Question

给出向量中的数字（例如1 5 10 12），我正在向量中寻找落在我选择的数字范围之间的那些数字（例如c（9,11））。我希望在这个小例子中返回vec = c(10)。

下面是一个更大的MWE，我使用dplyr::between对相关值进行子集化......但是，我正在寻找一种更快的方法（不使用并行化作为解决方案）。如果我能更好地解释一下，请告诉我。

# Data
set.seed(1)
targets <- sort(sample(1:1e8, 1e7, replace=FALSE))
vec <- c(1345706, 1405938)

# Function
dplyr_between <- function(vec, targets) {
            require(dplyr)
            targets <- targets[dplyr::between(targets, vec[1], vec[2])]
            return(targets)
        }

test <- dplyr_between(vec, targets)
# 1345732 1345761 1345779 1345780 1345797

编辑使用x < max & x > min根据评论添加功能（自删除后）

# More Functions
base_compare <- function(vec, targets) {
            targets <- targets[targets < vec[2] & targets > vec[1]]
            return(targets)
        }

base_compare(vec, targets)
# 1345732 1345761 1345779 1345780 1345797

使用@docendo

中的data.table::inrange

编辑

# inrange function
dt_inrange <- function(vec, targets) {
            require(data.table)
            targets <- targets[inrange(targets, vec[1], vec[2])]
            return(targets)
        }

dt_inrange(vec, targets)
# 1345732 1345761 1345779 1345780 1345797

基准

library(microbenchmark)
microbenchmark(dplyr_between(vec, targets), base_compare(vec, targets), dt_inrange(vec, targets), times=10L)
# Unit: milliseconds
                       # expr      min       lq     mean   median       uq      max
# dplyr_between(vec, targets) 265.5192 283.5998 296.0947 296.7552 309.4403 323.3634
#  base_compare(vec, targets) 303.4629 317.8389 343.6311 343.3765 354.6891 427.1962
#    dt_inrange(vec, targets) 129.3800 131.1634 142.8658 144.4569 149.3728 164.5824
 # neval
    # 10
    # 10
    # 10

谢谢！

Answer 1

简单的Rcpp实现：

temp.cpp中的C ++代码“

#include <Rcpp.h>
#include <vector>

using namespace Rcpp;
// [[Rcpp::plugins(cpp11)]]
// [[Rcpp::export]]
std::vector<int> betweenRcpp(IntegerVector vec, int lower, int upper) {
  std::vector<int> ret;
  for(int i=0; i<vec.size(); i++) {
    if((vec[i] > lower) & (vec[i] < upper)) {
      ret.push_back(vec[i]);
    } else if(vec[i] >= upper) {
      break;
    }
  }
  return ret;
}

R代码：

library(Rcpp)
library(microbenchmark)
setwd("~/Desktop")
# Data
set.seed(1)
targets <- sort(sample(1:1e8, 1e7, replace=FALSE))
vec <- c(1345706, 1405938)

# Function
dplyr_between <- function(vec, targets) {
  require(dplyr)
  targets <- targets[dplyr::between(targets, vec[1], vec[2])]
  return(targets)
}

sourceCpp("temp.cpp")

test <- dplyr_between(vec, targets)
test2 <- betweenRcpp(targets, vec[1], vec[2])

microbenchmark(dplyr_between(vec, targets), betweenRcpp(targets, vec[1], vec[2]), times=10)


Unit: microseconds
                                 expr       min        lq        mean      median        uq        max neval cld
          dplyr_between(vec, targets) 72066.027 77809.681 108023.3793 103723.4075 125280.89 173892.552    10   b
 betweenRcpp(targets, vec[1], vec[2])   439.124   464.475    502.7439    481.8025    543.12    594.578    10  a

解决方案之间的测试平等：

all(test == test2)

Answer 2

由于您的数据已排序，您可以使用键控数据表。我将预先排序数据等同于预先键入数据表，因此创建密钥的时间不是基准测试的一部分。我还从dt_inrange中移除了残骸，因此比较可以集中在手头的任务上。

key_dt = data.table(targets, key = "targets")
# note that `targets` does not need to be sorted beforehand
# the key = "targets" will sort it as the table is created.
# You can also use `setkey` to add a key to an existing data table.

dt_inrange <- function(vec, targets) {
            targets[inrange(targets, vec[1], vec[2])]
        }

key_dt_inrange <- function(vec, target_dt) {
  target_dt[inrange(targets, vec[1], vec[2]), targets]
}
print(microbenchmark(
  dt_inrange(vec, targets),
  key_dt_inrange(vec, key_dt),
  times = 10
), signif = 3, order = "mean")
# Unit: milliseconds
#                         expr  min   lq     mean median    uq   max neval cld
#  key_dt_inrange(vec, key_dt) 47.5 47.9 54.75557   50.4  52.2  98.6    10   a
#     dt_inrange(vec, targets) 48.8 49.8 99.18932   60.4 185.0 219.0    10   a

无论出于何种原因，看起来unkeyed方法有一些正确的偏斜，平均值比中位数大50％，但这在键控数据表方法中被阻止了。

Answer 3

microbenchmark(db = {
    x = findInterval(vec, targets)
    targets[(x[1]+1):x[2]]
},
dplyr_between(vec, targets))
#Unit: milliseconds
#                        expr       min        lq      mean    median        uq      max neval cld
#                          db  51.02101  58.43651  78.81237  70.51761  79.58609 410.3919   100  a 
# dplyr_between(vec, targets) 127.03341 148.65899 177.43284 156.37937 170.22009 431.5442   100   b


identical({x = findInterval(vec, targets)
          targets[(x[1]+1):x[2]]}, test)
#[1] TRUE

查找目标号码之间所有数字的最快方法

3 个答案: