string index matching alternate 1 and -1 - r

Question

I am trying to write a function that takes in a vector of integers and returns the indexes where 1 and -1 alternate. I thought this would be a simple function but it is proving devilishly hard to write!
Example:
index: 1 2 3 4 5 6 7 8 9 10
string: 1 0 0 0 1 -1 -1 0 1 -1 #this is input to function
returns index of -1 and 1 so that -1 index is listed first
[6,1] #from 1 to -1
[6,9] #from -1 to 1
[10,9] #from 1 to -1
My (non-working) attempt:
x<-c(1,0,0,0,1,-1,-1,0,1)
matchVals(x)
matchVals<-function(x){
current.index<-getStart(x)
#next VALUE to search for should be:
next.val<-x[current.index]*-1
next.index<-getNextVal(x[current.index:length(x)],next.val) + current.index #adding in offset
return(current.index,next.index)
}
#gets index of first value that isnt a 0
getStart<-function(x){
lapply(1:length(x),function(i){if(x[i]!=0)return(i)})
return(NA)
}
#gets FIRST index of the value specified (so dont feed it the entire string). Must add offset for truncated portion of string
getNextVal<-function(x,v){
lapply(1:length(x),function(i){if(x[i]==v)return(i)})
return(NA)
}

Answer 1

A way would to be to iterate, alternately, over the indices of 1 and -1 and save indices progressively.
Having:
x = c(1, 0, 0, 0, 1, -1, -1, 0, 1, -1)
and computing the indices:
i1 = which(x == 1)
i2 = which(x == -1)
find the progressive sequence recursively:
ff = function(x, y, acc = integer())
{
if(!length(x)) return(acc)
if(!length(y)) return(c(acc, x[[1L]]))
Recall(y[(findInterval(x[[1L]], y) + 1L):length(y)], x[-1L], c(acc, x[[1L]]))
}
ans = if(i1[[1]] < i2[[1]]) ff(i1, i2) else ff(i2, i1)
ans
#[1] 1 6 9 10
To get the exact desired output (among alternative ways):
tmp = embed(ans, 2)
i = (seq_len(nrow(tmp)) %% 2) == (if(i1[[1]] < i2[[1]]) 0 else 1)
tmp[i, ] = t(apply(tmp[i, , drop = FALSE], 1, rev))
tmp
# [,1] [,2]
#[1,] 6 1
#[2,] 6 9
#[3,] 10 9
Testing on other data:
X = c(0, 0, -1, 0, 0, -1, 1, 1, 1, -1, -1, 0, 0, -1, 0, 0, 1, -1,
-1, -1, 0, 1, 0, 1, 1, 1, -1, 0, 0, 1, 0, 1, 0, -1, 1, 1, 1,
-1, 0, 0, 1, 0, 1, 0, -1, 1, 1, 1)
i1 = which(X == 1)
i2 = which(X == -1)
if(i1[[1]] < i2[[1]]) ff(i1, i2) else ff(i2, i1)
# [1] 3 7 10 17 18 22 27 30 34 35 38 41 45 46
#..and proceed as necessary

Answer 2

This is an incomplete answer, but I think it's in the right direction.
Test case:
x <- c(1,0,0,0,1,-1,-1,0,1)
We basically want to ignore zero values, so let's (1) replace them by NAs and (2) use zoo::na.locf ("last observation carried forward") to replace them by the leading value
x2 <- x
x2[x2==0] <- NA
x2 <- zoo::na.locf(x2)
Now use rle() to identify runs/breakpoints:
(r <- rle(x2))
## Run Length Encoding
## lengths: int [1:3] 5 3 1
## values : num [1:3] 1 -1 1
The following statement gives (6,9,10), the locations you were interested in: r$values gives the corresponding information about the direction of the switch
cumsum(r$lengths)+1