我试图找到一种最佳方法来查找每行中最大值的索引。问题是我无法找到一种真正有效的方法。 一个例子：
Dummy <- matrix(runif(500000000,0,3), ncol = 10000)
> system.time(max.col(Dummy, "first"))
user system elapsed
5.532 0.075 5.599
> system.time(apply(Dummy,1,which.max))
user system elapsed
14.638 0.210 14.828
> system.time(rowRanges(Dummy))
user system elapsed
2.083 0.029 2.109
我的主要问题是，与使用rowRanges函数计算最大值和最小值相比，为什么计算最大值的索引的速度超过2倍。有没有办法如何提高计算每行最大索引的性能？
答案 0 :(得分：6)
扩展krlmlr的答案，一些基准：
在数据集上：
maxCol_R maxCol_col是一个R-by-row循环，maxCol_row是一个C-by-row循环，microbenchmark::microbenchmark(max.col(Dummy, "first"), maxCol_R(Dummy), maxCol_col(Dummy), maxCol_row(Dummy), times = 30)
#Unit: milliseconds
# expr min lq median uq max neval
# max.col(Dummy, "first") 1209.28408 1245.24872 1268.34146 1291.26612 1504.0072 30
# maxCol_R(Dummy) 1060.99994 1084.80260 1099.41400 1154.11213 1436.2136 30
# maxCol_col(Dummy) 86.52765 87.22713 89.00142 93.29838 122.2456 30
# maxCol_row(Dummy) 577.51613 583.96600 598.76010 616.88250 671.9191 30
all.equal(max.col(Dummy, "first"), maxCol_R(Dummy))
#[1] TRUE
all.equal(max.col(Dummy, "first"), maxCol_col(Dummy))
#[1] TRUE
all.equal(max.col(Dummy, "first"), maxCol_row(Dummy))
#[1] TRUE
是一个C-by-row循环。
maxCol_R = function(x)
{
ans = rep_len(1L, nrow(x))
mx = x[, 1L]
for(j in 2:ncol(x)) {
tmp = x[, j]
wh = which(tmp > mx)
ans[wh] = j
mx[wh] = tmp[wh]
}
ans
}
maxCol_col = inline::cfunction(sig = c(x = "matrix"), body = '
int nr = INTEGER(getAttrib(x, R_DimSymbol))[0], nc = INTEGER(getAttrib(x, R_DimSymbol))[1];
double *px = REAL(x), *buf = (double *) R_alloc(nr, sizeof(double));
for(int i = 0; i < nr; i++) buf[i] = R_NegInf;
SEXP ans = PROTECT(allocVector(INTSXP, nr));
int *pans = INTEGER(ans);
for(int j = 0; j < nc; j++) {
for(int i = 0; i < nr; i++) {
if(px[i + j*nr] > buf[i]) {
buf[i] = px[i + j*nr];
pans[i] = j + 1;
}
}
}
UNPROTECT(1);
return(ans);
', language = "C")
maxCol_row = inline::cfunction(sig = c(x = "matrix"), body = '
int nr = INTEGER(getAttrib(x, R_DimSymbol))[0], nc = INTEGER(getAttrib(x, R_DimSymbol))[1];
double *px = REAL(x), *buf = (double *) R_alloc(nr, sizeof(double));
for(int i = 0; i < nr; i++) buf[i] = R_NegInf;
SEXP ans = PROTECT(allocVector(INTSXP, nr));
int *pans = INTEGER(ans);
for(int i = 0; i < nr; i++) {
for(int j = 0; j < nc; j++) {
if(px[i + j*nr] > buf[i]) {
buf[i] = px[i + j*nr];
pans[i] = j + 1;
}
}
}
UNPROTECT(1);
return(ans);
', language = "C")
功能：
rangeCol = inline::cfunction(sig = c(x = "matrix"), body = '
int nr = INTEGER(getAttrib(x, R_DimSymbol))[0], nc = INTEGER(getAttrib(x, R_DimSymbol))[1];
double *px = REAL(x),
*maxbuf = (double *) R_alloc(nr, sizeof(double)),
*minbuf = (double *) R_alloc(nr, sizeof(double));
memcpy(maxbuf, &(px[0 + 0*nr]), nr * sizeof(double));
memcpy(minbuf, &(px[0 + 0*nr]), nr * sizeof(double));
SEXP ans = PROTECT(allocMatrix(INTSXP, nr, 2));
int *pans = INTEGER(ans);
for(int i = 0; i < LENGTH(ans); i++) pans[i] = 1;
for(int j = 1; j < nc; j++) {
for(int i = 0; i < nr; i++) {
if(px[i + j*nr] > maxbuf[i]) {
maxbuf[i] = px[i + j*nr];
pans[i] = j + 1;
}
if(px[i + j*nr] < minbuf[i]) {
minbuf[i] = px[i + j*nr];
pans[i + nr] = j + 1;
}
}
}
UNPROTECT(1);
return(ans);
', language = "C")
set.seed(007); m = matrix(sample(24) + 0, 6, 4)
m
# [,1] [,2] [,3] [,4]
#[1,] 24 7 23 6
#[2,] 10 17 21 11
#[3,] 3 22 20 14
#[4,] 2 18 1 15
#[5,] 5 19 12 8
#[6,] 16 4 9 13
rangeCol(m)
# [,1] [,2]
#[1,] 1 4
#[2,] 3 1
#[3,] 2 1
#[4,] 2 3
#[5,] 2 1
#[6,] 1 2
编辑 16年6月10日
稍微改变一下，找到max和min的指数：
{{1}}
答案 1 :(得分：5)
这是一个非常基本的Rcpp实现：
#include <Rcpp.h>
// [[Rcpp::export]]
Rcpp::NumericVector MaxCol(Rcpp::NumericMatrix m) {
R_xlen_t nr = m.nrow(), nc = m.ncol(), i = 0;
Rcpp::NumericVector result(nr);
for ( ; i < nr; i++) {
double current = m(i, 0);
R_xlen_t idx = 0, j = 1;
for ( ; j < nc; j++) {
if (m(i, j) > current) {
current = m(i, j);
idx = j;
}
}
result[i] = idx + 1;
}
return result;
}
/*** R
microbenchmark::microbenchmark(
"Rcpp" = MaxCol(Dummy),
"R" = max.col(Dummy, "first"),
times = 200L
)
#Unit: milliseconds
# expr min lq mean median uq max neval
# Rcpp 221.7777 224.7442 242.0089 229.6407 239.6339 455.9549 200
# R 513.4391 524.7585 562.7465 539.4829 562.3732 944.7587 200
*/
由于我的笔记本电脑没有足够的内存，我不得不将您的样本数据缩小一个数量级，但结果应该转换为原始样本数据：
Dummy <- matrix(runif(50000000,0,3), ncol = 10000)
all.equal(MaxCol(Dummy), max.col(Dummy, "first"))
#[1] TRUE
可以稍微更改一下，以返回每行min和max的索引：
// [[Rcpp::export]]
Rcpp::NumericMatrix MinMaxCol(Rcpp::NumericMatrix m) {
R_xlen_t nr = m.nrow(), nc = m.ncol(), i = 0;
Rcpp::NumericMatrix result(nr, 2);
for ( ; i < nr; i++) {
double cmin = m(i, 0), cmax = m(i, 0);
R_xlen_t min_idx = 0, max_idx = 0, j = 1;
for ( ; j < nc; j++) {
if (m(i, j) > cmax) {
cmax = m(i, j);
max_idx = j;
}
if (m(i, j) < cmin) {
cmin = m(i, j);
min_idx = j;
}
}
result(i, 0) = min_idx + 1;
result(i, 1) = max_idx + 1;
}
return result;
}
答案 2 :(得分：3)
R将矩阵存储在column-major order中。因此，迭代列通常会更快，因为一列的值在内存中彼此接近，并且将一次遍历缓存层次结构：
align-items: center;这应该接近最快的Rcpp解决方案能够获得的最短时间。使用nav的任何解决方案都必须复制每个列/行，这可以在使用Rcpp时保存。您可以决定潜在的加速速度是否值得您努力。
答案 3 :(得分：2)
通常，在R中执行操作的最快方法是调用C，C ++或FORTRAN。
似乎matrixStats::rowRanges是implemented in C，这解释了为什么它是最快的。
如果你想进一步提高性能，可能会有一点点速度来修改rowRanges.c代码以忽略最小值并获得最大值，但我认为增益会非常小
答案 4 :(得分：1)
尝试使用STL算法和RcppArmadillo。
microbenchmark::microbenchmark(MaxColArmadillo(Dummy), #Using RcppArmadillo
MaxColAlgorithm(Dummy), #Using STL algorithm max_element
maxCol_col(Dummy), #Column processing
maxCol_row(Dummy)) #Row processing
Unit: milliseconds
expr min lq mean median uq max neval
MaxColArmadillo(Dummy) 227.95864 235.01426 261.4913 250.17897 276.7593 399.6183 100
MaxColAlgorithm(Dummy) 292.77041 345.84008 392.1704 390.66578 433.8009 552.2349 100
maxCol_col(Dummy) 40.64343 42.41487 53.7250 48.10126 61.3781 128.4968 100
maxCol_row(Dummy) 146.96077 158.84512 173.0941 169.20323 178.7959 272.6261 100
STL实施
#include <Rcpp.h>
// [[Rcpp::export]]
// Argument is a matrix ansd returns a
// vector of max of each of the rows of the matrix
Rcpp::NumericVector MaxColAlgorithm(Rcpp::NumericMatrix m) {
//int numOfRows = m.rows();
//Create vector with 0 of size numOfRows
Rcpp::NumericVector total(m.rows());
for(int i = 0; i < m.rows(); ++i)
{
//Create vector of the rows of matrix
Rcpp::NumericVector rVec = m.row(i);
//Apply STL max of elemsnts on the vector and store in a vector
total(i) = *std::max_element(rVec.begin(), rVec.end());
}
return total;
}
RcppArmadillo实施
#include <RcppArmadillo.h>
// [[Rcpp::depends(RcppArmadillo)]]
using namespace Rcpp;
// [[Rcpp::export]]
arma::mat MaxColArmadillo(arma::mat x)
{
//RcppArmadillo max function where dim = 1 means max of each row
// of the matrix
return(max(x,1));
}