I have 2 nested for loops which i want to get rid of. Any thoughts? I am calculating distance between cities based on their longitude and latitude. There is a custom function earth.dist() that i am using in the loop.

for (i in 1:nrow(dat)) {
  #for each other airport
  for (j in 1:nrow(dat)) {
    #if both airport are different
    if (dat[i,3]!=dat[j,3]){
      airport1[k] <- dat[i,3]      
      airport2[k] <- dat[j,3]
      #find travel distance
      travdist[k] <- earth.dist(dat[i,5],dat[i,4],dat[j,5],dat[j,4])


function for distance calculation

earth.dist <- function (lon1, lat1, lon2, lat2){
  rad <- pi/180
  a1 <- lat1 * rad
  a2 <- lon1 * rad
  b1 <- lat2 * rad
  b2 <- lon2 * rad
  dlon <- b2 - a2
  dlat <- b1 - a1
  a <- (sin(dlat/2))^2 + cos(a1) * cos(b1) * (sin(dlon/2))^2
  c <- 2 * atan2(sqrt(a), sqrt(1 - a))
  R <- 6378.145
  d <- R * c
  • \$\begingroup\$ I would have loved to review your code but the code supplied is incomplete!!!! what's dat? - An array or matrix or dataframe? Please update code snippet . Welcome to Code Review! \$\endgroup\$
    – Tolani
    Jul 20, 2016 at 17:20
  • \$\begingroup\$ Out of curiosity, what's the formula that a and c variables are assigned to. Is it a standard formula or formulated one for this exercise. \$\endgroup\$
    – Tolani
    Jul 20, 2016 at 17:23
  • \$\begingroup\$ @TolaniJaiye-Tikolo dat is dataframe. and what c does, even i don't know. i got this function from internet as i searched for formula in R to calculate distance between different airports based on their longitude and latitude. Thanks \$\endgroup\$ Jul 21, 2016 at 2:14

1 Answer 1


First, let's download some data similar to yours (I assume). This csv available online has almost 7,000 airports:

url <- "https://commondatastorage.googleapis.com/ckannet-storage/2012-07-09T214020/global_airports.csv"
txt <- getURL(url)
data <- read.csv(textConnection(txt), stringsAsFactors = FALSE)

For illustration purposes, we will use a small sample: the six airports in Jamaica.

dat <- subset(data, country == "Jamaica",
              c("city", "country", "name", "latitude", "longitude"))
#              city country               name latitude longitude
# 1745    Ocho Rios Jamaica           Boscobel 18.40425 -76.96902
# 1746     Kingston Jamaica Norman Manley Intl 17.93567 -76.78750
# 1747  Montego Bay Jamaica      Sangster Intl 18.50372 -77.91336
# 1748 Port Antonio Jamaica          Ken Jones 18.19881 -76.53453
# 1749     Kingston Jamaica         Tinson Pen 17.98856 -76.82376
# 5878       Negril Jamaica   Negril Aerodrome 18.34000 -78.33556

Now let's have a look at your code. I will not review the math in earth.dist, I'll assume it is correct. One beautiful thing about that function is that it is vectorized, i.e., you could give it n-long vectors as inputs and it will compute n distances in a single call. Unfortunately, the rest of your code does not take advantage of it. Instead, your double loop only calls earth.dist with scalars at each time...

Instead of a double loop, you should be using the outer function. Have a look at the doc (?outer) if you are not familiar with it. The typical usage is outer(X, Y, FUN) where X and Y are vectors and FUN is a vectorized function. The output is a matrix Z where Z[i, j] is the result of FUN(X[i], Y[j]). But what's brilliant about outer is that it does not call FUN as many times as there are entries in Z (length(X) * length(Y)). No, it calls is only once. How? Because FUN is vectorized (a requirement) and outer knows how to take advantage of it.

So, here is how we can massage your data a bit so we can use outer. First, remember that outer loops on the pairwise combinations from two vectors. In our case, we could use the names of the airports:

airport.names <- dat$name

so we will be calling outer(airport.names, airport.names, FUN = airport.dist). All that is left is to write airport.dist: a vectorized function that will take as inputs two vectors of airport names and return their distances. We could first put the important data in a matrix with airport names as row names for easy access:

dat.mat <- as.matrix(dat[, c("latitude", "longitude")])
rownames(dat.mat) <- airport.names

Then define:

airport.dist <- function(name1, name2, data = dat.mat) {
    lon1 <- data[name1, "longitude"]
    lat1 <- data[name1, "latitude"]
    lon2 <- data[name2, "longitude"]
    lat2 <- data[name2, "latitude"]
    return(earth.dist(lon1, lat1, lon2, lat2))

Then run outer:

dist.mat <- outer(airport.names, airport.names, FUN = airport.dist)

and attach names to the columns and rows:

dimnames(dist.mat) <- list(airport.names, airport.names)

#                     Boscobel Norman Manley Intl Sangster Intl Ken Jones Tinson Pen Negril Aerodrome
# Boscobel             0.00000           55.58311     100.33081  51.30027   48.75737        144.54554
# Norman Manley Intl  55.58311            0.00000     134.79830  39.68379    7.02926        169.83797
# Sangster Intl      100.33081          134.79830       0.00000 149.58621  128.67955         48.17112
# Ken Jones           51.30027           39.68379     149.58621   0.00000   38.52863        191.03056
# Tinson Pen          48.75737            7.02926     128.67955  38.52863    0.00000        164.62138
# Negril Aerodrome   144.54554          169.83797      48.17112 191.03056  164.62138          0.00000

If you need to convince yourself that earth.dist was called a single time, you could add a cat("HELLO\n") somewhere inside its body (I did!). earth.dist having been called only once, there is no need to say how fast the computation will be.

Finally, if you want to store the distances in a three column (airport1, airport2, distance) data.frame rather than a matrix, you can do:

d <- dist.mat
dist.df <- data.frame(airport1 = rownames(d)[row(d)],
                      airport2 = colnames(d)[col(d)],
                      distance = c(dist.mat))

I hope it helps! Don't hesitate to comment below if you have questions.

  • \$\begingroup\$ What if i want to use a predictive modelling function? lets say auto.arima() or lm() instead of earth.dist()? \$\endgroup\$ Jul 26, 2016 at 13:29
  • \$\begingroup\$ These don't sound like vectorized functions, so you won't be able to use outer. You can gain a little bit of speed by running only half of the combinations if the distance is symmetric. You could also rely on parallelization to speed things up. But overall, that would be a very different question and answer... Please consider accepting my answer by clicking on the tick mark on the left if you think I properly addressed your initial request. Cheers. \$\endgroup\$
    – flodel
    Jul 26, 2016 at 22:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.