[R] inefficient for loop, is there a better way?
William Dunlap
wdunlap at tibco.com
Wed Dec 13 03:03:02 CET 2017
Try using stats::filter (not the unfortunately named dplyr::filter, which
is entirely different).
state>elev is a logical vector, but filter(), like most numerical
functions, treats TRUEs as 1s
and FALSEs as 0s.
E.g.,
> str( stats::filter( x=examp$stage>elev1, filter=rep(1,7),
method="convolution", sides=1) )
Time-Series [1:1000] from 1 to 1000: NA NA NA NA NA NA 3 3 2 2 ...
> str( stats::filter( x=examp$stage>elev2, filter=rep(1,7),
method="convolution", sides=1) )
Time-Series [1:1000] from 1 to 1000: NA NA NA NA NA NA 1 2 1 1 ...
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Tue, Dec 12, 2017 at 5:36 PM, Morway, Eric <emorway at usgs.gov> wrote:
> The code below is a small reproducible example of a much larger problem.
> While the script below works, it is really slow on the true dataset with
> many more rows and columns. I'm hoping to get the same result to examp,
> but with significant time savings.
>
> The example below is setting up a data.frame for an ensuing regression
> analysis. The purpose of the script below is to appends columns to 'examp'
> that contain values corresponding to the total number of days in the
> previous 7 ('per') above some stage ('elev1' or 'elev2'). Is there a
> faster method that leverages existing R functionality? I feel like the
> hack below is pretty clunky and can be sped up on the true dataset. I
> would like to run a more efficient script many times adjusting the value of
> 'per'.
>
> ts <- 1:1000
> examp <- data.frame(ts=ts, stage=sin(ts))
>
> hi1 <- list()
> hi2 <- list()
> per <- 7
> elev1 <- 0.6
> elev2 <- 0.85
> for(i in per:nrow(examp)){
> examp_per <- examp[seq(i - (per - 1), i, by=1),]
> stg_hi_cond1 <- subset(examp_per, examp_per$stage > elev1)
> stg_hi_cond2 <- subset(examp_per, examp_per$stage > elev2)
>
> hi1 <- c(hi1, nrow(stg_hi_cond1))
> hi2 <- c(hi2, nrow(stg_hi_cond2))
> }
> examp$days_abv_0.6_in_last_7 <- c(rep(NA, times=per-1), unlist(hi1))
> examp$days_abv_0.85_in_last_7 <- c(rep(NA, times=per-1), unlist(hi2))
>
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>
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