Replacement of loops with if statement in R

Question

Hi what would be the best way of doing following loops in R?

for (i in 1:nrow(df1)) {
  counter <- 0
  for (j in 1:nrow(df2)) {
    if (df2$x[j] >= df1$a[i] & df2$x[j] < df1$b[i]{counter = counter + 1}
  }
  df1$counter[i] <- counter
}

Answer 1

There are several ways to attack something like this. I'll demonstrate a few. Since you didn't provide data, look to the bottom for samples.

1. Fix the code you have (I think you are missing a close-paren):

   for (i in 1:nrow(df1)) {
     counter1 <- 0
     for (j in 1:nrow(df2)) {
       if (df2$x[j] >= df1$a[i] & df2$x[j] < df1$b[i]) { counter1 = counter1 + 1; }
     }
     df1$counter1[i] <- counter1
   }
   df1
   #    a  b counter1
   # 1  7 49        3
   # 2 18 87        4
   # 3 29  3        0
   # 4 89 21        0
   # 5 58 13        0
   # 6 22 66        4
   # 7 62 68        0
   # 8 97 98        0
   

   (From here on out, I will not show the output, rest assured it is the same. If you don't believe me, try it. I'll keep numbering the counter columns so you can see them side-by-side.)

2. We can capitalize on R's vectorizing of things. This means that instead of c(1+9, 2+9, 3+9), you can write c(1,2,3)+9 and do it all at once. Similarly, you can actually sum up a vector of boolean (logical) values, which should do what you would expect (sum(T,T,F) is 2). On those themes, let's remove the inner loop:

   for (i in 1:nrow(df1)) {
     df1$counter2[i] <- sum(df2$x >= df1$a[i] & df2$x < df1$b[i])
   }
   

3. This is still a little un-R-onic (adaptation of pythonic). Let's try one of the apply variants meant to operate on a simple vector and return a vector, which we'll capture as a counter:

   df1$counter3 <- sapply(seq_len(nrow(df1)),
                          function(i) sum(df2$x >= df1$a[i] & df2$x < df1$b[i]))
   

4. Another technique is a less-frequent one, but can be useful at times (depending on how/where you apply it). The outer function effectively gives you all combinations of two vectors (similar to but distinct from expand.grid).

   outer(seq_len(nrow(df1)), seq_len(nrow(df2)),
         function(i, j) df2$x[j] >= df1$a[i] & df2$x[j] < df1$b[i])
   #       [,1]  [,2]  [,3]  [,4]  [,5]
   # [1,] FALSE  TRUE  TRUE  TRUE FALSE
   # [2,]  TRUE  TRUE FALSE  TRUE  TRUE
   # [3,] FALSE FALSE FALSE FALSE FALSE
   # [4,] FALSE FALSE FALSE FALSE FALSE
   # [5,] FALSE FALSE FALSE FALSE FALSE
   # [6,]  TRUE  TRUE FALSE  TRUE  TRUE
   # [7,] FALSE FALSE FALSE FALSE FALSE
   # [8,] FALSE FALSE FALSE FALSE FALSE
   

   There is actually only one call to the function, where if you were to peek when it is called, you would see this:

   i
   #  [1] 1 2 3 4 5 6 7 8 1 2 3 4 5 6 7 8 1 2 3 4 5 6 7 8 1 2 3 4 5 6 7 8 1 2 3 4
   # [37] 5 6 7 8
   j
   #  [1] 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 3 3 3 3 3 3 3 3 4 4 4 4 4 4 4 4 5 5 5 5
   # [37] 5 5 5 5
   

   From here, that inner function unrolls to something like:

   # df2$x[j] >= df1$a[i] & df2$x[j] < df1$b[i] # i,j
     df2$x[1] >= df1$a[1] & df2$x[1] < df1$b[1] # 1,1
     df2$x[1] >= df1$a[2] & df2$x[1] < df1$b[2] # 2,1
     df2$x[1] >= df1$a[3] & df2$x[1] < df1$b[3] # 3,1
     # ...
     df2$x[1] >= df1$a[8] & df2$x[1] < df1$b[8] # 8,1
     df2$x[2] >= df1$a[1] & df2$x[2] < df1$b[1] # 1,2
     df2$x[2] >= df1$a[2] & df2$x[2] < df1$b[2] # 2,2
     # ...
     df2$x[5] >= df1$a[7] & df2$x[5] < df1$b[7] # 7,5
     df2$x[5] >= df1$a[8] & df2$x[5] < df1$b[8] # 8,5
   

   and then gets shaped like a matrix with the appropriate number of rows and columns depending on the lengths of the input vectors. (There are lots of matrix-esque things you can do with this outer-product function, this is warping it from mathematical to lookup/calculate.)

   Now that you have a matrix of logicals, it's easy enough to determine the sums of rows with colSums:

   rowSums(outer(seq_len(nrow(df1)), seq_len(nrow(df2)),
                 function(i, j) df2$x[j] >= df1$a[i] & df2$x[j] < df1$b[i]))
   # [1] 3 4 0 0 0 4 0 0
   

   (which could have been assigned with df1$counter4 <- rowSums(...))

---

Data:

set.seed(20181015)
n1 <- 5
n2 <- 8
df1 <- data.frame(a = sample(100, size=n2), b = sample(100, size=n2))
df1
#    a  b
# 1  7 49
# 2 18 87
# 3 29  3
# 4 89 21
# 5 58 13
# 6 22 66
# 7 62 68
# 8 97 98
df2 <- data.frame(x = sample(100, size=n1))
df2
#    x
# 1 51
# 2 31
# 3 17
# 4 41
# 5 49

---

Benchmarking, for the curious:

library(microbenchmark)
microbenchmark(
  c1 = {
    for (i in 1:nrow(df1)) {
      counter1 <- 0
      for (j in 1:nrow(df2)) {
        if (df2$x[j] >= df1$a[i] & df2$x[j] < df1$b[i]) { counter1 = counter1 + 1; }
      }
      df1$counter1[i] <- counter1
    }
  },
  c2 = {
    for (i in 1:nrow(df1)) {
      df1$counter2[i] <- sum(df2$x >= df1$a[i] & df2$x < df1$b[i])
    }
  },
  c3 = {
    sapply(seq_len(nrow(df1)),
           function(i) sum(df2$x >= df1$a[i] & df2$x < df1$b[i]))
  },
  c4 = {
    rowSums(outer(seq_len(nrow(df1)), seq_len(nrow(df2)),
                  function(i, j) df2$x[j] >= df1$a[i] & df2$x[j] < df1$b[i]))
  },
  times=100
)
# Unit: microseconds
#  expr    min      lq     mean median      uq     max neval
#    c1 7022.1 7669.45 9608.953 8301.4 8989.25 19038.8   100
#    c2 4168.5 4634.00 5698.094 4998.5 5405.45 15927.4   100
#    c3  153.7  182.60  237.050  194.1  216.40  3209.6   100
#    c4   35.2   48.30   62.348   61.5   70.95   141.0   100