问题

以下是您的翻译内容：

以下是我的数据示例（尽管我有数百行）。 每个ID都是唯一的，可能与一个ID关联多个个体（例如，个体A和D）。

我想要随机选择一个ID 1000次，有机会重新选择相同的ID，并记录在这个抽样方案中累积了多少个唯一的个体。

然后，我想生成一个图表，显示需要选择多少个ID才能累积所有唯一的个体，以便曲线在x轴上是ID，y轴上是个体时达到渐近线。

编辑：
我上面对所需图表的描述不清楚。编辑如下：

我想生成一个图表，显示选择（带替代）一个ID多少次才能积累所有唯一的个体... 而不是与唯一个体相关的唯一ID数量。曲线应该在选择了多少次ID后达到渐近线，而在每次样本后将ID放回池中。

例如，如果有500个与200个个体相关联的ID，我想要随机抽样这500个ID的池1000次（或者无论多少次），在每次样本后将ID放回池中，以查看我们需要多少次才能积累所有200个个体。

ID individual
1  A
2  B
3  A
4  C
5  D
6  D
7  D

答案1

得分: 1

I have translated the code parts for you:

我想要随机选择一个ID 1000次，有重新抽样相同ID的机会，并记录在这个抽样方案中累积了多少独特的个体。

您可以使用`dplyr`中的`slice_sample`从数据框中抽样行。例如：

```R
library(dplyr)

N_id <- 2000
N_individuals <- 50

df_full <- data.frame(id = 1:N_id,
                      individual = sample(1:N_individuals, N_id, replace = TRUE))

df_sample <- slice_sample(df_full, n = 1000, replace = TRUE)

unique_individuals <- length(unique(df_sample$individual))

然后，我想生成一个图表，显示需要选择多少个ID来积累所有的独特个体，以便曲线在X轴上显示ID，Y轴上显示个体时趋于渐近。

您可以将其包装成一个函数，以生成不同数量的ID和个体（以及抽样次数）的值，然后使用ggplot或其他绘图工具绘制不同的值。然而，这似乎是一个组合数学问题，也许更适合https://math.stackexchange.com/，因为这些值完全取决于ID和个体的数量。

在函数中编辑：

N_id <- 2000
N_individuals <- 50
N_draws <- 1000

sample_df_parameterized <- function(n_id, n_individuals, n_draws) {
  
  df_full <- data.frame(id = 1:n_id,
                        individual = sample(1:n_individuals, n_id, replace = TRUE))
  
  df_sample <- slice_sample(df_full, n = n_draws, replace = TRUE)
  
  unique_individuals <- length(unique(df_sample$individual))
  
  result_df <- data.frame(n_id = n_id,
                          n_individuals = n_individuals,
                          n_draws = n_draws,
                          unique_individuals = unique_individuals)
  
  return(result_df)
  
}

sample_df_parameterized(n_id = N_id,
                        n_individuals = N_individuals,
                        n_draws = N_draws)

Please note that the translation includes the code only, as you requested.

<details>
<summary>英文:</summary>

&gt; I would like to randomly select an ID 1000 times with the opportunity
&gt; to re-sample the same ID, and store how many unique individuals are
&gt; accumulated over this sampling scheme.

You can use `slice_sample` from `dplyr` to sample rows from a data frame. For example:

library(dplyr)

N_id <- 2000
N_individuals <- 50

df_full <- data.frame(id = 1:N_id,
individual = sample(1:N_individuals, N_id, replace = TRUE))

df_sample <- slice_sample(df_full, n = 1000, replace = TRUE)

unique_individuals <- length(unique(df_sample$individual))

> I would then like to generate a plot that shows how many ID's would
> need to be selected to accumulate all of the unique individuals, so
> that the curve reaches asymptote with ID's on the x-axis and
> individuals on the y-axis.

You can wrap that into a function to produce values for different numbers of ids and individuals (and draws for that matter), and then plot the different values using `ggplot` or another plotting tool. However, this strikes me as a combinatorics question perhaps better suited for https://math.stackexchange.com/, since these values will fully depend on the number of ids, individuals.

Edit: in a function:

N_id <- 2000
N_individuals <- 50
N_draws <- 1000

sample_df_parameterized <- function(n_id, n_individuals, n_draws) {

df_full <- data.frame(id = 1:n_id,
individual = sample(1:n_individuals, n_id, replace = TRUE))

df_sample <- slice_sample(df_full, n = n_draws, replace = TRUE)

unique_individuals <- length(unique(df_sample$individual))

result_df <- data.frame(n_id = n_id,
n_individuals = n_individuals,
n_draws = n_draws,
unique_individuals = unique_individuals)

return(result_df)

}

sample_df_parameterized(n_id = N_id,
n_individuals = N_individuals,
n_draws = N_draws)

</details>



# 答案2
**得分**: 1

以下是翻译好的部分：

这是使用基本的R语言尝试的示例，使用自定义函数来计算累积唯一值的数量：

更大的示例数据：

```R
set.seed(2)
dat <- data.frame(ID=1:250, individual=sample(1:100, 500, replace=TRUE))
length(unique(dat$individual))
##100

抽样、计数累积值，并绘图：

tmp <- dat[sample(seq_len(nrow(dat)), 1000, replace=TRUE),]

cumfun <- function(x) lengths(Reduce(union, x, accumulate=TRUE))
idcum <- cumfun(tmp$ID)
indcum <- cumfun(tmp$individual)

plot(idcum, indcum, type="l")

如果您愿意，您还可以稍微美化图表，添加您选择的最佳拟合线和一些更好的坐标轴：

plot(idcum, indcum, type="l", ylim=c(0,100), las=1,
     xlab="Cumulative ID count", ylab="Cumulative Individuals count",
     cex.lab=0.8, cex.axis=0.8, lty=2)
f <- function(x,a,b) {x/(a+b*x)}
fit <- nls(indcum ~ f(idcum,a,b), start=c(a=1,b=1))
curve(do.call(f, c(list(x), coef(fit))), add=TRUE, col="red")

set.seed(2)
dat <- data.frame(ID=1:250, individual=sample(1:100, 500, replace=TRUE))
length(unique(dat$individual))
##100

tmp <- dat[sample(seq_len(nrow(dat)), 1000, replace=TRUE),]

cumfun <- function(x) lengths(Reduce(union, x, accumulate=TRUE))
idcum <- cumfun(tmp$ID)
indcum <- cumfun(tmp$individual)

plot(idcum, indcum, type="l")

plot(idcum, indcum, type="l", ylim=c(0,100), las=1,
     xlab="Cumulative ID count", ylab="Cumulative Individuals count",
     cex.lab=0.8, cex.axis=0.8, lty=2)
f <- function(x,a,b) {x/(a+b*x)}
fit <- nls(indcum ~ f(idcum,a,b), start=c(a=1,b=1))
curve(do.call(f, c(list(x), coef(fit))), add=TRUE, col="red")