时间戳

首页 > 解决方案 > R - 创建多个场景组合

r - R - 创建多个场景组合

问题描述

我有四个（部分重叠）组，每组八个独特的申请人申请了我必须分配的工作的 20%、30%、40% 和 50%：

g20 <- c("a","b","c","d","e","f")
g30 <- c("a","b","c","d","e","f","g","h")
g40 <- c("c","d","e","f","g","h")
g50 <- c("e","f","g","h")

因为我只能按这四个增量来奖励作品，而且我必须选择不少于两个不超过四个的人，所以我有六个场景来奖励 100% 的作品：

  1. 50/50
  2. 50/30/20
  3. 40/40/20
  4. 40/30/30
  5. 40/20/20/20
  6. 30/30/20/20

对于每个场景，我需要找到所有可能的组合（无需替换），以将工作授予相应组中的申请人。

对于第一个场景，我可以很容易地完成这一点，t(combn(g50,2))但我不确定如何处理其他场景，我必须从不同的向量中提取组合，并确保在任何给定的组合中只选择一次申请人。输出需要是实际的组合，而不仅仅是组合的数量。

使用 R，我如何从四个不同的组中获取这些组合，并且（以场景 5 为例）确保“cdef”、“cedf”、“cfed”、“cfde”等都被视为相同的结果?

这可能吗？

标签： rcombinations

解决方案

还创建所有可能的组合，例如 Jon Spring 的解决方案，但使用包并删除欺骗申请人。

如果您的实际维度是每个 OP，您可以考虑扩展到所有可能的组合并删除重复申请人的行：

library(data.table)

g20 <- c("a","b","c","d","e","f")
g30 <- c("a","b","c","d","e","f","g","h")
g40 <- c("c","d","e","f","g","h")
g50 <- c("e","f","g","h")

scen <- paste0("g", c(30, 30, 20, 20))
allcombi <- do.call(CJ, mget(scen))
setnames(allcombi, paste0("V", 1L:length(allcombi)))

#remove rows with applicants that are repeated in different columns
nodupe <- allcombi[
    allcombi[, .I[anyDuplicated(unlist(.SD)) == 0L], 
        by=1:allcombi[,.N]]$V1]

#sort within columns with the same percentage of work
for(cols in split(names(nodupe), scen))
    nodupe[, (cols) := sort(.SD), by=seq_len(nodupe[,.N]), .SDcols=cols]

#remove identical combinations
ans <- unique(nodupe)
setnames(ans, scen)[]

输出：

     g30 g30 g20 g20
  1:   a   b   c   d
  2:   a   b   c   e
  3:   a   b   c   f
  4:   a   b   d   e
  5:   a   b   d   f
 ---                
221:   g   h   c   e
222:   g   h   c   f
223:   g   h   d   e
224:   g   h   d   f
225:   g   h   e   f

运行所有 6 个场景的代码和结果：

scenarios <- list(c(50,50), 
    c(50,30,20), 
    c(40,40,20), 
    c(40,30,30), 
    c(40,20,20,20), 
    c(30,30,20,20))

lapply(scenarios, 
    function(scen) {
        scen <- paste0("g", scen)
        allcombi <- do.call(CJ, mget(scen, envir=.GlobalEnv))
        setnames(allcombi, paste0("V", 1L:length(allcombi)))

        nodupe <- allcombi[
            allcombi[, .I[anyDuplicated(unlist(.SD)) == 0L], 
                by=1:allcombi[,.N]]$V1]

        for(cols in split(names(nodupe), scen))
            nodupe[, (cols) := sort(.SD), by=seq_len(nodupe[,.N]), .SDcols=cols]

        ans <- unique(nodupe)
        setnames(ans, scen)[]
})

输出：

[[1]]
   g50 g50
1:   e   f
2:   e   g
3:   e   h
4:   f   g
5:   f   h
6:   g   h

[[2]]
     g50 g30 g20
  1:   e   a   b
  2:   e   a   c
  3:   e   a   d
  4:   e   a   f
  5:   e   b   a
 ---            
128:   h   g   b
129:   h   g   c
130:   h   g   d
131:   h   g   e
132:   h   g   f

[[3]]
    g40 g40 g20
 1:   c   d   a
 2:   c   d   b
 3:   c   d   e
 4:   c   d   f
 5:   c   e   a
 6:   c   e   b
 7:   c   e   d
 8:   c   e   f
 9:   c   f   a
10:   c   f   b
11:   c   f   d
12:   c   f   e
13:   c   g   a
14:   c   g   b
15:   c   g   d
16:   c   g   e
17:   c   g   f
18:   c   h   a
19:   c   h   b
20:   c   h   d
21:   c   h   e
22:   c   h   f
23:   d   e   a
24:   d   e   b
25:   d   e   c
26:   d   e   f
27:   d   f   a
28:   d   f   b
29:   d   f   c
30:   d   f   e
31:   d   g   a
32:   d   g   b
33:   d   g   c
34:   d   g   e
35:   d   g   f
36:   d   h   a
37:   d   h   b
38:   d   h   c
39:   d   h   e
40:   d   h   f
41:   e   f   a
42:   e   f   b
43:   e   f   c
44:   e   f   d
45:   e   g   a
46:   e   g   b
47:   e   g   c
48:   e   g   d
49:   e   g   f
50:   e   h   a
51:   e   h   b
52:   e   h   c
53:   e   h   d
54:   e   h   f
55:   f   g   a
56:   f   g   b
57:   f   g   c
58:   f   g   d
59:   f   g   e
60:   f   h   a
61:   f   h   b
62:   f   h   c
63:   f   h   d
64:   f   h   e
65:   g   h   a
66:   g   h   b
67:   g   h   c
68:   g   h   d
69:   g   h   e
70:   g   h   f
    g40 g40 g20

[[4]]
     g40 g30 g30
  1:   c   a   b
  2:   c   a   d
  3:   c   a   e
  4:   c   a   f
  5:   c   a   g
 ---            
122:   h   d   f
123:   h   d   g
124:   h   e   f
125:   h   e   g
126:   h   f   g

[[5]]
    g40 g20 g20 g20
 1:   c   a   b   d
 2:   c   a   b   e
 3:   c   a   b   f
 4:   c   a   d   e
 5:   c   a   d   f
 6:   c   a   e   f
 7:   c   b   d   e
 8:   c   b   d   f
 9:   c   b   e   f
10:   c   d   e   f
11:   d   a   b   c
12:   d   a   b   e
13:   d   a   b   f
14:   d   a   c   e
15:   d   a   c   f
16:   d   a   e   f
17:   d   b   c   e
18:   d   b   c   f
19:   d   b   e   f
20:   d   c   e   f
21:   e   a   b   c
22:   e   a   b   d
23:   e   a   b   f
24:   e   a   c   d
25:   e   a   c   f
26:   e   a   d   f
27:   e   b   c   d
28:   e   b   c   f
29:   e   b   d   f
30:   e   c   d   f
31:   f   a   b   c
32:   f   a   b   d
33:   f   a   b   e
34:   f   a   c   d
35:   f   a   c   e
36:   f   a   d   e
37:   f   b   c   d
38:   f   b   c   e
39:   f   b   d   e
40:   f   c   d   e
41:   g   a   b   c
42:   g   a   b   d
43:   g   a   b   e
44:   g   a   b   f
45:   g   a   c   d
46:   g   a   c   e
47:   g   a   c   f
48:   g   a   d   e
49:   g   a   d   f
50:   g   a   e   f
51:   g   b   c   d
52:   g   b   c   e
53:   g   b   c   f
54:   g   b   d   e
55:   g   b   d   f
56:   g   b   e   f
57:   g   c   d   e
58:   g   c   d   f
59:   g   c   e   f
60:   g   d   e   f
61:   h   a   b   c
62:   h   a   b   d
63:   h   a   b   e
64:   h   a   b   f
65:   h   a   c   d
66:   h   a   c   e
67:   h   a   c   f
68:   h   a   d   e
69:   h   a   d   f
70:   h   a   e   f
71:   h   b   c   d
72:   h   b   c   e
73:   h   b   c   f
74:   h   b   d   e
75:   h   b   d   f
76:   h   b   e   f
77:   h   c   d   e
78:   h   c   d   f
79:   h   c   e   f
80:   h   d   e   f
    g40 g20 g20 g20

[[6]]
     g30 g30 g20 g20
  1:   a   b   c   d
  2:   a   b   c   e
  3:   a   b   c   f
  4:   a   b   d   e
  5:   a   b   d   f
 ---                
221:   g   h   c   e
222:   g   h   c   f
223:   g   h   d   e
224:   g   h   d   f
225:   g   h   e   f

推荐阅读