英文:
Creating result groups in R, using each element once (combination without repetition)
问题
我有一个包含6个个体的数据集：A，B，C，D，E，F。
我想将它们分成两组，每组包含三个个体，并已经在R中使用combn函数完成了这个任务：
m <- combn(n, 3)
这为我提供了所有20种可能的组合，其中个体出现在多个组中。然后，我想找到所有可能的结果组合，其中每个个体只能使用一次。
我想要使用不重复的组合来完成这个任务：
C(n,r) = n! / r!(n-r)!，因此我会得到10个结果，看起来像这样：
- abc + def
- abd + cef
- abe + cdf
- abf + cde
- acd + bef
- ace + bdf
- acf + bde
- ade + bcf
- adf + bce
- aef + bcd
我不太确定如何在R中编写这个代码，从我生成的组合列表中获取这些结果。
编辑：要生成我正在使用的数据集，我使用了以下代码：
individuals <- c("a","b","c","d","e","f")n <- length(individuals)x <- 3comb = function(n, x) {factorial(n) / factorial(n-x) / factorial(x)}comb(n,x)(m <- combn(n, 3))numbers <- mletters <- individualsfor (i in 1:length(numbers)) {m[i] <- letters[numbers[i]]}
英文:
I have a dataset of 6 individuals: A,B,C,D,E,F
I want to group these into two groups of three individuals and have done so with the combn function in R:
m <- combn(n, 3)
This gives me all 20 possible groups where individuals occur in multiple groups. From this set of groups I then went to find all possible combinations of results, where each individual can only be used once.
I would like to do this using combinations without repetition:
C(n,r) = n! / r!(n-r)! and would therefore get 10 results that would look like this:
- abc + def
- abd + cef
- abe + cdf
- abf + cde
- acd + bef
- ace + bdf
- acf + bde
- ade + bcf
- adf + bce
- aef + bcd
I am not sure how to code this in R, from the list of groups that I have generated.
Edit: to generate the dataset I am using I have used the following code:
individuals <- c("a","b","c","d","e","f")n <- length(individuals)x <- 3comb = function(n, x) {factorial(n) / factorial(n-x) / factorial(x)}comb(n,x)(m <- combn(n, 3))numbers <- mletters <- individualsfor (i in 1:length(numbers)) {m[i] <- letters[numbers[i]]}
答案1
得分: 1
以下是您要翻译的代码部分的内容：
In base R:
- 创建 3 个字母的组合，并将其存储在列表中（
asplit） - 创建 2 个组（每个组包含 3 个字母）的新组合
- 使用
Filter仅保留两部分没有共同元素的组合
individuals <- c("a","b","c","d","e","f")combn(individuals, 3, simplify = FALSE) |>combn(m = 2, simplify = FALSE) |>Filter(f = \(x) !any(x[[1]] %in% x[[2]]))
输出：
[[1]][[1]][[1]][1] "a" "b" "c"[[1]][[2]][1] "d" "e" "f"[[2]][[2]][[1]][1] "a" "b" "d"[[2]][[2]][1] "c" "e" "f"[[3]][[3]][[1]][1] "a" "b" "e"[[3]][[2]][1] "c" "d" "f"[[4]][[4]][[1]][1] "a" "b" "f"[[4]][[2]][1] "c" "d" "e"[[5]][[5]][[1]][1] "a" "c" "d"[[5]][[2]][1] "b" "e" "f"[[6]][[6]][[1]][1] "a" "c" "e"[[6]][[2]][1] "b" "d" "f"[[7]][[7]][[1]][1] "a" "c" "f"[[7]][[2]][1] "b" "d" "e"[[8]][[8]][[1]][1] "a" "d" "e"[[8]][[2]][1] "b" "c" "f"[[9]][[9]][[1]][1] "a" "d" "f"[[9]][[2]][1] "b" "c" "e"[[10]][[10]][[1]][1] "a" "e" "f"[[10]][[2]][1] "b" "c" "d"
英文:
In base R:
- Create
combnations of 3 letters and store it in a list (asplit) - Create new
combnations of 2 groups (of 3 letters) Filterthe list to only keep combinations where the both parts have no element in common
individuals <- c("a","b","c","d","e","f")combn(individuals, 3, simplify = FALSE) |>combn(m = 2, simplify = FALSE) |>Filter(f = \(x) !any(x[[1]] %in% x[[2]]))
output
[[1]][[1]][[1]][1] "a" "b" "c"[[1]][[2]][1] "d" "e" "f"[[2]][[2]][[1]][1] "a" "b" "d"[[2]][[2]][1] "c" "e" "f"[[3]][[3]][[1]][1] "a" "b" "e"[[3]][[2]][1] "c" "d" "f"[[4]][[4]][[1]][1] "a" "b" "f"[[4]][[2]][1] "c" "d" "e"[[5]][[5]][[1]][1] "a" "c" "d"[[5]][[2]][1] "b" "e" "f"[[6]][[6]][[1]][1] "a" "c" "e"[[6]][[2]][1] "b" "d" "f"[[7]][[7]][[1]][1] "a" "c" "f"[[7]][[2]][1] "b" "d" "e"[[8]][[8]][[1]][1] "a" "d" "e"[[8]][[2]][1] "b" "c" "f"[[9]][[9]][[1]][1] "a" "d" "f"[[9]][[2]][1] "b" "c" "e"[[10]][[10]][[1]][1] "a" "e" "f"[[10]][[2]][1] "b" "c" "d"
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