freemarker怎么遍历map集合

 我对你的代码做了一些小调整而没有改变它的整体风格。 注意Yoong Kim的建议并尝试将代码分解成更小的部分，以使其更具可读性和可维护性，这将是一件好事。 
 
 
 
 您的函数现在获得两个“n”参数，表示每行中有多少个样本，以及您需要多少次迭代（列）。 
 
 
 你在循环中增加了数组Median和Mean ，这需要很多关于重新分配内存和复制内容的麻烦，这会减慢一切。 我已经预定义X_after并在循环后移动平均值和中值计算以避免这种情况。 （作为奖励， mean和median仅被调用一次而不是n_iteration次。） 
 
 
 对ifelse的调用并不是真的需要。 
 
 
 调用rlnorm一次，为x和lod生成足够的值比调用它两次要快一点。 
 
 
 这是更新的功能。 
 
dset2 <- function (mu, sigma, n_samples, n_iterations, p) {    
  X_after <- matrix(NA_real_, nrow = n_iterations, ncol = n_samples)
  pct_cens <- numeric(n_iterations)
  count <- 1
  while(count <= n_iterations) {     
    random_values <- rlnorm(2L * n_samples, log(mu), log(sigma))
    lod <- quantile(random_values[1:n_samples], p = p)
    X_before <- random_values[(n_samples + 1L):(2L * n_samples)]
    X_after[count, ] <- pmax(X_before, lod)
    delta <- X_before <= lod
    pct_cens[count] <- mean(delta)
    if (pct_cens > 0 && pct_cens < 1 ) count <- count + 1
  }

  Median <- apply(X_after, 1, median)
  Mean <- rowMeans(X_after)
  data.frame(Pct_cens_array=pct_cens, Mean=Mean, Median=Median) 
}
 
 比较时间，例如， 
 
mu=1
sigma=3
n_samples=10L
n_iterations = 1000L
p=0.10
system.time(dset(mu,sigma, n_samples, n_iterations, p))
system.time(dset2(mu,sigma, n_samples, n_iterations, p))
 
 在我的机器上，有3倍加速。 

I've made a few little tweaks to your code without changing the whole style of it. It would be good to heed Yoong Kim's advice and try to break up the code into smaller pieces, to make it more readable and maintainable.
 
 
 
Your function now gets two "n" arguments, for how many samples you have in each row, and how many iterations (columns) you want.
 
 
You were growing the arrays Median and Mean in the loop, which requires a lot of messing about reallocating memory and copying things, which slows everything down. I've predefined X_after and moved the mean and median calculations after the loop to avoid this. (As a bonus, mean and median only get called once instead of n_iteration times.)
 
 
The calls to ifelse weren't really needed.
 
 
It is a little quicker to call rlnorm once, generating enough values for x and the lod, than to call it twice.
 
 
Here's the updated function.
 
dset2 <- function (mu, sigma, n_samples, n_iterations, p) {    
  X_after <- matrix(NA_real_, nrow = n_iterations, ncol = n_samples)
  pct_cens <- numeric(n_iterations)
  count <- 1
  while(count <= n_iterations) {     
    random_values <- rlnorm(2L * n_samples, log(mu), log(sigma))
    lod <- quantile(random_values[1:n_samples], p = p)
    X_before <- random_values[(n_samples + 1L):(2L * n_samples)]
    X_after[count, ] <- pmax(X_before, lod)
    delta <- X_before <= lod
    pct_cens[count] <- mean(delta)
    if (pct_cens > 0 && pct_cens < 1 ) count <- count + 1
  }

  Median <- apply(X_after, 1, median)
  Mean <- rowMeans(X_after)
  data.frame(Pct_cens_array=pct_cens, Mean=Mean, Median=Median) 
}
 
Compare timings with, for example,
 
mu=1
sigma=3
n_samples=10L
n_iterations = 1000L
p=0.10
system.time(dset(mu,sigma, n_samples, n_iterations, p))
system.time(dset2(mu,sigma, n_samples, n_iterations, p))
 
On my machine, there is a factor of 3 speedup.