R: How to filter/smooth binary signal

Question

I have a binary vector which represents a time series. I'd like to filter out fast switches like 00000001100000000 should be zeros and likewise 11111111111011111 should be just ones.

What kind of filter/function would be appropriate for that task?

Answer 1

Maybe this is a stupid approach but rle/inverse.rle seem to be good candidates. E.g. if you define a fast switch as a period of less than 3 equal values:

b1 <- c(rep(0, 7), rep(1, 2), rep(0, 7))
b2 <- c(rep(1, 10), 0, rep(1, 4))

binaryFilter <- function(x, threshold=3) {
  r <- rle(x)
  isBelowThreshold <- r$lengths < threshold
  r$values[isBelowThreshold] <- abs(1-r$values[isBelowThreshold])
  return(inverse.rle(r))
}

binaryFilter(b1)
# [1] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

binaryFilter(b2)
# [1] 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

Answer 2

How about taking the neighbouring values into account using a weighted average? In this case the 2 neighbours of every value (which has 2 neighbours on both sides) are considered. Of course this can be adjusted.

> v <- sample(c(0,1),30,replace=TRUE)

> v
 [1] 0 1 1 1 0 0 0 0 1 1 0 1 0 0 1 0 1 1 0 0 0 0 1 1 1 0 1 0 0 0

# embed(v,5) is a short version for this:
# cbind(v[1:26],v[2:27],v[3:28],v[4:29],v[5:30])

> m <- embed(v,5)

> c(round(m %*% c(.1,.2,.4,.2,.1)))
 [1] 1 1 0 0 0 0 1 1 0 0 0 0 0 0 1 1 0 0 0 0 1 1 1 0 0 0

---

before: 0 1 1 1 0 0 0 0 1 1 0 1 0 0 1 0 1 1 0 0 0 0 1 1 1 0 1 0 0 0
after:  . . 1 1 0 0 0 0 1 1 0 0 0 0 0 0 1 1 0 0 0 0 1 1 1 0 0 0 . .

as you can see, the loners are gone.

---

As suggested by sgibb, the whole fuzz can be boiled down to:

round(filter(v, c(.1,.2,.4,.2,.1)))

(But I guess the above written out version makes it clear what is done, which is why I leave it)