在 sapply R 中使用函数

Question

我需要用#2 中的函数创建一个直方图，有人有想法吗？我收到警告信息： 在 hist(as.integer(project.timeV)) 中：强制引入的 NAs

#1.
roundNorm <- function(m, sd) { #m is Mean, sd is Standard Deviation
  x <- rnorm(1,m,sd)
  roundUp.X <- ceiling(x)
  return(roundUp.X)
}

#2.
project <- function(x) { # route 1: A-B-C-F, route 2: A-D-E-F
  a.time <- roundNorm(10,3)
  b.time <- roundNorm(10,3)
  c.time <- 5
  d.time <- 10
  e.time <- roundNorm(5,2)
  f.time <- 4
  if(b.time+c.time > d.time+e.time){
    projectLength <- a.time+b.time+c.time+f.time
    criRoute <- "A,B,C,F"
    answer <- c(projectLength,criRoute)
  }
  else {
    projectLength <- a.time+d.time+e.time+f.time
    criRoute <- "A,D,E,F"
    answer <- c(projectLength,criRoute)
  }
  return(answer)
}

#3.
samp <- 1:10000
project.timeV <- sapply(samp,project)
hist(as.integer(project.timeV))

Answer 1

如果我这样做，我不会收到任何错误。

project.timeV <- sapply(samp, project, simplify = FALSE)
project.timeV <- do.call(rbind, project.timeV)
hist(as.integer(project.timeV[, 1]))

这是结果的预览。请注意，我不希望 sapply 简化我的结果。这是我对do.call 所做的事情，因为我想按行而不是按列进行展平。

> head(project.timeV)
     [,1] [,2]     
[1,] "34" "A,D,E,F"
[2,] "34" "A,B,C,F"
[3,] "26" "A,D,E,F"
[4,] "24" "A,B,C,F"
[5,] "29" "A,D,E,F"
[6,] "30" "A,D,E,F"