英文:
R: Verifying The Results Of Coin Flips
问题
Here is the translated portion of your text:
"我正在使用R编程语言工作。
假设我有以下问题：
- 有一枚硬币，如果它正面朝上，那么下一次翻转为正面的概率为0.6（如果是反面，下一次翻转也是0.6）
- 一个班级里有100名学生
- 每个学生随机翻转这枚硬币的次数
- 第n个学生的最后一次翻转不会影响第n+1个学生的第一次翻转（即，当下一个学生翻转硬币时，第一次翻转有0.5的正反面概率，但这个学生的下一次翻转取决于上一次翻转）
我试图编写R代码来表示这个问题。
首先，我定义了变量：
library(dplyr)library(stringr)# 生成数据set.seed(123)ids <- 1:100student_id <- sample(ids, 100000, replace = TRUE)coin_result <- character(1000)coin_result[1] <- sample(c("H", "T"), 1)
接下来，我尝试编写翻转过程：
for (i in 2:length(coin_result)) {if (student_id[i] != student_id[i-1]) {coin_result[i] <- sample(c("H", "T"), 1)} else if (coin_result[i-1] == "H") {coin_result[i] <- sample(c("H", "T"), 1, prob = c(0.6, 0.4))} else {coin_result[i] <- sample(c("H", "T"), 1, prob = c(0.4, 0.6))}}# 整理数据my_data <- data.frame(student_id, coin_result)my_data <- my_data[order(my_data$student_id),]
最后，我尝试验证结果：
my_data %>%group_by(student_id) %>%summarize(Sequence = str_c(coin_result, lead(coin_result)), .groups = 'drop') %>%filter(!is.na(Sequence)) %>%count(Sequence)
尽管代码运行了，但我不认为我的代码是正确的 - 当我看结果时：
# A tibble: 4 x 2Sequence n<chr> <int>1 HH 238102 HT 250433 TH 250424 TT 26005
我认为如果我是正确的，HH 应该明显大于 HT，而 TT 应该明显大于 TH。
请问有人可以告诉我是否我做对了，以及如何纠正它吗？
谢谢！
英文:
I am working with the R programming language.
Suppose I have the following problem:
- There is a coin where if it lands head then the probability of the next flip being heads is 0.6 (and if tails then the next flip being tails is also 0.6)
- There are 100 students in a class
- Each student flips this coin a random number of times
- The last flip of student_n does not influence the first flip of student_n+1 (i.e. when the next student flips the coin, the first flip has 0.5 probability of heads or tails, but the next flip for this student depends on the previous flip)
I am trying to write R code to represent this problem.
First I defined the variables:
library(dplyr)library(stringr)# generate dataset.seed(123)ids <- 1:100student_id <- sample(ids, 100000, replace = TRUE)coin_result <- character(1000)coin_result[1] <- sample(c("H", "T"), 1)
Next, I tried to write the flipping process:
for (i in 2:length(coin_result)) {if (student_id[i] != student_id[i-1]) {coin_result[i] <- sample(c("H", "T"), 1)} else if (coin_result[i-1] == "H") {coin_result[i] <- sample(c("H", "T"), 1, prob = c(0.6, 0.4))} else {coin_result[i] <- sample(c("H", "T"), 1, prob = c(0.4, 0.6))}}#tidy upmy_data <- data.frame(student_id, coin_result)my_data <- my_data[order(my_data$student_id),]
Finally, I tried to verify the results:
my_data %>%group_by(student_id) %>%summarize(Sequence = str_c(coin_result, lead(coin_result)), .groups = 'drop') %>%filter(!is.na(Sequence)) %>%count(Sequence)
Even though the code ran, I don't think my code is correct - when I look at the results:
# A tibble: 4 x 2Sequence n<chr> <int>1 HH 238102 HT 250433 TH 250424 TT 26005
I think if I was correct, HH should have been significantly greater than HT , and TT should have been significantly greater than TH.
Can someone please tell me if I have done this correctly and how to correct it?
Thanks!
答案1
得分: 1
I think you need to sort the student_id vector before the loop, so that your comparison of student_id[i] != student_id[i-1] would be valid. Otherwise, it's not catching consecutive flips from the same student.
结果似乎合理，其中HH和TT一起占总翻转的60.4%。
library(tidyverse)set.seed(123)ids <- 1:100# 仅以下一行已更改，所有其他行与您的代码相同student_id <- sort(sample(ids, 100000, replace = TRUE))coin_result <- character(1000)coin_result[1] <- sample(c("H", "T"), 1)for (i in 2:length(coin_result)) {if (student_id[i] != student_id[i-1]) {coin_result[i] <- sample(c("H", "T"), 1)} else if (coin_result[i-1] == "H") {coin_result[i] <- sample(c("H", "T"), 1, prob = c(0.6, 0.4))} else {coin_result[i] <- sample(c("H", "T"), 1, prob = c(0.4, 0.6))}}# 整理数据my_data <- data.frame(student_id, coin_result)my_data <- my_data[order(my_data$student_id),]my_data %>%group_by(student_id) %>%summarize(Sequence = str_c(coin_result, lead(coin_result)), .groups = 'drop') %>%filter(!is.na(Sequence)) %>%count(Sequence)# 一个tibble: 4 × 2Sequence n1 HH 297632 HT 197823 TH 197754 TT 30580
英文:
I think you need to sort the student_id vector before the loop, so that your comparison of student_id[i] != student_id[i-1] would be valid. Otherwise, it's not catching consecutive flips from the same student.
The result seems to make sense, where HH and TT together occupies 60.4% of the total flips.
library(tidyverse)set.seed(123)ids <- 1:100# only the following line was changed, all other lines are same as your codestudent_id <- sort(sample(ids, 100000, replace = TRUE))coin_result <- character(1000)coin_result[1] <- sample(c("H", "T"), 1)for (i in 2:length(coin_result)) {if (student_id[i] != student_id[i-1]) {coin_result[i] <- sample(c("H", "T"), 1)} else if (coin_result[i-1] == "H") {coin_result[i] <- sample(c("H", "T"), 1, prob = c(0.6, 0.4))} else {coin_result[i] <- sample(c("H", "T"), 1, prob = c(0.4, 0.6))}}#tidy upmy_data <- data.frame(student_id, coin_result)my_data <- my_data[order(my_data$student_id),]my_data %>%group_by(student_id) %>%summarize(Sequence = str_c(coin_result, lead(coin_result)), .groups = 'drop') %>%filter(!is.na(Sequence)) %>%count(Sequence)# A tibble: 4 × 2Sequence n<chr> <int>1 HH 297632 HT 197823 TH 197754 TT 30580
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