嵌套循環-標記奪回

Question

感謝您抽時間閱讀。

下面的代碼創建了一個圖形，該圖形將抽取100個樣本，這些樣本在總人口（400）的5％和15％之間。

但是，我想做的是在圖中添加另外兩個部分。 它看起來像這樣：

從1-100個樣本中抽取100個樣本，它們占總人口的5％至15％（400）。 從101-200個樣本中抽取100個樣本，樣本數量占總人口的5％至15％（800）。 從201-300抽取100個樣本，樣本數量占總人口的5％至15％（300）。

我認為這將需要嵌套的for循環。 有人對如何執行此操作有意見嗎？

謝謝你的時間。 基爾斯滕

N <- 400
pop <- c(1:N)

lower.bound <- round(x = .05 * N, digits = 0)
lower.bound ## Smallest possible sample size

upper.bound <- round(x = .15 * N, digits = 0)
upper.bound ## Largest possible sample size

length.ss.interval <- length(c(lower.bound:upper.bound))
length.ss.interval ## total possible sample sizes, ranging form lower.bound
to upper.bound
sample(x = c(lower.bound:upper.bound),
       size = 1,
       prob = c(rep(1/length.ss.interval, length.ss.interval)))

n.samples <- 100

dat <- matrix(data = NA,
              nrow = length(pop),
              ncol = n.samples + 1)

dat[,1] <- pop

for(i in 2:ncol(dat)) {
  a.sample <- sample(x = pop,
                     size = sample(x = c(lower.bound:upper.bound),
                                   size = 1,
                                   prob = c(rep(1/length.ss.interval,
length.ss.interval))),
                     replace = FALSE)
  dat[,i] <- dat[,1] %in% a.sample
}
schnabel.comp <- data.frame(sample = 1:n.samples,
                            n.sampled = apply(X = dat, MARGIN = 2, FUN =
sum)[2:length(apply(X = dat, MARGIN = 2, FUN = sum))]
)
n.prev.sampled <- c(0, rep(NA, n.samples-1))
n.prev.sampled

n.prev.sampled[2] <- sum(ifelse(test = dat[,3] == 1 & dat[,2] == 1,
                                yes = 1,
                                no = 0))

for(i in 4:ncol(dat)) {
  n.prev.sampled[i-1] <- sum(ifelse(test = dat[,i] == 1 &
rowSums(dat[,2:(i-1)]) > 0,
                                    yes = 1,
                                    no = 0))
}

schnabel.comp$n.prev.sampled <- n.prev.sampled
schnabel.comp$n.newly.sampled <- with(schnabel.comp,
                                      n.sampled - n.prev.sampled)
schnabel.comp$cum.sampled <- c(0,
cumsum(schnabel.comp$n.newly.sampled)[2:n.samples-1])
schnabel.comp$numerator <- with(schnabel.comp,
                                n.sampled * cum.sampled)
schnabel.comp$pop.estimate <- NA

for(i in 1:length(schnabel.comp$pop.estimate)) {
  schnabel.comp$pop.estimate[i] <- sum(schnabel.comp$numerator[1:i]) /
sum(schnabel.comp$n.prev.sampled[1:i])
}

if (!require("ggplot2")) {install.packages("ggplot2"); require("ggplot2")}
if (!require("scales")) {install.packages("scales"); require("scales")}


small.sample.dat <- schnabel.comp

small.sample <- ggplot(data = small.sample.dat,
                       mapping = aes(x = sample, y = pop.estimate)) +
  geom_point(size = 2) +
  geom_line() +
  geom_hline(yintercept = N, col = "red", lwd = 1) +
  coord_cartesian(xlim = c(0:100), ylim = c(300:500)) +
  scale_x_continuous(breaks = pretty_breaks(11)) +
  scale_y_continuous(breaks = pretty_breaks(11)) +
  labs(x = "\nSample", y = "Population estimate\n",
       title = "Sample sizes are between 5% and 15%\nof the population") +
  theme_bw(base_size = 12) +
  theme(aspect.ratio = 1)

我的想法是使用以下命令創建嵌套的ifelse語句：

N.2 <- 800
N.3 <- 300
pop.2 <- c(401:N.2)
pop.3 <- c(801:N)

lower.bound.2 <- round(x = .05 * N.2, digits = 0)
upper.bound.2 <- round(x = .15 * N.2, digits = 0)

lower.bound.3 <- round(x = .05 * N.3, digits = 0)
upper.bound.3 <- round(x = .15 * N.3, digits = 0)

也許...的一些排列

dat <- imatrix(ifelse(n.samples ,= 100),
              yes = nrow = length(pop),
              no = ifelse(n.samples > 100 & > 201),
              yes = nrow = length(pop.2),
              no = nrow = length(pop.3),
              ncol = n.samples + 1)

Answer 1

這是您想要的嗎？ 我在下面編寫的函數mark_recapture接受四個參數（樣本數，樣本的上下限和總體大小），並輸出一個矩陣，其中的行代表總體中的個體，而列則代表樣本。 如果在給定的樣本中捕獲了一個個體，則該個體為1，否則為0。定義該函數后，您可以使用3種不同的總體大小對其運行3次，以獲得3種不同的矩陣。

# define variables
num_samp <- 100
lower_sampsize <- 0.05
upper_sampsize <- 0.15

# define sampling function that outputs matrix
mark_recapture <- function (num_samp, pop_size, lower_sampsize, upper_sampsize) {

    # empty matrix
    mat <- matrix(0, pop_size, num_samp)

    # min and max sample size
    min <- ceiling(lower_sampsize*pop_size)
    max <- floor(upper_sampsize*pop_size)

    # vector of random sample sizes between min and max
    samp_sizes <- sample(min:max, num_samp, replace=TRUE)

    # draw the samples and fill in the matrix
    for (i in 1:num_samp) {mat[sample(1:pop_size, samp_sizes[i]),i] <- 1}

    # return matrix
    return(mat)
}

# do the sampling from the 3 populations
mat1 <- mark_recapture(num_samp=num_samp, pop_size=400, lower_sampsize=lower_sampsize, upper_sampsize=upper_sampsize)
mat2 <- mark_recapture(num_samp=num_samp, pop_size=800, lower_sampsize=lower_sampsize, upper_sampsize=upper_sampsize)
mat3 <- mark_recapture(num_samp=num_samp, pop_size=300, lower_sampsize=lower_sampsize, upper_sampsize=upper_sampsize)

盡管這超出了這個問題的范圍，但我只會提到有專用的R包來分析和模擬標記捕獲數據，例如multimark 。 只是Google“重新獲得CRAN商標”，您會發現許多選擇。 我建議您仔細研究一下這些內容，並仔細考慮您要在此處實現的目標，因為您可能正在嘗試重新發明輪子。