英文:
Permute the position of a subset of a vector
问题
不要回答我要翻译的问题。
以下是已翻译的内容：
我想对向量的子集进行排列。
例如，假设我有一个向量（x），然后我选择向量的一个随机子集（例如，40% 的值）。
我想要的是生成一个新的向量（x2），与（x）相同，除了随机子集中的值的位置被随机交换。
例如：
- x = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10
- 随机子集 = 1, 4, 5, 8
- x2 可能是 = 4, 2, 3, 8, 1, 6, 7, 5, 9, 10
这是一个示例向量（x）以及如何选择一个占其值40%的随机子集的索引的方法。帮助生成（x2）将不胜感激！
x <- seq(1,10,1)
which(x%in%sample(x)[seq_len(length(x)*0.40)])
英文:
I want to permute a subset of a vector.
For example, say I have a vector (x) and I select a random subset of the vector (e.g., 40% of its values).
What I want to do is output a new vector (x2) that is identical to (x) except the positions of the values within the random subset are randomly swapped.
For example:
- x = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10
- random subset = 1, 4, 5, 8
- x2 could be = 4, 2, 3, 8, 1, 6, 7, 5, 9, 10
Here's an an example vector (x) and how I'd select the indices of a random subset of 40% of its values. Any help making (x2) would be appreciated!
x <- seq(1,10,1)
which(x%in%sample(x)[seq_len(length(x)*0.40)])
答案1
得分: 3
首先从索引中抽取比例为 p 的 sample，然后对具有该索引的元素进行采样和重新分配。
f <- function(x, p=0.4) {
r <- sample(seq_along(x), length(x)*p)
x[r] <- sample(x[r])
`attr<-`(x, 'subs', r) ## 添加具有被抽样的索引的属性
}
set.seed(42)
f(x)
# [1] 8 2 3 4 1 5 7 10 6 9
# attr(,"subs")
# [1] 1 5 10 8
数据:
x <- 1:10
英文:
First draw a sample of proportion p from the indices, then sample and re-assign elements with that indices.
f <- \(x, p=0.4) {
r <- sample(seq_along(x), length(x)*p)
x[r] <- sample(x[r])
`attr<-`(x, 'subs', r) ## add attribute w/ indices that were sampled
}
set.seed(42)
f(x)
# [1] 8 2 3 4 1 5 7 10 6 9
# attr(,"subs")
# [1] 1 5 10 8
Data:
x <- 1:10
答案2
得分: 1
以下是翻译好的代码部分：
# 确保有更快的代码来完成你的要求，但这是一个解决方案：
x <- seq(1,10,1)
y <- which(x %in% sample(x)[seq_len(length(x) * 0.40)]) # 定义 "y" 为随机子集的向量
# 需要的库
library(combinat)
permutation <- permn(y) # permn() 函数在R中生成元素x的所有排列的列表。
# https://www.geeksforgeeks.org/calculate-combinations-and-permutations-in-r/
permutation_sampled <- sample(permutation, 1) # 从排列中随机抽取一个。
x[y] <- permutation_sampled[[1]] # 使用y作为应替代元素的索引，将所选排列替换到x中。
希望这对你有所帮助！如果你有任何其他问题，请随时提出。
英文:
For sure there is a faster code to do what you are asking, but, a solution would be:
x <- seq(1,10,1)
y <- which(x%in%sample(x)[seq_len(length(x)*0.40)]) # Defined as "y" the vector of the random subset
# required libraries
library(combinat)
permutation <- permn(y) # permn() function in R generates a list of all permutations of the elements of x.
# https://www.geeksforgeeks.org/calculate-combinations-and-permutations-in-r/
permutation_sampled <- sample(permutation,1) # Sample one of the permutations.
x[y] <- permutation_sampled[[1]] # Substitute the selected permutation in x using y as the index of the elements that should be substituted.
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