How to construct this binary variable in R? - r

Question

The aim is check if value at index i is 1 and then make the previous six entries as 1.
x <- c(0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1)
## Required output
y <- c(1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1)
## Attempt
for(j in seq_along(x)){
if(x[j] == 1){
for(i in (j-6):j)
x[i] = 1
}}
Could you help solve this or better approach ?
Thanks.

Answer 1

A fully vectorized solution using filter:
as.integer( #turn logical value into numbers
as.logical( #coerce to logical --> 0 becomes FALSE, everything else TRUE
rev( #reverse order
filter( #linear filtering
c(rep(0, 6), #pad with zeros in the beginning to avoid NAs
rev(x)), #revers order of input vector
c(rep(1, 7)), sides=1 #y_i = x_i * 1 + x_(i-1) * 1 +...+ x_(i-6) * 1
)[-(1:6)]))) #remove NA values
#[1] 1 1 1 1 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1

Answer 2

You could try the following options (though don't forget to initialize x when trying each option as I'm overriding it)
indx <- mapply(function(x, y) x:y, which(x == 1) - 6 , which(x == 1))
x[indx[indx > 0]] <- 1
x
## [1] 1 1 1 1 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1
Or even simpler
indx <- sapply(which(x == 1) - 6, function(x) x:(x + 6))
x[indx[indx > 0]] <- 1
x
## [1] 1 1 1 1 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1
Or
indx <- apply(cbind(which(x == 1) - 6 , which(x == 1)), 1, function(x) x[1]:x[2])
x[indx[indx > 0]] <- 1
x
## [1] 1 1 1 1 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1
Or
indx <- seq_len(6)
indx <- sapply(which(x == 1), function(x) x - indx)
x[indx[indx > 0]] <- 1
x
## [1] 1 1 1 1 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1

Answer 3

Using 'for' loop:
ddf = data.frame(x,y=0)
for(i in 1:nrow(ddf)){
if(ddf[i,'x']==1){
j = i-5
if(j<1) j=1
ddf[j:i,'y'] = 1
}
}
ddf
x y
1 0 1
2 0 1
3 0 1
4 1 1
5 0 0
6 0 0
7 0 0
8 0 0
9 0 0
10 0 0
11 0 0
12 0 1
13 0 1
14 0 1
15 0 1
16 0 1
17 1 1
18 0 1
19 1 1
y = ddf$y
y
[1] 1 1 1 1 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1

Answer 4

y<-x
y[unlist(sapply(which(x==1),
function(val){
val:(max(val-6,1))
}
)
)
]<-1
> y
[1] 1 1 1 1 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1
Explanation :
I first look for indices of x=1 with which(x==1). Then, for each of the indices I get the indices from the one with x=1 to the 6th before that with sapply(...) then I unlist the result to only have a vector of indices for which y must be 1.
I then assigned 1 to the corresponding y values.
another writing, in 2 steps :
y<-x
ind<-unlist(sapply(which(x==1),function(val){val:(max(val-6,1))}))
y[ind]<-1
> y
[1] 1 1 1 1 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1