满足条件的所有数字组合

问题描述 投票：0回答：6

我正在使用 R 编程语言。

假设我有数字：2010,2011,2012,2013,2014,2015,2016,2017,2018,2019,2020

我的问题：我想找出这些数字的所有可能的差值大于1的对。 (2013-2010)、(2019-2011)等

目前，我正在使用一种非常笨拙的方法来做到这一点：

x = 2010:2020
grid = expand.grid(x,x)
grid$diff = grid$Var1 - grid$Var2
grid$condition = ifelse(grid$Var1 - grid$Var2 > 1, "yes", "no")

final = grid[grid$condition == "yes",]

有更有效的方法吗？

例如，这里我首先生成所有可能的组合，然后消除无效的组合。当数量很多时，这可能无效。有更好的方法吗？

谢谢！

r subset combinations combn
6个回答
4
投票

在你的问题中，目标是获得向量中数字的组合，使得该对的差值大于 1，但你不想进行不必要的比较，浪费处理时间。虽然我喜欢 Ritchie Sacramento's answer 对于连续序列，我想提供一个可以处理序列中的间隙、未排序序列和非整数序列的答案。我认为要使其更“高效”的唯一方法是迭代序列并收集比当前迭代数至少大 1 的所有数字。

在我的方法中，我首先强制序列排序。然后，我循环遍历序列中的每个元素，并仅比较当前元素之后的索引。如果这些满足差异标准，我会将它们添加到运行的配对列表中。这假设您没有重复的数字，否则您可能需要先运行 

unique()
。

#' Find pairs in a sequence with a difference greater than a specified offset.
#'
#' This function sorts the input sequence, and then iterates through the sorted sequence 
#' to find pairs of numbers with a difference greater than the specified difference offset. 
#' Pairs are returned in a matrix where pair[1] > pair[2].
#'
#' @param x A numeric vector representing the input sequence.
#' @param differenceOffset A numeric value representing the minimum difference between pairs. Default is 1.
#' @return A matrix where each row represents a pair with a difference greater than the difference offset.
#' @examples
#' find_pairs(c(1,2,3,4,5))  # uses default differenceOffset of 1
#' find_pairs(c(2,4,6,8,10), 2)
find_pairs <- function(x, differenceOffset = 1) {
  # sort the sequence
  x_sorted <- sort(x)
  len <- length(x_sorted)

  # initialize a list to store the pairs
  pairs <- list()

  # iterate through the sorted sequence
  for(current in 1:(len - 1)) {
    # iterate and test the next index forward of the current
    for(test in (current + 1):len) {
      # if the difference is > differenceOffset, add the pair to the list
      if(x_sorted[test] - x_sorted[current] > differenceOffset) {
        # flip the order to put pair[1] > pair[2] in the output
        pairs <- c(pairs, list(c(x_sorted[test], x_sorted[current])))
      }
    }
  }

  # convert the list to a matrix
  do.call(rbind, pairs)
}

这里有几个用法示例。

# unsorted example
> find_pairs(c(4,3,2,5,1))
     [,1] [,2]
[1,]    3    1
[2,]    4    1
[3,]    5    1
[4,]    4    2
[5,]    5    2
[6,]    5    3

# use difference offset of >2
> find_pairs(c(1,2,3,4,5), 2)
     [,1] [,2]
[1,]    4    1
[2,]    5    1
[3,]    5    2

# your example
> find_pairs(2010:2020)
      [,1] [,2]
 [1,] 2012 2010
 [2,] 2013 2010
 [3,] 2014 2010
 [4,] 2015 2010
 [5,] 2016 2010
 [6,] 2017 2010
 [7,] 2018 2010
 [8,] 2019 2010
 [9,] 2020 2010
[10,] 2013 2011
...
[40,] 2018 2016
[41,] 2019 2016
[42,] 2020 2016
[43,] 2019 2017
[44,] 2020 2017
[45,] 2020 2018

# gap example
> find_pairs(c(1:2, 6:8))
     [,1] [,2]
[1,]    6    1
[2,]    7    1
[3,]    8    1
[4,]    6    2
[5,]    7    2
[6,]    8    2
[7,]    8    6


# non-integer example
> find_pairs(seq(1, 3, by = 0.25), 1)
      [,1] [,2]
 [1,] 2.25 1.00
 [2,] 2.50 1.00
 [3,] 2.75 1.00
 [4,] 3.00 1.00
 [5,] 2.50 1.25
 [6,] 2.75 1.25
 [7,] 3.00 1.25
 [8,] 2.75 1.50
 [9,] 3.00 1.50
[10,] 3.00 1.75

编辑

就时间复杂度而言，上述解决方案的总时间复杂度可能 [O(n log n + n)] 小于使用 

expand.grid
[O(n^2)] 处理长向量的解决方案。

jblood94 的答案 与基准测试表明上述方法根本不是真正有效。在原始函数

find_pairs
中，我试图尽可能明确地阐明该过程，但这导致了相当大的计算开销。编写快速而华丽的复杂单行通常是一项有趣的挑战，但这几乎总是以几乎完全难以辨认为代价。当然，使用像data.table
这样专门为优化此类问题而设计的包将是最好的解决方案。但是，为了保持清晰简洁的代码的完整性，并使用基本 R，我在下面提交了修订版
find_pairs.2。
在此版本中，我遵循类似的思路，但做了一些优化。首先，我为这些对分配一个大矩阵而不是一个空列表。虽然这内存效率较低，但它似乎会影响时间。其次，我利用 while 循环来查找第一个值超过 
differenceOffset
 的索引。一旦我们找到这个索引，并且因为我们已经对数组进行了排序，我们知道包括和超出它的所有值都必须大于 
differenceOffset
。然后，我只需将当前值（通过标量扩展）与大于测试条件的值一起直接设置到矩阵中的下一个可用位置（由 
pair_count
er 确定）。最后，我返回修剪后的矩阵。
find_pairs.2 <- function(x, differenceOffset = 1) {
  # sort the sequence
  x <- sort(x)
  len <- length(x)

  # initialize a matrix to store the pairs
  # assume that in the worst case, we have len*(len-1)/2 pairs
  max_pairs <- len * (len - 1) / 2
  pairs <- matrix(nrow = max_pairs, ncol = 2)

  # initialize a counter for the number of pairs found
  pair_count <- 0

  # iterate through the sorted sequence
  for (current in 1:(len - 1)) {
    # find the first index (test) that is strictly greater than differenceOffset
    test <- current + 1
    while (test <= len && x[test] - x[current] <= differenceOffset) {
      test <- test + 1
    }
    
    # if test is within bounds, add all pairs from test to len to the matrix
    if (test <= len) {
      num_new_pairs <- len - test + 1
      pairs[(pair_count + 1):(pair_count + num_new_pairs), 1] <- x[test:len]
      pairs[(pair_count + 1):(pair_count + num_new_pairs), 2] <- x[current]
      pair_count <- pair_count + num_new_pairs
    }
  }

  # trim unused slots in the matrix
  return(pairs[1:pair_count, ])
}



如果您确实想加快计算速度，可以使用 
Rcpp
 包并编译新解决方案的 C++ 版本（基准如下）。
find_pairs.cpp

#include <Rcpp.h>
using namespace Rcpp;

// [[Rcpp::export]]
NumericMatrix find_pairs_cpp(NumericVector x, int differenceOffset = 1) {
  // Sort the sequence
  std::sort(x.begin(), x.end());

  int len = x.size();

  // Initialize a matrix to store the pairs
  // Assume that in the worst case, we have len*(len-1)/2 pairs
  int max_pairs = len * (len - 1) / 2;
  NumericMatrix pairs(max_pairs, 2);

  // Initialize a counter for the number of pairs found
  int pair_count = 0;

  // Iterate through the sorted sequence
  for (int current = 0; current < len - 1; ++current) {
    // Find the first index (test) that is strictly greater than differenceOffset
    int test = current + 1;
    while (test < len && x[test] - x[current] <= differenceOffset) {
      ++test;
    }
    
    // If test is within bounds, add all pairs from test to len to the matrix
    if (test < len) {
      int num_new_pairs = len - test;
      for (int j = 0; j < num_new_pairs; ++j) {
        pairs(pair_count + j, 0) = x[test + j];
        pairs(pair_count + j, 1) = x[current];
      }
      pair_count += num_new_pairs;
    }
  }

  // Trim unused slots in the matrix
  NumericMatrix result = pairs(Range(0, pair_count - 1), _);
  return result;
}



然后，像这样获取 C++ 函数：


# Load Rcpp package
library(Rcpp)

# Source the C++ script
sourceCpp("find_pairs.cpp")

# Now you can use your C++ function in R
x <- 2010:2020
find_pairs_cpp(x)



基准测试


作为旁注。我将其与 
jblood94 的答案
中定义的函数进行了基准测试，并看到了显着的改进，特别是因为原始的 find_pairs 函数甚至没有在更大的测试中幸存下来！
# "small" vector
x <- runif(1e2, 0, 10)

bm_small <- microbenchmark::microbenchmark(
  expand.grid = nrow(f0(x)),
  find_pairs.2 = nrow(find_pairs.2(x)),
  find_pairs_cpp = nrow(find_pairs_cpp(x)),
  f1 = nrow(f1(x)),
  f2 = nrow(f2(x)),
  f3 = nrow(f3(x)),
  f4 = nrow(f4(x)),
  check = "equal",
  times = 1000
)



> print(bm_small)
Unit: microseconds
           expr    min      lq      mean  median      uq     max neval
    expand.grid  776.5  827.80  878.1540  853.60  895.10  1559.9  1000
   find_pairs.2  394.3  432.10  462.8021  449.40  476.60  1042.6  1000
 find_pairs_cpp   28.1   39.60   48.8424   43.30   48.20   138.5  1000    
             f1  769.2  823.30  903.5507  847.10  885.60 21644.9  1000
             f2  657.8  700.10  832.8963  721.30  748.95 32129.7  1000
             f3 3387.0 3508.05 3768.5233 3565.00 3655.65 50499.4  1000
             f4 2511.5 2608.00 2770.2623 2650.95 2720.60 26553.3  1000



x <- runif(1e4, 0, 10)

bm_large <- microbenchmark::microbenchmark(
  expand.grid = nrow(f0(x)),
  find_pairs.2 = nrow(find_pairs.2(x)), # runs for several minutes without completing
  f1 = nrow(f1(x)),
  f2 = nrow(f2(x)),
  f3 = nrow(f3(x)),
  f4 = nrow(f4(x)),
  check = "equal",
  times = 10
)



> print(bm_large)
Unit: milliseconds
         expr       min        lq      mean    median        uq      max neval
  expand.grid   5344.5752 5551.6006 5640.7089 5689.3884 5750.9637 5767.5464 10
find_pairs.2    2531.3152 2574.1561 2679.6633 2643.0467 2788.3917 2879.4604 10
 find_pairs_cpp  405.0732  458.4121  505.8300  494.3591  531.5002  640.3842 10
             f1 2933.1794 2973.3883 3061.8112 3063.7269 3151.2178 3226.2590 10
             f2  906.5848  927.4680  979.4814  943.6236  962.3983 1177.1214 10
             f3  654.4493  699.1284  798.1042  787.8151  905.0556  958.2739 10
             f4  946.0565 1006.1900 1108.1919 1122.6784 1145.9541 1301.9947 10



干杯！

    





      

      
      
      

      

        

          

            

              
3
投票

              
            

          

        

        


nums <- 2010:2020

combn(nums, 2)[,combn(nums, 2, diff) > 1]



更新：从头开始创建对：


library(purrr)

f <- function(d, start, end){
    map(start:(end-d), \(x) c(x, x + d))
}

pairs <- function(min_diff, start, end){
  map(min_diff:(end-start), \(d) f(d, start, end)) |> 
    flatten()
}

pairs(2, 2010, 2020)


    



      

      
      
      

      

        

          

            

              
3
投票

              
            

          

        

        
data.table
选项。
我简化了您的选择以使其进行更公平的比较，因为不需要 ifelse
 子步骤
library(data.table)
x = 2010:4500

system.time({
    xd <- data.table(x)
    xd[, xp1 := x + 1]
    out1 <- xd[xd, on=.(x>xp1), nomatch=0L, .(Var1=x.x, Var2=i.x, diff=x.x-i.x)]
})
##   user  system elapsed 
##   0.04    0.01    0.07 

system.time({
    grid = expand.grid(x,x)
    out2 <- grid[grid$Var1 - grid$Var2 > 1,]
    out2$diff <- out2$Var1 - out2$Var2
})
##   user  system elapsed 
##   0.23    0.08    0.32 



每一行的结果都匹配：


all(mapply(\(x,y) all(x==y), out1, out2))
## [1] TRUE


    



      

      
      
      

      

        

          

            

              
3
投票

              
            

          

        

        
dist
解决方案、使用循环的高效基本解决方案以及高效
data.table
解决方案，所有这些都以迄今为止提出的一些解决方案为基准。
dist
解决方案：
library(parallelDist)

f1 <- function(x, mindiff = 1) {
  n <- 2*length(x)
  d <- parDist(as.matrix(x))
  y <- which(d > mindiff)
  i <- (n + 1 - sqrt((n - 1)^2 - 8*(y - 1)))%/%2
  data.frame(Var1 = x[i], Var2 = x[i + y - (n - i)*(i - 1)/2], diff = d[y])
}



内存高效的基本解决方案，可打破 
for
 循环：
f2 <- function(x, mindiff = 1) {
  x <- sort(x)
  n <- length(x)
  n1 <- n + 1L
  idx <- rep(n1, n - 1L)
  j <- 2L
  
  for (i in 1:(n - 1L)) {
    for (j in j:n) {
      if (x[j] - x[i] > mindiff) {
        idx[i] <- j
        break
      }
    }
    
    if (idx[i] == n1) break
  }
  
  span <- n - idx + 1L
  within(
    out <- data.frame(
      Var1 = x[rep.int(1:(n - 1L), span)], Var2 = x[sequence(span, idx)]
    ),
    diff <- Var2 - Var1
  )
}



先前解决方案的矢量化版本：


library(data.table)

f3 <- function(x, mindiff = 1) {
  x <- sort(x)
  n <- length(x)

  setorder(
    data.table(
      x = c(x, x + mindiff), idx1 = rep(1:n, 2), keep = rep(0:1, each = n)
    ), x
  )[,idx2 := cummax(idx1)][
    keep == 1L, {
      span <- n - idx2
      .(
        Var1 = x[rep.int(idx1, span)],
        Var2 = x[sequence(span, idx2 + 1L)]
      )
    }
  ][,diff := Var2 - Var1]
}



为目前提出的其他解决方案定义函数：


f0 <- function(x, mindiff = 1) {
  # modified from OP
  grid = expand.grid(x, x)
  out <- grid[grid$Var1 - grid$Var2 > mindiff,]
  out$diff <- out$Var1 - out$Var2
  out
}

f4 <- function(x, mindiff = 1) {
  # from @thelatemail
  xd <- data.table(x)
  xd[, xp1 := x + mindiff]
  xd[xd, on=.(x>xp1), nomatch=0L, .(Var1=x.x, Var2=i.x, diff=x.x-i.x)]
}

find_pairs <- function(x, differenceOffset = 1) {
  # from @Khlick
  # sort the sequence
  x_sorted <- sort(x)
  len <- length(x_sorted)
  
  # initialize a list to store the pairs
  pairs <- list()
  
  # iterate through the sorted sequence
  for(current in 1:(len - 1)) {
    # iterate and test the next index forward of the current
    for(test in (current + 1):len) {
      # if the difference is > differenceOffset, add the pair to the list
      if(x_sorted[test] - x_sorted[current] > differenceOffset) {
        # flip the order to put pair[1] > pair[2] in the output
        pairs <- c(pairs, list(c(x_sorted[test], x_sorted[current])))
      }
    }
  }
  
  # convert the list to a matrix
  do.call(rbind, pairs)
}



使用较小数据集进行基准测试：


x <- runif(1e2, 0, 10)

microbenchmark::microbenchmark(
  expand.grid = nrow(f0(x)),
  loop1 = nrow(find_pairs(x)),
  dist = nrow(f1(x)),
  loop2 = nrow(f2(x)),
  data.table1 = nrow(f3(x)),
  data.table2 = nrow(f4(x)),
  check = "equal"
)
#> Unit: microseconds
#>         expr     min       lq      mean   median       uq      max neval
#>  expand.grid   606.9   673.15   746.301   705.05   754.60   3534.9   100
#>        loop1 66948.0 68710.55 71555.309 70563.85 73099.55 113492.4   100
#>         dist   534.5   578.20   634.820   620.00   671.80    898.3   100
#>        loop2   319.0   393.90   466.185   435.85   514.00    920.0   100
#>  data.table1  1870.9  1980.05  2158.900  2079.15  2302.05   3229.2   100
#>  data.table2  1804.9  2032.10  2746.983  2719.85  3284.35   5810.3   100



具有更大数据集的基准：


x <- runif(1e4, 0, 10)

microbenchmark::microbenchmark(
  expand.grid = nrow(f0(x)),
  # loop1 = nrow(find_pairs(x)), # runs for several minutes without completing
  dist = nrow(f1(x)),
  loop2 = nrow(f2(x)),
  data.table1 = nrow(f3(x)),
  data.table2 = nrow(f4(x)),
  check = "equal",
  times = 10
)
#> Unit: milliseconds
#>         expr       min        lq      mean    median        uq       max neval
#>  expand.grid 4333.9824 4526.9577 4578.6956 4580.5573 4625.4831 4792.6571    10
#>         dist 1996.4856 2121.0837 2134.5839 2161.4953 2168.5389 2215.8524    10
#>        loop2  667.5903  748.8159  838.5044  788.3667  930.6512 1043.5914    10
#>  data.table1  432.8631  570.3818  635.3953  633.7162  706.0260  791.4238    10
#>  data.table2 1592.9779 1619.0466 1736.3997 1764.8470 1826.4533 1872.4085    10


    



      

      
      
      

      

        

          

            

              
2
投票

              
            

          

        

        
rep()
 和 
sequence()
 来构建向量。这应该和使用基本 R 一样高效。
f <- function(start, end, difference) {
  ds <- ((end - start) - difference + 1):1
  data.frame(
    v1 = sequence(ds, (start + difference):end),
    v2 = rep(start:(end - difference), ds)
    )
}

f(2010, 2020, 2)

     v1   v2
1  2012 2010
2  2013 2010
3  2014 2010
4  2015 2010
5  2016 2010
6  2017 2010
7  2018 2010
8  2019 2010
9  2020 2010
10 2013 2011
...
43 2019 2017
44 2020 2017
45 2020 2018


    

                


      

      
      
      

      

        

          

            

              
2
投票

              
            

          

        

        
lapply
do.call(
    rbind,
    Filter(
        length,
        lapply(
            x,
            \(i) if (any(lg <- (i - x > 1))) cbind(i, j = x[lg])
        )
    )
)



给予


         i    j
 [1,] 2012 2010
 [2,] 2013 2010
 [3,] 2013 2011
 [4,] 2014 2010
 [5,] 2014 2011
 [6,] 2014 2012
 [7,] 2015 2010
 [8,] 2015 2011
 [9,] 2015 2012
[10,] 2015 2013
[11,] 2016 2010
[12,] 2016 2011
[13,] 2016 2012
[14,] 2016 2013
[15,] 2016 2014
[16,] 2017 2010
[17,] 2017 2011
[18,] 2017 2012
[19,] 2017 2013
[20,] 2017 2014
[21,] 2017 2015
[22,] 2018 2010
[23,] 2018 2011
[24,] 2018 2012
[25,] 2018 2013
[26,] 2018 2014
[27,] 2018 2015
[28,] 2018 2016
[29,] 2019 2010
[30,] 2019 2011
[31,] 2019 2012
[32,] 2019 2013
[33,] 2019 2014
[34,] 2019 2015
[35,] 2019 2016
[36,] 2019 2017
[37,] 2020 2010
[38,] 2020 2011
[39,] 2020 2012
[40,] 2020 2013
[41,] 2020 2014
[42,] 2020 2015
[43,] 2020 2016
[44,] 2020 2017
[45,] 2020 2018



另一个基本 R 选项可以使用 
combn
do.call(
    rbind,
    Filter(
        length,
        combn(x[order(-x)],
            2,
            \(...) if (diff(...) < -1) data.frame(t(...)),
            simplify = FALSE
        )
    )
)



这给出了


     X1   X2
1  2020 2018
2  2020 2017
3  2020 2016
4  2020 2015
5  2020 2014
6  2020 2013
7  2020 2012
8  2020 2011
9  2020 2010
10 2019 2017
11 2019 2016
12 2019 2015
13 2019 2014
14 2019 2013
15 2019 2012
16 2019 2011
17 2019 2010
18 2018 2016
19 2018 2015
20 2018 2014
21 2018 2013
22 2018 2012
23 2018 2011
24 2018 2010
25 2017 2015
26 2017 2014
27 2017 2013
28 2017 2012
29 2017 2011
30 2017 2010
31 2016 2014
32 2016 2013
33 2016 2012
34 2016 2011
35 2016 2010
36 2015 2013
37 2015 2012
38 2015 2011
39 2015 2010
40 2014 2012
41 2014 2011
42 2014 2010
43 2013 2011
44 2013 2010
45 2012 2010


    

      

      
    

  

  

    
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