我正在编写一个函数来查找一组值的中位数。数据表示为唯一值的向量（称为＆＃39;值＆＃39;）和它们的频率向量（＆＃39; freqs＆＃39;）。频率通常非常高，因此将它们粘贴出来会占用大量内存。我有一个缓慢的R实现，它是我的代码中的主要瓶颈，所以我正在编写一个自定义Rcpp函数用于R / Bioconductor包。 Bioconductor的网站建议不要使用C ++ 11，这对我来说是一个问题。
我的问题在于尝试根据值的顺序将两个向量排序在一起。在R中，我们可以使用order（）函数。尽管遵循了关于这个问题的建议，我似乎无法使其发挥作用：C++ sorting and keeping track of indexes
以下几行是问题所在：
// sort vector based on order of values
IntegerVector idx_ord = std::sort(idx.begin(), idx.end(),
bool (int i1, int i2) {return values[i1] < values[i2];});
这是完整的功能，对任何人都有兴趣。任何进一步的提示将不胜感激：
#include <Rcpp.h>
using namespace Rcpp;
// [[Rcpp::export]]
double median_freq(NumericVector values, IntegerVector freqs) {
int len = freqs.size();
if (any(freqs!=0)){
int med = 0;
return med;
}
// filter out the zeros pre-sorting
IntegerVector non_zeros;
for (int i = 0; i < len; i++){
if(freqs[i] != 0){
non_zeros.push_back(i);
}
}
freqs = freqs[non_zeros];
values = values[non_zeros];
// find the order of values
// create integer vector of indices
IntegerVector idx(len);
for (int i = 0; i < len; ++i) idx[i] = i;
// sort vector based on order of values
IntegerVector idx_ord = std::sort(idx.begin(), idx.end(),
bool (int i1, int i2) {return values[i1] < values[i2];});
//apply to freqs and values
freqs = freqs[idx_ord];
values=values[idx_ord];
IntegerVector cum_freqs(len);
cum_freqs[0] = freqs[0];
for (int i = 1; i < len; ++i) cum_freqs[i] = freqs[i] + cum_freqs[i-1];
int total_freqs = cum_freqs[len-1];
// split into odd and even frequencies and calculate the median
if (total_freqs % 2 == 1) {
int med_ind = (total_freqs + 1)/2 - 1; // C++ indexes from 0
int i = 0;
while ((i < len) && cum_freqs[i] < med_ind){
i++;
}
double ret = values[i];
return ret;
} else {
int med_ind_1 = total_freqs/2 - 1; // C++ indexes from 0
int med_ind_2 = med_ind_1 + 1; // C++ indexes from 0
int i = 0;
while ((i < len) && cum_freqs[i] < med_ind_1){
i++;
}
double ret_1 = values[i];
i = 0;
while ((i < len) && cum_freqs[i] < med_ind_2){
i++;
}
double ret_2 = values[i];
double ret = (ret_1 + ret_2)/2;
return ret;
}
}
对于使用RUnit测试框架的任何人，这里有一些基本的单元测试：
test_median_freq <- function(){
checkEquals(median_freq(1:10,1:10),7)
checkEquals(median_freq(1:10,rep(1,10)),5.5)
checkEquals(median_freq(2:6,c(1,2,1,45,2)),5)
}
谢谢！
答案 0 :(得分：5)
我实际上会将值和频率合并到std::pair<double, int>中，然后只用std::sort对它们进行排序;通过这种方式，您始终可以将值和频率保持在一起。这使您可以编写更清晰的代码，因为没有其他一组索引浮动：
#include <Rcpp.h>
using namespace Rcpp;
// [[Rcpp::export]]
double median_freq(NumericVector values, IntegerVector freqs) {
const int len = freqs.size();
std::vector<std::pair<double, int> > allDat;
int freqSum = 0;
for (int i=0; i < len; ++i) {
allDat.push_back(std::pair<double, int>(values[i], freqs[i]));
freqSum += freqs[i];
}
std::sort(allDat.begin(), allDat.end());
int accum = 0;
for (int i=0; i < len; ++i) {
accum += allDat[i].second;
if (freqSum % 2 == 0) {
if (accum > freqSum / 2) {
return allDat[i].first;
} else if (accum == freqSum / 2) {
return (allDat[i].first + allDat[i+1].first) / 2;
}
} else {
if (accum >= (freqSum+1)/2) {
return allDat[i].first;
}
}
}
return NA_REAL; // Should not be reached
}
在R中尝试：
median_freq(1:10, 1:10)
# [1] 7
median_freq(1:10,rep(1,10))
# [1] 5.5
median_freq(2:6,c(1,2,1,45,2))
# [1] 5
我们还可以编写一个简单的R实现来确定我们使用Rcpp获得的效率增益：
med.freq.r <- function(values, freqs) {
ord <- order(values)
values <- values[ord]
freqs <- freqs[ord]
s <- sum(freqs)
cs <- cumsum(freqs)
idx <- min(which(cs >= s/2))
if (s %% 2 == 0 && cs[idx] == s/2) {
(values[idx] + values[idx+1]) / 2
} else {
values[idx]
}
}
med.freq.r(1:10, 1:10)
# [1] 7
med.freq.r(1:10,rep(1,10))
# [1] 5.5
med.freq.r(2:6,c(1,2,1,45,2))
# [1] 5
要进行基准测试，让我们看一组非常大的值：
set.seed(144)
values <- rnorm(1000000)
freqs <- sample(1:100, 1000000, replace=TRUE)
all.equal(median_freq(values, freqs), med.freq.r(values, freqs))
# [1] TRUE
library(microbenchmark)
microbenchmark(median_freq(values, freqs), med.freq.r(values, freqs))
# Unit: milliseconds
# expr min lq mean median uq max neval
# median_freq(values, freqs) 128.5322 131.6095 146.8360 145.6389 159.6117 165.0306 10
# med.freq.r(values, freqs) 715.2187 744.5709 776.0539 765.9178 817.7157 855.1898 10
对于100万个条目，Rcpp解决方案比R解决方案快约5倍;考虑到编译开销，如果您正在处理真正巨大的向量或者这是一个经常重复的选项，那么该性能才有吸引力。
一般情况下，我们知道如何在不排序的情况下计算中位数（详细信息请查看http://www.cc.gatech.edu/~mihail/medianCMU.pdf）。虽然算法比排序和迭代更精细，但它可以产生显着的加速：
double fast_median_freq(NumericVector values, IntegerVector freqs) {
const int len = freqs.size();
std::vector<std::pair<double, int> > allDat;
int freqSum = 0;
for (int i=0; i < len; ++i) {
allDat.push_back(std::pair<double, int>(values[i], freqs[i]));
freqSum += freqs[i];
}
int target = freqSum / 2;
int low = 0;
int high = len-1;
while (true) {
// Random pivot; move to the end
int rnd = low + (rand() % (high-low+1));
std::swap(allDat[rnd], allDat[high]);
// In-place pivot
int highPos = low; // Start of values higher than pivot
int lowSum = 0; // Sum of frequencies of elements below pivot
for (int pos=low; pos < high; ++pos) {
if (allDat[pos].first <= allDat[high].first) {
lowSum += allDat[pos].second;
std::swap(allDat[highPos], allDat[pos]);
++highPos;
}
}
std::swap(allDat[highPos], allDat[high]); // Move pivot to "highPos"
// If we found the element then return; o/w recurse on proper side
if (lowSum >= target) {
// Recurse on lower elements
high = highPos - 1;
} else if (lowSum + allDat[highPos].second >= target) {
// Return
if (target < lowSum + allDat[highPos].second || freqSum % 2 == 1) {
return allDat[highPos].first;
} else {
double nextHighest = std::min_element(allDat.begin() + highPos+1, allDat.begin() + len-1)->first;
return (allDat[highPos].first + nextHighest) / 2;
}
} else {
// Recurse on higher elements
low = highPos + 1;
target -= (lowSum + allDat[highPos].second);
}
}
}
基准：
all.equal(median_freq(values, freqs), fast_median_freq(values, freqs))
[1] TRUE
microbenchmark(median_freq(values, freqs), med.freq.r(values, freqs), fast_median_freq(values, freqs), times=10)
# Unit: milliseconds
# expr min lq mean median uq max neval
# median_freq(values, freqs) 119.57989 122.48622 130.47841 130.48811 132.75421 146.36136 10
# med.freq.r(values, freqs) 665.72803 690.15016 708.05729 702.65885 731.83936 749.36834 10
# fast_median_freq(values, freqs) 24.37572 29.39641 31.86144 31.77459 34.88418 36.81606 10
线性方法提供了比排序迭代Rcpp解决方案快4倍的速度和比基础R解决方案高20倍的速度。