[R] inefficient for loop, is there a better way?

[Bert Gunter]

I believe ?filter will do what you want.

I used  n = 100 instead of 1000:

ts <- 1:100
examp <- data.frame(ts=ts, stage=sin(ts))
examp <- within(examp, {
  abv_1 <- filter(stage > 0.6, rep(1,7),sides =1)
  abv_2 <- filter(stage > .85, rep(1,7), sides =1)
   })
examp

I think this should be fairly fast, but let us know if not. There may be
other alternatives that might be faster.
Assuming it does what you wanted, of course.

Cheers,
Bert


Bert Gunter

"The trouble with having an open mind is that people keep coming along and
sticking things into it."
-- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )

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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