Question

正如标题所述，我在将计算值添加到空向量时遇到问题，并且在具有1000次迭代的FOR循环的情况下，我只是插入相同的值1000次，而不是100次中的每一个的不同值迭代。我逐行筛选代码，将i和j设置为某些值以强制停止迭代。请参阅下面的代码：

# Population Size
N <- 100000
# Iterations per Sample Size
n.sim <- 1000
# Various Sample Sizes
samples.sim <- seq(from=100, to=5000, by=50)

cap.recap.partdeux <- function(N, n.sim, samples.sim){
n <- NA            # sample size
bias.of.est <- NA  # bias of estimator
sd.of.est <- NA    # standard deviation of estimators by sample
l.sim.chap <- NA   # temporary list of estimators to use for later
  # For Loop: Various Sample Sizes
  for(i in 1:length(samples.sim)){
    # For Loop: Iterations per Sample Size
    for (j in 1:n.sim){
      i <- 1:1
      j <- 1:2
      # Catch One
      sim.one <- sample(N, samples.sim, replace=T, prob=NULL)
      sim.one <- as.numeric(sim.one) # Convert to numeric
      # Catch Two
      sim.two <- sample(N, samples.sim, replace=T, prob=NULL)
      sim.two <- as.numeric(sim.two) # Convert to numeric
      # Find Common Elements
      sim.m.two <- intersect(sim.one, sim.two)
      # Amount of Common Elements
      sim.l.m.two <- length(sim.m.two)
      # Calculate Chapman Estimator
      sim.chap <- ((samples.sim[i]+1)*(samples.sim[i]+1)/(sim.l.m.two+1))-1
      l.sim.chap[j] <- list(sim.chap)
    } # End For Loop: Iterations per Sample Size
    # Calculate bias of estimator for each sample
    sum.b.est <- sum(unlist(l.sim.chap), na.rm=T)
    bias.est <- (sum.b.est/n.sim)-N
    bias.of.est[i] <- bias.est
    # Calculate standard deviation of estimator for each sample
    sd.est <- sd(unlist(l.sim.chap), na.rm = TRUE)
    sd.of.est[i] <- sd.est
    # Sample Size
    n[i] <- samples.sim[i]
  } # End For Loop: Various Sample Sizes
  # Return Three Columns and make Data Frame
  Three <- (data.frame(n, bias.of.est, sd.of.est))
  # List of Data Frame with True Population
  Output <- (list(Three, "POP"=N))
  return(Output)
} # End Function
Output <- cap.recap.partdeux(N=100000, n.sim=1000, samples.sim)
Test <- data.frame(Output)

基本上，在l.sim.chap [i]＆lt; - list（sim.chap）的行上，向量l.sim.chap加载了第一次sim.chap迭代的重复值，而不是sim.chap来自每次和后续迭代的值。

Answer 1

删除了i＆lt; -1和j＆lt; - 1.代码现在可以使用了。太奇怪了。

# Population Size
N <- 100000
# Iterations per Sample Size
n.sim <- 1000
# Various Sample Sizes
samples.sim <- seq(from=100, to=5000, by=50)

cap.recap.partdeux <- function(N, n.sim, samples.sim){
n <- NA            # sample size
bias.of.est <- NA  # bias of estimator
sd.of.est <- NA    # standard deviation of estimators by sample
l.sim.chap <- NA   # temporary list of estimators to use for later
  # For Loop: Various Sample Sizes
  for(i in 1:length(samples.sim)){
    # For Loop: Iterations per Sample Size
    for (j in 1:n.sim){
      # Catch One
      sim.one <- sample(N, samples.sim, replace=T, prob=NULL)
      sim.one <- as.numeric(sim.one) # Convert to numeric
      # Catch Two
      sim.two <- sample(N, samples.sim, replace=T, prob=NULL)
      sim.two <- as.numeric(sim.two) # Convert to numeric
      # Find Common Elements
      sim.m.two <- intersect(sim.one, sim.two)
      # Amount of Common Elements
      sim.l.m.two <- length(sim.m.two)
      # Calculate Chapman Estimator
      sim.chap <- ((samples.sim[i]+1)*(samples.sim[i]+1)/(sim.l.m.two+1))-1
      l.sim.chap[j] <- list(sim.chap)
    } # End For Loop: Iterations per Sample Size
    # Calculate bias of estimator for each sample
    sum.b.est <- sum(unlist(l.sim.chap), na.rm=T)
    bias.est <- (sum.b.est/n.sim)-N
    bias.of.est[i] <- bias.est
    # Calculate standard deviation of estimator for each sample
    sd.est <- sd(unlist(l.sim.chap), na.rm = TRUE)
    sd.of.est[i] <- sd.est
    # Sample Size
    n[i] <- samples.sim[i]
  } # End For Loop: Various Sample Sizes
  # Return Three Columns and make Data Frame
  Three <- (data.frame(n, bias.of.est, sd.of.est))
  # List of Data Frame with True Population
  Output <- (list(Three, "POP"=N))
  return(Output)
} # End Function
Output <- cap.recap.partdeux(N=100000, n.sim=1000, samples.sim)
Test <- data.frame(Output)

问题在FOR循环中添加输入/计算

1 个答案: