
I'll like to transform it into:
a.nodup = 2, 1.01, 1.02, 3.01, 3.02, 3.03, 4
basically, find the duplicates and assign a unique value by
adding a small
amount and keep it in order.
I come up with the following codes, but it runs slow if t is
large. Is there
a better way to do it?
nodup = function(t)
{
t.index=0
t.dup=duplicated(t)
for (i in 2:length(t))
{
if (t.dup[i]==T)
t.index=t.index+0.01
else t.index=0
t[i]=t[i]+t.index
}
return(t)
}
--
View this message in context:
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9p3059079.html
Sent from the R help mailing list archive at Nabble.com.
______________________________________________
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______________________________________________
R-help@r-project.org mailing list
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PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
Romain Francois
Professional R Enthusiast
+33(0) 6 28 91 30 30
http://romainfrancois.blog.free.fr
|- http://bit.ly/9VOd3l : ZAT! 2010
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______________________________________________
R-help@r-project.org mailing list
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
