Question

我想找到R中的fastes方式，以识别最接近给定Xtimes值的Ytimes数组中元素的索引。

到目前为止，我一直在使用一个简单的for循环，但是必须有一种更好的方法：

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])

Answer 1

强制性Rcpp溶液。利用您的向量已排序且不包含重复项的事实，可以将O(n^2)转换为O(n)。可能对您的应用程序不切实际；）

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

Answer 2

R是矢量化的，因此跳过for循环。这样可以节省脚本和计算时间。只需将for循环替换为apply函数。由于我们要返回一维向量，因此我们使用sapply。

YmatchIndex <- sapply(Xtimes, function(x){which.min(abs(Ytimes - x))})

证明apply更快：

library(microbenchmark)
library(ggplot2)

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

# time it
mbm <- microbenchmark(
  for_loop = for (i in 1:length(Xtimes)) {
    YmatchIndex[i] = which.min(abs(Ytimes - Xtimes[i]))
  },
  apply    = sapply(Xtimes, function(x){which.min(abs(Ytimes - x))}),
  times = 100
)

# plot
autoplot(mbm)

请参见?apply for more。

Answer 3

我们可以使用findInterval有效地做到这一点。（cut也可以使用，但需要做更多的工作）。

首先，让我们偏移Ytimes偏移量，以便我们可以找到最近的而不是下一个较小的。我将首先演示伪造数据：

y <- c(1,3,5,10,20)
y2 <- c(-Inf, y + c(diff(y)/2, Inf))
cbind(y, y2[-1])
#       y     
# [1,]  1  2.0
# [2,]  3  4.0
# [3,]  5  7.5
# [4,] 10 15.0
# [5,] 20  Inf
findInterval(c(1, 1.9, 2.1, 8), y2)
# [1] 1 1 2 4

第二列（以-Inf开头）将给我们带来突破。请注意，每个都位于对应值与其跟随者之间的中间位置。

好的，让我们将其应用于矢量：

Y2 <- Ytimes + c(diff(Ytimes)/2, Inf)
head(cbind(Ytimes, Y2))
#         Ytimes         Y2
# [1,] 0.0000000 0.06006006
# [2,] 0.1201201 0.18018018
# [3,] 0.2402402 0.30030030
# [4,] 0.3603604 0.42042042
# [5,] 0.4804805 0.54054054
# [6,] 0.6006006 0.66066066

Y2 <- c(-Inf, Ytimes + c(diff(Ytimes)/2, Inf))
cbind(Xtimes, Y2[ findInterval(Xtimes, Y2) ])
#       Xtimes            
#  [1,]      1   0.9009009
#  [2,]      5   4.9849850
#  [3,]      8   7.9879880
#  [4,]     10   9.9099099
#  [5,]     15  14.9549550
#  [6,]     19  18.9189189
#  [7,]     23  22.8828829
#  [8,]     34  33.9339339
#  [9,]     45  44.9849850
# [10,]     51  50.9909910
# [11,]     55  54.9549550
# [12,]     57  56.9969970
# [13,]     78  77.8978979
# [14,]    120 119.9399399

（我使用cbind只是为了进行并排演示，而不必这样做。）

基准：

mbm <- microbenchmark::microbenchmark(
  for_loop = {
    YmatchIndex <- array(0,length(Xtimes))
    for (i in 1:length(Xtimes)) {
      YmatchIndex[i] = 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) ]
  },
  times = 100
)
mbm
# Unit: microseconds
#      expr    min     lq     mean  median      uq    max neval
#  for_loop 2210.5 2346.8 2823.678 2444.80 3029.45 7800.7   100
#     apply   48.8   58.7  100.455   65.55   91.50 2568.7   100
#  fndIntvl   18.3   23.4   34.059   29.80   40.30   83.4   100
ggplot2::autoplot(mbm)

在R中找到数组中最接近元素的最快方法

3 个答案: