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我想监控算法的进度,以了解算法将工作多长时间我想看到类似的东西

for(i in 1:100) print( paste("iter number",i))
[1] "iter number 1"
[1] "iter number 2"
[1] "iter number 3"
[1] "iter number 4"
[1] "iter number 5"
[1] "iter number 6"
[1] "iter number 7"

我试图做这样的事情,但我的计数器没有更新。

i <- 0
fit <- function(x) {
  
  ##/// my count
  i <- i+1
  print( paste("iter number",i))
  ##////
  
  
  y <- numeric(2)
  y[1] <- crossprod(x, x)
  y[2] <- crossprod(x - 5, x - 5)

  return (y)
}

当我运行优化算法时,我得到以下结果

library(mco)
ns <- nsga2(fit, 2, 2,
            generations=150, popsize=100,
            lower.bounds=rep(-5, 2),
            upper.bounds=rep(10, 2))

[1] "iter number 1"
[1] "iter number 1"
[1] "iter number 1"
[1] "iter number 1"
[1] "iter number 1"
[1] "iter number 1"
[1] "iter number 1"
[1] "iter number 1"
[1] "iter number 1"
[1] "iter number 1"
[1] "iter number 1"
[1] "iter number 1"

如何正确地做到这一点?

4

1 回答 1

1

正如评论中所解释的,这里有一个小演示。

这是因为 i 只会在您的功能内更新而不是您的功能之外。这就是为什么您的函数不打印 0 而是一直打印 1 的原因。因为环境。您可以做的是更改i <- i+1i <<- i+1并且 i 将在函数之外进行更新。

简单的演示

i <- 0
z <- 0

fit <- function() {
  i <- i + 1
  print(paste("i iter number:", i))
  z <<- z + 1 # note the <<- instead of <-
  print(paste("z iter number:", z))
}

for(x in 1:3) fit()

[1] "i iter number: 1"
[1] "z iter number: 1"
[1] "i iter number: 1"
[1] "z iter number: 2"
[1] "i iter number: 1"
[1] "z iter number: 3"

i # is still 0
[1] 0
z # is now 3
[1] 3
于 2022-01-13T14:09:08.987 回答