Big data ways to calculate sets of distances in R?

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Problem: We need a big data method for calculating distances between points. We outline what we'd like to do below with a five-observation dataframe. However, this particular method is infeasible as the number of rows gets large (> 1 million). In the past, we've used SAS to do this kind of analysis, but we'd prefer R if possible. (Note: I'm not going to show code because, while I outline a way to do this on smaller datasets below, this is basically an impossible method to use with data on our scale.)

We start with a dataframe of stores, each of which has a latitude and longitude (though this is not a spatial file, nor do we want to use a spatial file).

# you can think of x and y in this example as Cartesian coordinates
stores <- data.frame(id = 1:5,
                     x = c(1, 0, 1, 2, 0),
                     y = c(1, 2, 0, 2, 0))

stores
  id x y
1  1 1 1
2  2 0 2
3  3 1 0
4  4 2 2
5  5 0 0

For each store, we want to know the number of stores within x distance. In a small dataframe, this is straightforward. Create another dataframe of all coordinates, merge back in, calculate distances, create an indicator if the distance is less than x and add up the indicators (minus one for the store itself, which is at distance 0). This would result in a dataset that looks like this:

   id x y  s1.dist  s2.dist  s3.dist  s4.dist  s5.dist
1:  1 1 1 0.000000 1.414214 1.000000 1.414214 1.414214
2:  2 0 2 1.414214 0.000000 2.236068 2.000000 2.000000
3:  3 1 0 1.000000 2.236068 0.000000 2.236068 1.000000
4:  4 2 2 1.414214 2.000000 2.236068 0.000000 2.828427
5:  5 0 0 1.414214 2.000000 1.000000 2.828427 0.000000

When you count (arbitrarily) under 1.45 as "close," you end up with indicators that look like this:

# don't include the store itself in the total
   id x y s1.close s2.close s3.close s4.close s5.close total.close
1:  1 1 1        1        1        1        1        1           4
2:  2 0 2        1        1        0        0        0           1
3:  3 1 0        1        0        1        0        1           2
4:  4 2 2        1        0        0        1        0           1
5:  5 0 0        1        0        1        0        1           2

The final product should look like this:

   id total.close
1:  1           4
2:  2           1
3:  3           2
4:  4           1
5:  5           2

All advice appreciated.

Thank you very much

bigdata dataframe coordinates r matrix
2 Answers

Any reason you can't loop instead of making it one big calculation?

stores <- data.frame(id = 1:5,
                     x = c(1, 0, 1, 2, 0),
                     y = c(1, 2, 0, 2, 0))

# Here's a Euclidean distance metric, but you can drop anything you want in here
distfun <- function(x0, y0, x1, y1){
  sqrt((x1-x0)^2+(y1-y0)^2)
}

# Loop over each store
t(sapply(seq_len(nrow(stores)), function(i){
  distances <- distfun(x0 = stores$x[i], x1 = stores$x,
                       y0 = stores$y[i], y1 = stores$y)
  # Calculate number less than arbitrary cutoff, subtract one for self
  num_within <- sum(distances<1.45)-1
  c(stores$id[i], num_within)
}))

Produces:

     [,1] [,2]
[1,]    1    4
[2,]    2    1
[3,]    3    2
[4,]    4    1
[5,]    5    2

This will work with a data set of any size that you can bring into R, but it'll just get slower as the size increases. Here's a test on 10,000 entries that runs in a couple seconds on my machine:

stores <- data.frame(id=1:10000, 
                     x=runif(10000, max = 10), 
                     y=runif(10000, max = 10))

          [,1] [,2]
    [1,]     1  679
    [2,]     2  698
    [3,]     3  618
    [4,]     4  434
    [5,]     5  402
...
 [9995,]  9995  529
 [9996,]  9996  626
 [9997,]  9997  649
 [9998,]  9998  514
 [9999,]  9999  667
[10000,] 10000  603

It get slower with more calculations (because it has to run between every pair of points, this will always be O(n^2)) but without knowing the actual distance metric you'd like to calculate we can't optimize the slow part any further.

user_78354350

Did you really already try the classic dist() function? The core is implemented in C and should thus be fast.

Probably the coercion to a matrix (which takes place in dist anyway) already costs a lot of time, maybe it could be read in immediately as a matrix and not first as a data frame.

M <- as.matrix(stores[-1])

dist(M, diag=TRUE, upper=TRUE)
#          1        2        3        4        5
# 1 0.000000 1.414214 1.000000 1.414214 1.414214
# 2 1.414214 0.000000 2.236068 2.000000 2.000000
# 3 1.000000 2.236068 0.000000 2.236068 1.000000
# 4 1.414214 2.000000 2.236068 0.000000 2.828427
# 5 1.414214 2.000000 1.000000 2.828427 0.000000

Otherwise you could try this C++ implementation which is basically a copy of @coatless's code. However, I used Rcpp package for use in an R script.

library(Rcpp)
cppFunction('Rcpp::NumericMatrix calcPWD1 (const Rcpp::NumericMatrix & x){
  unsigned int outrows = x.nrow(), i = 0, j = 0;
  double d;
  Rcpp::NumericMatrix out(outrows,outrows);

  for (i = 0; i < outrows - 1; i++){
    Rcpp::NumericVector v1 = x.row(i);
    for (j = i + 1; j < outrows ; j ++){
      d = sqrt(sum(pow(v1-x.row(j), 2.0)));
      out(j,i)=d;
      out(i,j)=d;
    }
  }

  return out;
}')

calcPWD1(M)
#          [,1]     [,2]     [,3]     [,4]     [,5]
# [1,] 0.000000 1.414214 1.000000 1.414214 1.414214
# [2,] 1.414214 0.000000 2.236068 2.000000 2.000000
# [3,] 1.000000 2.236068 0.000000 2.236068 1.000000
# [4,] 1.414214 2.000000 2.236068 0.000000 2.828427
# [5,] 1.414214 2.000000 1.000000 2.828427 0.000000

However, the benchmark is yet clearly in favor of dist, so you should give it a try:

M_big <- M[sample(nrow(M), 1e4, replace=TRUE), ]  ## inflate to 10k rows
microbenchmark::microbenchmark(
  dist=dist(M_big, diag=TRUE, upper=TRUE),
  calcPWD1=calcPWD1(M_big),
  control=list(warmup=10L),
  times=3L
)
# Unit: milliseconds
#     expr       min        lq     mean   median        uq       max neval cld
#     dist  640.1861  660.1396  765.881  680.093  828.7284  977.3638     3  a 
# calcPWD1 1419.4106 1439.1353 1505.253 1458.860 1548.1736 1637.4873     3   b

Be sure to read @coatless's and Dirk Eddelbuettel's answers where they write some more about C, C++, and R and have other versions of the function.

user_65740380
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