All-to-all setdiff on two numeric vectors with a numeric threshold for accepting matches

Question

What I want to do is more or less a combination of the problems discussed in the two following threads:

• Perform non-pairwise all-to-all comparisons between two

Answer 1

If you are happy to use a non-base package, data.table::inrange is a convenient function.

x1[!inrange(x1, x2 - 0.045, x2 + 0.045)]
# [1] 1002.570  301.569

x2[!inrange(x2, x1 - 0.045, x1 + 0.045)]
# [1]   22.12   53.00 5666.31  100.10

---

inrange is also efficient on larger data sets. On e.g. 1e5 vectors, inrange is > 700 times faster than the two other alternatives:

n <- 1e5
b1 <- runif(n, 0, 10000)
b2 <- b1 + runif(n, -1, 1)

microbenchmark(
  f1 = f(b1, b2, 0.045, 5000),
  f2 = list(in_b1_not_in_b2 = b1[sapply(b1, function(x) !any(abs(x - b2) <= 0.045))],
       in_b2_not_in_b1 = b2[sapply(b2, function(x) !any(abs(x - b1) <= 0.045))]),
  f3 = list(in_b1_not_in_b2 = b1[!inrange(b1, b2 - 0.045, b2 + 0.045)],
       in_b2_not_in_b1 = b2[!inrange(b2, b1 - 0.045, b1 + 0.045)]),
  unit = "relative", times = 10)
# Unit: relative
#  expr      min       lq     mean   median        uq       max neval
#    f1 1976.931 1481.324 1269.393 1103.567 1173.3017 1060.2435    10
#    f2 1347.114 1027.682  858.908  766.773  754.7606  700.0702    10
#    f3    1.000    1.000    1.000    1.000    1.0000    1.0000    10

---

And yes, they give the same result:

n <- 100
b1 <- runif(n, 0, 10000)
b2 <- b1 + runif(n, -1, 1)

all.equal(f(b1, b2, 0.045, 5000),
          list(in_b1_not_in_b2 = b1[sapply(b1, function(x) !any(abs(x - b2) <= 0.045))],
               in_b2_not_in_b1 = b2[sapply(b2, function(x) !any(abs(x - b1) <= 0.045))]))
# TRUE

all.equal(f(b1, b2, 0.045, 5000),
          list(in_b1_not_in_b2 = b1[!inrange(b1, b2 - 0.045, b2 + 0.045)],
               in_b2_not_in_b1 = b2[!inrange(b2, b1 - 0.045, b1 + 0.045)]))
# TRUE

---

Several related, potentially useful answers when searching for inrange on SO.

Answer 2

Here is an alternative approach

in_b1_not_in_b2 <- b_1[sapply(b_1, function(x) !any(abs(x - b_2) <= 0.045))]
in_b1_not_in_b2
#[1] 1002.570  301.569

in_b2_not_in_b1 <- b_2[sapply(b_2, function(x) !any(abs(x - b_1) <= 0.045))]
in_b2_not_in_b1
#[1]   22.12   53.00 5666.31  100.10

Answer 3

A vectorized beast:

D <- abs(outer(b_1, b_2, "-")) > 0.045

in_b1_not_in_b2 <- b_1[rowSums(D) == length(b_2)]
#[1] 1002.570  301.569

in_b2_not_in_b1 <- b_2[colSums(D) == length(b_1)]
#[1]   22.12   53.00 5666.31  100.10

---

hours later...

Henrik shared a question complaining the memory explosion when using outer for long vectors: Matching two very very large vectors with tolerance (fast! but working space sparing). However, the memory bottleneck for outer can be easily killed by blocking.

f <- function (b1, b2, threshold, chunk.size = 5000) {

  n1 <- length(b1)
  n2 <- length(b2)
  chunk.size <- min(chunk.size, n1, n2)

  RS <- numeric(n1)  ## rowSums, to be accumulated
  CS <- numeric(n2)  ## colSums, to be accumulated

  j <- 0
  while (j < n2) {
    chunk.size_j <- min(chunk.size, n2 - j)
    ind_j <- (j + 1):(j + chunk.size_j)
    b2_j <- b2[ind_j]
    i <- 0
    while (i < n1) {
      chunk.size_i <- min(chunk.size, n1 - i)
      ind_i <- (i + 1):(i + chunk.size_i)
      M <- abs(outer(b1[ind_i], b2_j, "-")) > threshold
      RS[ind_i] <- RS[ind_i] + rowSums(M)
      CS[ind_j] <- CS[ind_j] + colSums(M)
      i <- i + chunk.size_i
      }
    j <- j + chunk.size_j
    }

  list(in_b1_not_in_b2 = b1[RS == n2], in_b2_not_in_b1 = b2[CS == n1])
  }

With this function, outer never uses more memory than storing two chunk.size x chunk.size matrices. Now let's do something crazy.

b1 <- runif(1e+5, 0, 10000)
b2 <- b1 + runif(1e+5, -1, 1)

If we do a simple outer, we need memory to store two 1e+5 x 1e+5 matrices, which is up to 149 GB. However, on my Sandy Bridge (2011) laptop with only 4 GB RAM, computation is feasible.

system.time(oo <- f(b1, b2, 0.045, 5000))
#   user  system elapsed 
#365.800 167.348 533.912 

The performance is actually good enough, given that we have been using a very poor algorithm.

All answers here do exhausted search, that has complexity length(b1) x length(b2). We could reduce this to length(b1) + length(b2) if we work on sorted arrays. But such deeply optimized algorithm can only be implemented with compiled language to obtain efficiency.