Is there a vectorized parallel max() and min()?
Asked Answered
A

4

44

I have a data.frame with columns "a" and "b". I want to add columns called "high" and "low" that contain the highest and the lowest among columns a and b.

Is there a way of doing this without looping over the lines in the dataframe?

edit: this is for OHLC data, and so the high and low column should contain the highest and lowest element between a and b on the same line, and not among the whole columns. sorry if this is badly worded.

Arlinda answered 8/4, 2011 at 7:1 Comment(0)
M
56

Sounds like you're looking for pmax and pmin ("parallel" max/min):

Extremes                 package:base                  R Documentation

Maxima and Minima

Description:

     Returns the (parallel) maxima and minima of the input values.

Usage:

     max(..., na.rm = FALSE)
     min(..., na.rm = FALSE)

     pmax(..., na.rm = FALSE)
     pmin(..., na.rm = FALSE)

     pmax.int(..., na.rm = FALSE)
     pmin.int(..., na.rm = FALSE)

Arguments:

     ...: numeric or character arguments (see Note).

   na.rm: a logical indicating whether missing values should be
          removed.

Details:

     ‘pmax’ and ‘pmin’ take one or more vectors (or matrices) as
     arguments and return a single vector giving the ‘parallel’ maxima
     (or minima) of the vectors.  The first element of the result is
     the maximum (minimum) of the first elements of all the arguments,
     the second element of the result is the maximum (minimum) of the
     second elements of all the arguments and so on.  Shorter inputs
     are recycled if necessary.  ‘attributes’ (such as ‘names’ or
     ‘dim’) are transferred from the first argument (if applicable).
Misalliance answered 8/4, 2011 at 7:7 Comment(0)
M
4

Here's a version I implemented using Rcpp. I compared pmin with my version, and my version is roughly 3 times faster.

library(Rcpp)

cppFunction("
  NumericVector min_vec(NumericVector vec1, NumericVector vec2) {
    int n = vec1.size();
    if(n != vec2.size()) return 0;
    else {
      NumericVector out(n);
      for(int i = 0; i < n; i++) {
        out[i] = std::min(vec1[i], vec2[i]);
      }
      return out;
    }
  }
")

x1 <- rnorm(100000)
y1 <- rnorm(100000)

microbenchmark::microbenchmark(min_vec(x1, y1))
microbenchmark::microbenchmark(pmin(x1, y1))

x2 <- rnorm(500000)
y2 <- rnorm(500000)

microbenchmark::microbenchmark(min_vec(x2, y2))
microbenchmark::microbenchmark(pmin(x2, y2))

The microbenchmark function output for 100,000 elements is:

> microbenchmark::microbenchmark(min_vec(x1, y1))
Unit: microseconds
            expr     min       lq     mean  median       uq
 min_vec(x1, y1) 215.731 222.3705 230.7018 224.484 228.1115
     max neval
 284.631   100
> microbenchmark::microbenchmark(pmin(x1, y1))
Unit: microseconds
         expr     min       lq     mean  median      uq      max
 pmin(x1, y1) 891.486 904.7365 943.5884 922.899 954.873 1098.259
 neval
   100

And for 500,000 elements:

> microbenchmark::microbenchmark(min_vec(x2, y2))
Unit: milliseconds
            expr      min       lq     mean   median       uq
 min_vec(x2, y2) 1.493136 2.008122 2.109541 2.140318 2.300022
     max neval
 2.97674   100
> microbenchmark::microbenchmark(pmin(x2, y2))
Unit: milliseconds
         expr      min       lq     mean   median       uq
 pmin(x2, y2) 4.652925 5.146819 5.286951 5.264451 5.445638
      max neval
 6.639985   100

So you can see the Rcpp version is faster.

You could make it better by adding some error checking in the function, for instance: check that both vectors are the same length, or that they are comparable (not character vs. numeric, or boolean vs. numeric).

Mercedes answered 26/10, 2016 at 18:38 Comment(0)
E
0

If your data.frame name is dat.

dat$pmin <- do.call(pmin,dat[c("a","b")])
dat$pmax <- do.call(pmax,dat[c("a","b")])
Equipment answered 8/4, 2011 at 12:52 Comment(0)
T
0

Another possible solution:

set.seed(21)
Data <- data.frame(a=runif(10),b=runif(10))
Data$low <- apply(Data[,c("a","b")], 1, min)
Data$high <- apply(Data[,c("a","b")], 1, max)
Theoretics answered 8/4, 2011 at 14:38 Comment(1)
This is still sequential.Knighterrant

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