R：从列表中随机选择项目

Question

我正在使用 R 编程语言。我想创建一个问题：

• 有4个人："person 1", "person 2", "person 3" ,"person 4"

• 有一份食物清单："pizza", "apples", "tacos", "ice cream", "grapes", "olives", "sushi", "chocolate", "cake", "nachos", "pasta", "cookies", "popcorn", "soup"

• 一年中有天的列表：1 to 365

在这个问题中，人员列表是随机排序的。此列表的顺序决定了谁“先选”。

根据列表的顺序，列表开头的人将被分配4个随机数：

• Rand_1_1（第一个随机数，第一个随机数）：第一个随机数将决定第一个人可以选择的“食物”数量（例如 3 个）

• Rand_1_2 (Person 1, Second Random Number) : 第二个随机数将根据“Rand_1_1”对应的“食物”数量决定第一个人选择哪些“食物”（例如“炸玉米饼”、“玉米片”、“汤”）

• Rand_1_3（第一个人，第三个随机数）：第三个随机数将决定第一个人的天数

• Rand_1_4（第 1 个人，第四个随机数）：第四个随机数将决定第一个人的天数的另一个界限（例如，第 1 个人可能被分配“41 到 160”天） .

这是为所有人完成的随机数分配过程：Rand_1_1、Rand_1_2、Rand_1_3、Rand_1_4、Rand_2_1、Rand_2_2、Rand_2_3、Rand_2_4、Rand_3_1、Rand_3_2、Rand_3_3、Rand_3_4、Rand_4_1、Rand_4_2、Rand_4_3、

但是，有一个逻辑约束：

• “人”不能选择之前“人”已经选择的“食物”（例如，如果人 1 在人 2 之前选择了选择，如果人 1 选择了“披萨”，那么人 2不能选择“披萨”）

• “人”不能选择前面“人”已经选择的“天数范围”

• 在所有 4 人都完成选择后，有可能某些“食品”和某些“日期范围”可以保持未选择状态。另一方面，假设如果第 1 个人首先选择并且他碰巧选择了所有“食物”——那么当然，其他人将没有“食物”。同样的逻辑也适用于“日期范围”。

现在，我正在尝试在 R 中编写代码：

首先，我加载了库：

#load libraries

library(dplyr)
library(gtools)

其次，我创建了数据：

#create data

#variable 1: days of the year 

days <- 1:365

#variable 2 : food

food <- c("pizza", "apples", "tacos", "ice cream", "grapes", "olives", "sushi", "chocolate", "cake", "nachos", "pasta", "cookies", "popcorn", "soup")
food_data <- data.frame(food)
food_data$id <-  1:nrow(food_data)

# people 
people <- c("person 1", "person 2", "person 3" ,"person 4")

第三，我创建了人们选择的顺序：

# randomize order of people : this decides the order of who picks "days" and "food"

set.seed(123)

order_of_people = permute(people)

# in this example, "person 3" will pick first, "person 4" will pick second, "person 1" will pick third and "person 2" will pick last

order_of_people
[1] "person 3" "person 4" "person 1" "person 2"

我的问题：我知道如何为每个人分配一个随机数，但我不知道如何分配随机数以遵守逻辑约束。例如：

#choices for person 3 (according to the random seed, person 3 will pick 5 food items)

set.seed(120)

dim = dim(food_data)

#number of items selected by person 3

Rand_3_1 <- sample.int(dim[1], 1)

#which food items selected by person 3 (corresponding to the food_id : "3, 9, 6, 7, 4")
set.seed(120)

Rand_3_2 = c( sample.int(dim[1], 1), sample.int(dim[1], 1), sample.int(dim[1], 1),  sample.int(dim[1], 1), sample.int(dim[1], 1))

#which days selected by person 3 (according to this random seed, "65 to 87")

set.seed(120)

Rand_3_3 <- sample.int(365, 1)
Rand_3_4 <- sample.int(365, 1)

因此，我可以为每个人创建“选择框架”：

#Person 1
Rand_1_1 <- sample.int(dim[1], 1)

Rand_1_2 = c( #fill randomly with amount of items specified by Rand_1_1)

Rand_1_3 <- sample.int(365, 1)
Rand_1_4 <- sample.int(365, 1)

#Person 2
Rand_2_1 <- sample.int(dim[1], 1)

Rand_2_2 = c( #fill randomly with amount of items specified by Rand_2_1)

Rand_2_3 <- sample.int(365, 1)
Rand_2_4 <- sample.int(365, 1)


#Person 3
Rand_3_1 <- sample.int(dim[1], 1)

Rand_3_2 = c( #fill randomly with amount of items specified by Rand_3_1)

Rand_3_3 <- sample.int(365, 1)
Rand_3_4 <- sample.int(365, 1)

#Person 4
Rand_4_1 <- sample.int(dim[1], 1)

Rand_4_2 = c( #fill randomly with amount of items specified by Rand_4_1)

Rand_4_3 <- sample.int(365, 1)
Rand_4_4 <- sample.int(365, 1)

但我不知道如何做到这一点，以便遵守逻辑约束。最后，我试图产生这样的东西：

#desired results
Person 1 : "apples, tacos" and "4-51"
Person 2: "cookies, nachos, cake, olives", and "56-180"
Person 3: "ice cream", and "200-214"
Person 4: "sushi, popcorn" and "350-365"

有人可以告诉我怎么做吗？

谢谢

Answer 1

这似乎很简单，除非我错过了什么，你有一份食物和日子的清单，所以每次你遇到一个人时，只要跟踪可用的选项，然后只从这些选项中选择：

## initialize the inputs
set.seed(1)
foods <- letters[1:20]
nfoods <- sample(5, 4)
tmp <- list(unpicked = list(food = foods, days = 1:365))

## run it
tmp <- lapply(nfoods, function(x) tmp <<- f(tmp$unpicked$food, x, tmp$unpicked$days))

## summarize it
res <- sapply(seq_along(tmp), function(ii) {
  x <- tmp[[ii]]
  x <- sprintf('person %s: "%s" and "%s"', ii, toString(x$picked$food),
               paste0(x$picked$days, collapse = '-'))
  cat(x, sep = '\n')
  invisible(x)
})
# person 1: "b" and "270-299"
# person 2: "s, a, k, g" and "306-336"
# person 3: "m, t, h" and "42-89"
# person 4: "j, f, q, o, l" and "129-210"


f <- function(food, n, days) {
  # f(1:4, 3, 1:10)
  food1 <- sample(food, n)
  
  days1 <- split(days, cumsum(is.na(days)))
  days1 <- sample(Filter(function(x) length(x) > 2, days1), 1)[[1]]
  days1 <- sort(sample(sort(days1), 2))
  
  days[do.call('seq', as.list(days1))] <- NA
  
  list(
    picked = list(food = food1, days = days1),
    unpicked = list(food = setdiff(food, food1), days = days)
  )
}