如何在R中的负（和正）值上使用for循环？答案

【问题标题】：How to use a for loop over negative (and positive) values in R?如何在R中的负（和正）值上使用for循环？
【发布时间】：2021-12-17 22:47:55
【问题描述】：

我正在尝试在一系列正负值上使用 for 循环，然后绘制结果。但是，我无法让 R 不绘制正确的值，因为负值似乎搞砸了索引。

更准确地说，我正在运行的代码是：

# Setup objects
R = (1:20)
rejection = rep(NA, 20)
t = seq(from = -10, to = 10, by = 1)
avg_rej_freq = rep(NA, 21)

# Test a hypothesis for each possible value of x and each replication
for (x in t) {
  for (r in R) {
    # Generate 1 observation from N(x,1)
    y =  rnorm(1, x, 1)
    # Take the average of this observation
    avg_y = mean(y)
    # Test this observation using the test we found in part a
    if (avg_y >= 1 + pnorm(.95))
    {rejection[r] = 1}
    if (y < 1 + pnorm(.95))
    {rejection[r] = 0}
  }
  # Calculate the average rejection frequency across the 20 samples
  avg_rej_freq[x] = mean(rejection)
}

# Plot the different values of x against the average rejection frequency
plot(t, avg_rej_freq)

生成的图表应如下所示

# Define the rejection probability for n=1
rej_prob = function(x)(1-pnorm(1-x+qnorm(0.95)))

# Plot it
curve(rej_prob,from = -10, to = 10, xlab = expression(theta), 
      ylab = "Rejection probability")

...但是我的代码显然有问题，将图表上的正值向左移动。

任何有关如何解决此问题的帮助将不胜感激！

【问题讨论】：

标签： r

【解决方案1】：

是的，因为您怀疑负指数会导致问题。 R 不知道如何将某些东西作为“负优先”对象存储在向量中，所以它只是丢弃它们。相反，请尝试使用 seq_along 生成所有正索引的向量并循环遍历它们：

# Setup objects
R = (1:20)
rejection = rep(NA, 20)
t = seq(from = -10, to = 10, by = 1)
avg_rej_freq = rep(NA, 21)

# Test a hypothesis for each possible value of x and each replication
for (x in seq_along(t)) {
  for (r in R) {
    # Generate 1 observation from N(x,1)
    # Now we ask for the value of t at index x rather than t directly
    y =  rnorm(1, t[x], 1)
    # Take the average of this observation
    avg_y = mean(y)
    # Test this observation using the test we found in part a
    if (avg_y >= 1 + pnorm(.95))
    {rejection[r] = 1}
    if (y < 1 + pnorm(.95))
    {rejection[r] = 0}
  }
  # Calculate the average rejection frequency across the 20 samples
  avg_rej_freq[x] = mean(rejection)
}

# Plot the different values of x against the average rejection frequency
plot(t, avg_rej_freq)

产生以下情节：

【讨论】：

【解决方案2】：

不知道为什么要使用 for 循环来模拟矢量化函数 pnrom()，但仍然纠正代码中的错误（检查 cmets）：

# Test a hypothesis for each possible value of x and each replication
for (x in t) {
  for (r in R) {
    # Generate 1 observation from N(x,1)
    y =  rnorm(1, x, 1)
    # no need to take average since you have a single observation
    # Test this observation using the test we found in part a
    rejection[r] = ifelse(y >= 1 + pnorm(.95), 1, 0)
  }
  # Calculate the average rejection frequency across the 20 samples
  # `R` vector index starts from 1, transform your x values s.t., negative values become positive
  avg_rej_freq[x-min(t)+1] = mean(rejection)
}


# Define the rejection probability for n=1
rej_prob = function(x)(1-pnorm(1-x+qnorm(0.95)))

# Plot it
curve(rej_prob,from = -10, to = 10, xlab = expression(theta), 
      ylab = "Rejection probability")

# plot your points
points(t, avg_rej_freq, pch=19, col='red')

【讨论】：

【解决方案3】：

不知道为什么 for 循环等，你在做什么可以折叠成一行。其余代码取自@Sandipan Dey：

R <- 20
t <- seq(from = -10, to = 10, by = 1)

#All the for-loops collapsed into this one line:
avg_rej_freq <- rowMeans(matrix(rnorm(R * length(t), t), 21) >= 1 + pnorm(.95))

rej_prob <- function(x) 1 - pnorm(1 - x + qnorm(0.95))
curve(rej_prob,from = -10, to = 10, xlab = expression(theta), 
               ylab = "Rejection probability")

# plot your points
points(t, avg_rej_freq, pch=19, col='red')

【讨论】：