Please don't post in HTML... you may not recognize it, but the receiving
end does not necessarily (and in this case did not) look like the sending
end, and the cleanup can impede answers you are hoping to get.
In many cases, loops can be vectorized. However, near as I can tell this
is an example of an algorithm that simply needs a loop [1].
One bit of advice: the coredata function is horribly slow. Just converting
your time series objects to numeric vectors for the purpose of this
computation sped up the algorithm by 500x on 10000 point series.
Converting it to inline C++ as below sped it up by yet another factor of
40x. 20000x is nothing to sneeze at.
#######
## optional temporary setup for windows
## assumes you have installed Rtools
gcc <- "C:\\Rtools\\bin"
rtools <- "C:\\Rtools\\gcc-4.6.3\\bin"
path <- strsplit(Sys.getenv("PATH"), ";")[[1]]
new_path <- c(rtools, gcc, path)
new_path <- new_path[!duplicated(tolower(new_path))]
Sys.setenv(PATH = paste(new_path, collapse = ";"))
## end of optional
library(Rcpp)
cppFunction(
"DataFrame EvapSimRcpp( NumericVector RR
, NumericVector ETmax
, const double Smax
, const double initialStorage ) {
int n = RR.size();
// create empty time-series to fill
// effective rainfall (i.e. rainfall minus intercepted rainfall)
NumericVector RReff( n );
// intercepted rainfall( n );
NumericVector Rint( n );
// residual potential evapotranspiration (ie ETmax minus
// evaporation from interception)
NumericVector ETres( n, NA_REAL );
double evap;
// volume of water in interception storage at start of
// computation
double storage = initialStorage;
for ( int i=0; i<n; i++ ) {
// compute interception capacity for time step i (maximum
// interception capacity minus any water intercepted but not
// evaporated during previous time-step).
Rint[ i ] = Smax - storage;
// compute intercepted rainfall: equal to rainfall if smaller
// than interception capacity, and to interception capacity if
// larger.
if ( RR[ i ] < Rint[ i ] ) Rint[ i ] = RR[ i ];
// compute effective rainfall (rainfall minus intercepted
// rainfall).
RReff[ i ] = RR[ i ] - Rint[ i ];
// update interception storage: initial interception storage +
// intercepted
// rainfall.
storage = storage + Rint[ i ];
// compute evaporation from interception storage: equal to
// potential evapotranspiration if the latter is smaller than
// interception storage, and to interception storage if larger.
if ( storage > ETmax[ i ] )
evap = ETmax[ i ];
else
evap = storage;
// compute residual potentiel evapotranspiration: potential
// evapotranspiration minus evaporation from interception
// storage.
ETres[ i ] = ETmax[ i ] - evap;
// update interception storage, to be carried over to next
// time-step: interception storage minus evaporation from
// interception storage.
storage = storage - evap;
}
DataFrame DF = DataFrame::create( Named( \"int\" ) = Rint
, Named( \"RReff\" ) = RReff
, Named( \"ETres\" ) = ETres
);
return DF;
}
")
# Assumes your initial variables are already defined
EvapSimRcpp( RR, ETmax, Smax, 0 )
#######
[1]
http://stackoverflow.com/questions/7153586/can-i-vectorize-a-calculation-which-depends-on-previous-elements
On Sat, 6 Dec 2014, A Duranel wrote:
Hello
I use R to run a simple model of rainfall interception by vegetation:
rainfall falls on vegetation, some is retained by the vegetation (part of
which can evaporate), the rest falls on the ground (quite crude but very
similar to those used in SWAT or MikeSHE, for the hydrologists among you).
It uses a loop on zoo time-series of rainfall and potential
evapotranspiration. Unfortunately I did not find a way to vectorize it and
it takes ages to run on long datasets. Could anybody help me to make it run
faster?
library(zoo)
set.seed(1)
# artificial potential evapotranspiration time-series
ETmax<-zoo(runif(10, min=1, max=6), c(1:10))
# artificial rainfall time-series
RR<-zoo(runif(10, min=0, max=6), c(1:10))
## create empty time-series to fill
# effective rainfall (i.e. rainfall minus intercepted rainfall)
RReff<-zoo(NA, c(1:10))
# intercepted rainfall
int<-zoo(NA, c(1:10))
# residual potential evapotranspiration (ie ETmax minus evaporation from
interception)
ETres<-zoo(NA, c(1:10))
# define maximum interception storage capacity (maximum volume of rainfall
that can be intercepted per time step, provided the interception store is
empty at start of time-step)
Smax<-3
# volume of water in interception storage at start of computation
storage<-0
for (i in 1:length(ETmax)) {
# compute interception capacity for time step i (maximum interception
capacity minus any water intercepted but not evaporated during previous
time-step).
int[i]<-Smax-storage
# compute intercepted rainfall: equal to rainfall if smaller than
interception capacity, and to interception capacity if larger.
if(RR[i]<int[i]) int[i]<-RR[i]
# compute effective rainfall (rainfall minus intercepted rainfall).
RReff[i]<-RR[i]-int[i]
# update interception storage: initial interception storage + intercepted
rainfall.
storage<-storage+coredata(int[i])
# compute evaporation from interception storage: equal to potential
evapotranspiration if the latter is smaller than interception storage, and
to interception storage if larger.
if(storage>coredata(ETmax[i])) evap<-coredata(ETmax[i]) else evap<-storage
# compute residual potentiel evapotranspiration: potential
evapotranspiration minus evaporation from interception storage.
ETres[i]<-ETmax[i]-evap
# update interception storage, to be carried over to next time-step:
interception storage minus evaporation from interception storage.
storage<-storage-evap
}
Many thanks for your help!
Arnaud
UCL Department of Geography, UK
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