Alternative to for loop and indexing? - r

Question

I have a large data set of 3 columns, Order, Discharge, Date (numeric). There are 20 years of daily Discharge values for each Order, which can extend beyond 100.
> head(dat)
Order Discharge date
1 0.04712 6574
2 0.05108 6574
3 0.00000 6574
4 0.00000 6574
5 3.54100 6574
6 3.61500 6574
For a given Order x, I would like to replace the Discharge value with the average of the Discharge at x+1 and x-1 for that date. I have been doing this in a crude manner with a for loop and indexing, but it takes over an hour to process. I know there has to be a better way.
x <- 4
for(i in min(dat[,3]):max(dat[,3]))
dat[,2][dat[,3] == i & dat[,1] == x ] <-
mean(c(dat[,2][dat[,3] == i & dat[,1] == x + 1],
dat[,2][dat[,3] == i & dat[,1] == x - 1]))
Gives
> head(dat)
Order Discharge date
1 0.04712 6574
2 0.05108 6574
3 0.00000 6574
4 1.77050 6574
5 3.54100 6574
6 3.61500 6574
Where the Discharge at Order 4, for date 6574 has been replaced with 1.77050. It works, but it's ridiculously slow.
I should specify that I don't need to do this calculation on every Order, but only a select few (only 8 out of a total of 117). Based on the answer, I have the following.
dat$NewDischarge <- by(dat$Discharge,dat$date,function(x)
colMeans(cbind(c(x[-1],NA), x,
c(NA, x[-length(x)])), na.rm=T))
I am trying to figure out a way still to only have the values of the select Orders to be calculated and am stuck in the rut of a for loop and indexing on date and Orders.

Answer 1

I would go by it as following:
Ensure that Order is a factor.
For each Order, you now have a sub-problem:
Sort the sub-data-frame by date.
Each Discharge-mean can be produced "vectorally" as:
colMeans(cbind(c(Discharge[-1], NA), Discharge, c(NA, Discharge[-length(Discharge)])))
The sub-problem can be dealt with a simple for-loop or the function by. I would prefer by.
Your data has been rearranged, but you can easily reorder it.
For point 2.2, imagine it (or try it) with a simple vector and see the effects of the cbind operation. It also forces you to consider the limit-situations; how is the first and last Discharge-value calculated (no preceding or proceeding dates).

Answer 2

There are several ways to solve your particular dilemma, but the basic question to ask when confronted with a slow for loop is, "How do I use vectorization to replace this loop?" (Well, maybe you should ask "Should I...?" first.) In your case, you're looping across dates, but there's no need to explicitly do that, since just grabbing all of the rows where dat$Order==x will implicitly grab all the dates.
The dataset you posted only has one date, but I can generate some fake data to illustrate:
generate.data <- function(n.order, n.date){
dat <- expand.grid(Order=seq_len(n.order), date=seq_len(n.date))
dat$Discharge <- rlnorm(n.order * n.date)
dat[, c("Order", "Discharge", "date")]
}
dat <- generate.data(10, 5)
head(dat)
# Order Discharge date
# 1 1 2.1925563 1
# 2 2 0.4093022 1
# 3 3 2.5525497 1
# 4 4 1.9274013 1
# 5 5 1.1941986 1
# 6 6 1.2407451 1
tail(dat)
# Order Discharge date
# 45 5 1.4344575 5
# 46 6 0.5757580 5
# 47 7 0.4986190 5
# 48 8 1.2076292 5
# 49 9 0.3724899 5
# 50 10 0.8288401 5
Here's all the rows where dat$Order==4, across all dates:
dat[dat$Order==4, ]
# Order Discharge date
# 4 4 1.9274013 1
# 14 4 3.5319072 2
# 24 4 0.2374532 3
# 34 4 0.4549798 4
# 44 4 0.7654059 5
You can just take the Discharge column, and you'll have the left-hand side of your assignment:
dat[dat$Order==4, ]$Discharge
# [1] 1.9274013 3.5319072 0.2374532 0.4549798 0.7654059
Now you just need the right side, which has two components: the x-1 discharges and the x+1 discharges. You can grab these the same way you grabbed the x discharges:
dat[dat$Order==4-1, ]$Discharge
# [1] 2.5525497 1.9143963 0.2800546 8.3627810 7.8577635
dat[dat$Order==4+1, ]$Discharge
# [1] 1.1941986 4.6076114 0.3963693 0.4190957 1.4344575
To obtain the new values, you need the parallel mean. R doesn't have a pmean function, but you can cbind these and take the rowMeans:
rowMeans(cbind(dat[dat$Order==4-1, ]$Discharge, dat[dat$Order==4+1, ]$Discharge))
# [1] 1.8733741 3.2610039 0.3382119 4.3909383 4.6461105
So, in the end you have:
dat[dat$Order==4, ]$Discharge <- rowMeans(cbind(dat[dat$Order==4-1, ]$Discharge,
dat[dat$Order==4+1, ]$Discharge))
You can even use %in% to make this work across all of your x values.
Note that this assumes your data is ordered.