R fill vector efficiently - r

Question

I have a fairly big vector (>500,000 in length). It contains a bunch of NA interspersed with 1 and it is always guaranteed that it begins with 1.
I would like to replace some of the NA in v1 with 1, based on a comparison operation on contiguous indices of another vector v2 (of same length as v1).
Is there an efficient way of doing this in vectorized notation so that looping is done in a low-level implementation? Maybe using ifelse?
Reproducible example below:
v1<-c(1,NA,NA,NA,1,NA,NA,NA,NA,NA,1,NA,NA,1,NA,1,NA,NA,NA,NA,NA,NA,NA,NA,NA,1)
v2<-c(10,10,10,9,10,9,9,9,9,9,10,10,10,11,8,12,12,12,12,12,12,12,12,12,12,13)
# goal is to fill through v1 in such a way that whenever
# v1[i] == NA and v1[i-1] == 1 and v2[i] == v2[i-1], then v1[i] == 1
MM<-data.frame(v1,v2)
for (i in 2:length(v1)){
# conditions: v1[i-1] == 1; v1[i]==NA; v2[i]==v2[i-1]
if (!is.na(v1[i-1]) && is.na(v1[i]) && v2[i]==v2[i-1]){
v1[i]<-1
}
}
MM$v1_altered<-v1
MM

Answer 1

There's probably a faster solution, but this is the best I could come up with in a couple minutes. My solution is slower than the OPs for small vectors, but increasingly faster for larger vectors.
library(zoo) # for na.locf
library(rbenchmark)
v1<-c(1,NA,NA,NA,1,NA,NA,NA,NA,NA,1,NA,NA,1,NA,1,NA,NA,NA,NA,NA,NA,NA,NA,NA,1)
v2<-c(10,10,10,9,10,9,9,9,9,9,10,10,10,11,8,12,12,12,12,12,12,12,12,12,12,13)
V1 <- rep(v1, each=20000) # 520,000 observations
V2 <- rep(v2, each=20000) # 520,000 observations
fun1 <- function(v1,v2) {
for (i in 2:length(v1)){
if (!is.na(v1[i-1]) && is.na(v1[i]) && v2[i]==v2[i-1]){
v1[i]<-1
}
}
v1
}
fun2 <- function(v1,v2) {
# create groups in which we need to assess missing values
d <- cumsum(as.logical(c(0,diff(v2))))
# for each group, carry the first obs forward
ave(v1, d, FUN=function(x) na.locf(x, na.rm=FALSE))
}
all.equal(fun1(V1,V2), fun2(V1,V2))
# [1] TRUE
benchmark(fun1(V1,V2), fun2(V1,V2))
# test replications elapsed relative user.self sys.self
# 1 fun1(V1, V2) 100 194.29 6.113593 192.72 0.17
# 2 fun2(V1, V2) 100 31.78 1.000000 30.74 0.95

Answer 2

A vectorised solution would look like:
v1[-1] <- ifelse(diff(v2), 0, v1[-length(v1)])
But the above won't work, and I don't think you can avoid an explicit loop since, if I understand correctly, you want to propagate new values. So, how about:
cmp <- diff(v2)
for (i in 2:length(v1)){
v1[i] <- if(cmp[i-1]) 0 else v1[i-1]
}

Answer 3

It may not be faster, but v1[i] <- v1[i-1] * (cmp[i-1] == 0) avoids all explicit "if" calls. I can't test it right now, but you might try #James solution vs. looping over this form for, say a vector of 1e4 length to see which executes faster.

Answer 4

The function fun1 can be speeded up considerably by using the compiler package.
Using the code provided by Joshua and extending it with the compiler package:
library(zoo) # for na.locf
library(rbenchmark)
library(compiler)
v1 <- c(1,NA,NA,NA,1,NA,NA,NA,NA,NA,1,NA,NA,1,NA,1,NA,NA,NA,NA,NA,NA,NA,NA,NA,1)
v2 <- c(10,10,10,9,10,9,9,9,9,9,10,10,10,11,8,12,12,12,12,12,12,12,12,12,12,13)
fun1 <- function(v1,v2) {
for (i in 2:length(v1)){
if (!is.na(v1[i-1]) && is.na(v1[i]) && v2[i]==v2[i-1]){
v1[i]<-1
}
}
v1
}
fun2 <- function(v1,v2) {
# create groups in which we need to assess missing values
d <- cumsum(as.logical(c(0,diff(v2))))
# for each group, carry the first obs forward
ave(v1, d, FUN=function(x) na.locf(x, na.rm=FALSE))
}
fun3 <- cmpfun(fun1)
fun1(v1,v2)
fun2(v1,v2)
all.equal(fun1(v1,v2), fun2(v1,v2))
all.equal(fun1(v1,v2), fun3(v1,v2))
Nrep <- 1000
V1 <- rep(v1, each=Nrep)
V2 <- rep(v2, each=Nrep)
all.equal(fun1(V1,V2), fun2(V1,V2))
all.equal(fun1(V1,V2), fun3(V1,V2))
benchmark(fun1(V1,V2), fun2(V1,V2), fun3(V1,V2))
we get the following result
benchmark(fun1(V1,V2), fun2(V1,V2), fun3(V1,V2))
test replications elapsed relative user.self sys.self user.child
1 fun1(V1, V2) 100 12.252 5.706567 12.190 0.045 0
2 fun2(V1, V2) 100 2.147 1.000000 2.133 0.013 0
3 fun3(V1, V2) 100 3.702 1.724266 3.644 0.023 0
So the compiled fun1 is a lot faster than the original fun1 but still slower than fun2.