在 R 中查找数组中最接近元素的最快方法

我想找到 R 中最快的方法来识别 Ytimes 数组中最接近给定 Xtimes 值的元素索引。

到目前为止，我一直在使用一个简单的 for 循环，但必须有更好的方法来做到这一点：

Xtimes <- c(1,5,8,10,15,19,23,34,45,51,55,57,78,120)
Ytimes <- seq(0,120,length.out = 1000)

YmatchIndex = array(0,length(Xtimes))
for (i in 1:length(Xtimes)) {
  YmatchIndex[i] = which.min(abs(Ytimes - Xtimes[i]))
}

print(Ytimes[YmatchIndex])

强制性 Rcpp 解决方案。利用矢量已排序并且不包含重复项的事实来将O(n^2)进入一个O(n)。对于您的应用来说可能实用，也可能不实用；）

C++:

#include <Rcpp.h>
#include <cmath>
using namespace Rcpp;

// [[Rcpp::export]]
IntegerVector closest_pts(NumericVector Xtimes, NumericVector Ytimes) {
  int xsize = Xtimes.size();
  int ysize = Ytimes.size();
  int y_ind = 0;
  double minval = R_PosInf;
  IntegerVector output(xsize);
  for(int x_ind = 0; x_ind < xsize; x_ind++) {
    while(std::abs(Ytimes[y_ind] - Xtimes[x_ind]) < minval) {
      minval = std::abs(Ytimes[y_ind] - Xtimes[x_ind]);
      y_ind++;
    }
    output[x_ind] = y_ind;
    minval = R_PosInf;
  }
  return output;
}

R:

microbenchmark::microbenchmark(
  for_loop = {
    for (i in 1:length(Xtimes)) {
      which.min(abs(Ytimes - Xtimes[i]))
    }
  },
  apply    = sapply(Xtimes, function(x){which.min(abs(Ytimes - x))}),
  fndIntvl = {
    Y2 <- c(-Inf, Ytimes + c(diff(Ytimes)/2, Inf))
    Ytimes[ findInterval(Xtimes, Y2) ]
  },
  rcpp = closest_pts(Xtimes, Ytimes),
  times = 100
)

Unit: microseconds
     expr      min      lq     mean   median       uq      max neval cld
 for_loop 3321.840 3422.51 3584.452 3492.308 3624.748 10458.52   100   b
    apply   68.365   73.04  106.909   84.406   93.097  2345.26   100  a 
 fndIntvl   31.623   37.09   50.168   42.019   64.595   105.14   100  a 
     rcpp    2.431    3.37    5.647    4.301    8.259    10.76   100  a 

identical(closest_pts(Xtimes, Ytimes), findInterval(Xtimes, Y2))
# TRUE