Displaying 1 result from an estimated 1 matches for "examp_p".
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example
2017 Dec 13
3
inefficient for loop, is there a better way?
...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 &...