R中是否有一个函数可以有效地检查一个值是否大于一个并且小于另一个数字？它也应该适用于向量。
本质上，我正在寻找以下功能的更快版本：
> in.interval <- function(x, lo, hi) (x > lo & x < hi)
> in.interval(c(2,4,6), 3, 5)
[1] FALSE TRUE FALSE
这里的问题是x必须触摸两次，与更有效的方法相比，计算消耗了两倍的内存。在内部，我会假设它是这样工作的：
tmp1 <- (x > lo)tmp2 <- (x < hi)retval <- tmp1 & tmp2现在，在第 2 步之后，内存中有两个布尔向量，x必须查看两次。我的问题是：是否有一个（内置？）函数可以一步完成所有这些操作，而无需分配额外的内存？
跟进这个问题：R：从范围内的数据表中选择值
编辑：我已经根据 CauchyDistributedRV 在https://gist.github.com/4344844的回答设置了一个要点
正如@James 在评论中所说，诀窍是从 x 中减去 low 和 high 之间的中间值，然后检查该差异是否小于 low 和 high 之间距离的一半。或者，在代码中：
in.interval2 <- function(x, lo, hi) {
abs(x-(hi+lo)/2) < (hi-lo)/2
}
这与 hack 一样快.bincode，并且是您正在寻找的算法的实现。你可以把它翻译成 C 或 C++ 并尝试如果你得到加速。
与其他解决方案的比较：
x <- runif(1e6,1,10)
require(rbenchmark)
benchmark(
in.interval(x, 3, 5),
in.interval2(x, 3, 5),
findInterval(x, c(3, 5)) == 1,
!is.na(.bincode(x, c(3, 5))),
order='relative',
columns=c("test", "replications", "elapsed", "relative")
)
给
test replications elapsed relative
4 !is.na(.bincode(x, c(3, 5))) 100 1.88 1.000
2 in.interval2(x, 3, 5) 100 1.95 1.037
3 findInterval(x, c(3, 5)) == 1 100 3.42 1.819
1 in.interval(x, 3, 5) 100 3.54 1.883
findInterval比in.interval长 x 快。
library(microbenchmark)
set.seed(123L)
x <- runif(1e6, 1, 10)
in.interval <- function(x, lo, hi) (x > lo & x < hi)
microbenchmark(
findInterval(x, c(3, 5)) == 1L,
in.interval(x, 3, 5),
times=100)
和
Unit: milliseconds
expr min lq median uq max
1 findInterval(x, c(3, 5)) == 1L 23.40665 25.13308 25.17272 25.25361 27.04032
2 in.interval(x, 3, 5) 42.91647 45.51040 45.60424 45.75144 46.38389
== 1L如果不需要，则更快，如果要找到的“间隔”大于 1，则很有用
> system.time(findInterval(x, 0:10))
user system elapsed
3.644 0.112 3.763
如果速度至关重要，那么这个 C 实现虽然不能容忍整数而不是数字参数，但速度很快
library(inline)
in.interval_c <- cfunction(c(x="numeric", lo="numeric", hi="numeric"),
' int len = Rf_length(x);
double lower = REAL(lo)[0], upper = REAL(hi)[0],
*xp = REAL(x);
SEXP out = PROTECT(NEW_LOGICAL(len));
int *outp = LOGICAL(out);
for (int i = 0; i < len; ++i)
outp[i] = (xp[i] - lower) * (xp[i] - upper) <= 0;
UNPROTECT(1);
return out;')
其他答案中提出的一些解决方案的时间安排是
microbenchmark(
findInterval(x, c(3, 5)) == 1L,
in.interval.abs(x, 3, 5),
in.interval(x, 3, 5),
in.interval_c(x, 3, 5),
!is.na(.bincode(x, c(3, 5))),
times=100)
和
Unit: milliseconds
expr min lq median uq
1 findInterval(x, c(3, 5)) == 1L 23.419117 23.495943 23.556524 23.670907
2 in.interval.abs(x, 3, 5) 12.018486 12.056290 12.093279 12.161213
3 in.interval_c(x, 3, 5) 1.619649 1.641119 1.651007 1.679531
4 in.interval(x, 3, 5) 42.946318 43.050058 43.171480 43.407930
5 !is.na(.bincode(x, c(3, 5))) 15.421340 15.468946 15.520298 15.600758
max
1 26.360845
2 13.178126
3 2.785939
4 46.187129
5 18.558425
在 bin.cpp 文件中重新讨论速度问题
#include <Rcpp.h>
using namespace Rcpp;
// [[Rcpp::export]]
SEXP bin1(SEXP x, SEXP lo, SEXP hi)
{
const int len = Rf_length(x);
const double lower = REAL(lo)[0], upper = REAL(hi)[0];
SEXP out = PROTECT(Rf_allocVector(LGLSXP, len));
double *xp = REAL(x);
int *outp = LOGICAL(out);
for (int i = 0; i < len; ++i)
outp[i] = (xp[i] - lower) * (xp[i] - upper) <= 0;
UNPROTECT(1);
return out;
}
// [[Rcpp::export]]
LogicalVector bin2(NumericVector x, NumericVector lo, NumericVector hi)
{
NumericVector xx(x);
double lower = as<double>(lo);
double upper = as<double>(hi);
LogicalVector out(x);
for( int i=0; i < out.size(); i++ )
out[i] = ( (xx[i]-lower) * (xx[i]-upper) ) <= 0;
return out;
}
// [[Rcpp::export]]
LogicalVector bin3(NumericVector x, const double lower, const double upper)
{
const int len = x.size();
LogicalVector out(len);
for (int i=0; i < len; i++)
out[i] = ( (x[i]-lower) * (x[i]-upper) ) <= 0;
return out;
}
有时间
> library(Rcpp)
> sourceCpp("bin.cpp")
> microbenchmark(bin1(x, 3, 5), bin2(x, 3, 5), bin3(x, 3, 5),
+ in.interval_c(x, 3, 5), times=1000)
Unit: milliseconds
expr min lq median uq max
1 bin1(x, 3, 5) 1.546703 2.668171 2.785255 2.839225 144.9574
2 bin2(x, 3, 5) 12.547456 13.583808 13.674477 13.792773 155.6594
3 bin3(x, 3, 5) 2.238139 3.318293 3.357271 3.540876 144.1249
4 in.interval_c(x, 3, 5) 1.545139 2.654809 2.767784 2.822722 143.7500
大约相等的部分加速来自使用常量len而不是out.size()循环边界，并且分配逻辑向量而不初始化它（LogicalVector(len)因为它将在循环中初始化）。
如果你可以处理NAs，你可以使用.bincode：
.bincode(c(2,4,6), c(3, 5))
[1] NA 1 NA
library(microbenchmark)
set.seed(42)
x = runif(1e8, 1, 10)
microbenchmark(in.interval(x, 3, 5),
findInterval(x, c(3, 5)),
.bincode(x, c(3, 5)),
times=5)
Unit: milliseconds
expr min lq median uq max
1 .bincode(x, c(3, 5)) 930.4842 934.3594 955.9276 1002.857 1047.348
2 findInterval(x, c(3, 5)) 1438.4620 1445.7131 1472.4287 1481.380 1551.419
3 in.interval(x, 3, 5) 2977.8460 3046.7720 3075.8381 3182.013 3288.020
我能找到的主要加速是通过对函数进行字节编译。即使是 Rcpp 解决方案（尽管使用 Rcpp 糖，而不是更深入的 C 解决方案）也比编译的解决方案慢。
library( compiler )
library( microbenchmark )
library( inline )
in.interval <- function(x, lo, hi) (x > lo & x < hi)
in.interval2 <- cmpfun( in.interval )
in.interval3 <- function(x, lo, hi) {
sapply( x, function(xx) {
xx > lo && xx < hi }
)
}
in.interval4 <- cmpfun( in.interval3 )
in.interval5 <- rcpp( signature(x="numeric", lo="numeric", hi="numeric"), '
NumericVector xx(x);
double lower = Rcpp::as<double>(lo);
double upper = Rcpp::as<double>(hi);
return Rcpp::wrap( xx > lower & xx < upper );
')
x <- c(2, 4, 6)
lo <- 3
hi <- 5
microbenchmark(
in.interval(x, lo, hi),
in.interval2(x, lo, hi),
in.interval3(x, lo, hi),
in.interval4(x, lo, hi),
in.interval5(x, lo, hi)
)
给我
Unit: microseconds
expr min lq median uq max
1 in.interval(x, lo, hi) 1.575 2.0785 2.5025 2.6560 7.490
2 in.interval2(x, lo, hi) 1.035 1.4230 1.6800 2.0705 11.246
3 in.interval3(x, lo, hi) 25.439 26.2320 26.7350 27.2250 77.541
4 in.interval4(x, lo, hi) 22.479 23.3920 23.8395 24.3725 33.770
5 in.interval5(x, lo, hi) 1.425 1.8740 2.2980 2.5565 21.598
library( compiler )
library( inline )
library( microbenchmark )
in.interval.oldRcpp <- rcpp(
signature(x="numeric", lo="numeric", hi="numeric"), '
NumericVector xx(x);
double lower = Rcpp::as<double>(lo);
double upper = Rcpp::as<double>(hi);
return Rcpp::wrap( (xx > lower) & (xx < upper) );
')
in.interval.abs <- rcpp(
signature(x="numeric", lo="numeric", hi="numeric"), '
NumericVector xx(x);
double lower = as<double>(lo);
double upper = as<double>(hi);
LogicalVector out(x);
for( int i=0; i < out.size(); i++ ) {
out[i] = ( (xx[i]-lower) * (xx[i]-upper) ) <= 0;
}
return wrap(out);
')
in.interval.abs.sugar <- rcpp(
signature( x="numeric", lo="numeric", hi="numeric"), '
NumericVector xx(x);
double lower = as<double>(lo);
double upper = as<double>(hi);
return wrap( ((xx-lower) * (xx-upper)) <= 0 );
')
x <- runif(1E5)
lo <- 0.5
hi <- 1
microbenchmark(
in.interval.oldRcpp(x, lo, hi),
in.interval.abs(x, lo, hi),
in.interval.abs.sugar(x, lo, hi)
)
all.equal( in.interval.oldRcpp(x, lo, hi), in.interval.abs(x, lo, hi) )
all.equal( in.interval.oldRcpp(x, lo, hi), in.interval.abs.sugar(x, lo, hi) )
给我
1 in.interval.abs(x, lo, hi) 662.732 666.4855 669.939 690.6585 1580.707
2 in.interval.abs.sugar(x, lo, hi) 722.789 726.0920 728.795 742.6085 1671.093
3 in.interval.oldRcpp(x, lo, hi) 1870.784 1876.4890 1892.854 1935.0445 2859.025
> all.equal( in.interval.oldRcpp(x, lo, hi), in.interval.abs(x, lo, hi) )
[1] TRUE
> all.equal( in.interval.oldRcpp(x, lo, hi), in.interval.abs.sugar(x, lo, hi) )
[1] TRUE