来自列表的函数输入

Question

如何运行一个函数（在 R 中），其中一些输入是从列表（或数据框）中提取的？我认为这比运行一个 for 循环更有效吗？

我正在运行宁模拟并想要更改变量值，但由于它们需要很长时间才能运行我希望它们运行一夜之间并通过滴答自动不同的值。

函数代码如下：

n = 10000         
mu = 0                          
sd = 1                          
n.sub = 100                       
iboot = 100
isim = 1000     ### REDUCED FOR THIS EXAMPLE ###
var.values <- NULL         
var.values.pop <- NULL  

hist.fn <- function(n,mu,sd,n.sub,iboot)
{ 
Pop <- rnorm(n,mu,sd)
var.pop <- var(Pop)                       
Samp <- sample(Pop, n.sub, replace = FALSE)  
var.samp <- var(Samp)                                         
for(i in 1:isim) {            
  for(j in 1:iboot) {         
    Boot <- sample(Samp, n.sub, replace = TRUE)           
    var.values[j] <- var(Boot)                                  
    }                                                               
  Samp <- sample(Pop, n.sub, replace = FALSE)         
  var.values.pop[i] <- var(Samp)                                   
  }                                                               
hist.pop <- hist(var.values.pop,plot=F)
hist.boot <- hist(var.values,plot=F)
#mypath = file.path("C:", "Output", paste("hist.boot_n.", n.sub, "_var.", sd^2, "_isim.", isim, "_iboot.", iboot, ".wmf", sep=""))
#win.metafile(file=mypath)
plot.new()   #### ADDED FOR THIS EXAMPLE INSTEAD OF OUTPUTTING TO FILE ####
plot(hist.pop, freq=FALSE, xlim=range(var.values.pop, var.values), ylim=range(hist.pop$density, hist.boot$density), main = paste("Histogram of variances \n n=",n.sub," mu=",mu,"var=",sd^2,"\n n.sim=",isim,"n.boot=",iboot,"\n"), cex.main=0.8, xlab="Variance", col="red") 
plot(hist.boot, freq=FALSE, col="blue", border="blue", add=T, density=20, angle=45) 
abline(v=var.pop, lty=2, col="black", lwd=2)
 legend("topright", legend=c("sample","bootstrap"),col=c("red","blue"),lty=1,lwd=2,bty="n",cex=0.7)
#dev.off()
}

hist.fn(n,mu,sd,n.sub,iboot)

然后我希望 sd、n.sub 和 iboot 通过运行ning 更改为以下值：

sd <- c(1,10,100,1000)
n.sub <- c(4,10,100,1000)
iboot <- c(100,1000,10000)

Answer 1

您应该使用 do.call，它将使用列表中的参数应用函数。我已将您的示例简化为运行个示例循环。您可以修改脚本的 print 行以监控更大作业的进度：

# The function
hist.fn <- function(n,mu,isim,sd,n.sub,iboot)
{ 
  Pop <- rnorm(n,mu,sd)
  var.pop <- var(Pop)                       
  Samp <- sample(Pop, n.sub, replace = FALSE)  
  var.samp <- var(Samp)
  var.values <- NaN*seq(isim) # sets up an empty vector for results
  var.values.pop <- NaN*seq(isim) # sets up an empty vector for results
  for(i in seq(isim)) {            
    for(j in seq(iboot)) {         
      Boot <- sample(Samp, n.sub, replace = TRUE)           
      var.values[j] <- var(Boot) 
      print(paste("i =", i, "; j =", j))
    }                                                               
    Samp <- sample(Pop, n.sub, replace = FALSE)         
    var.values.pop[i] <- var(Samp)                                   
  }                                                               
  list(var.values=var.values, var.values.pop=var.values.pop) #returns results in the form of a list
}

# Global variables
n = 100         
mu = 0                          
isim = 10

# Changing variables
sd <- c(1,10,20,30)
n.sub <- c(4,10,20,30)
iboot <- c(100,200,300,400)

df <- data.frame(sd=sd, n.sub=n.sub, iboot=iboot)
res <- vector(mode="list", nrow(df)) # sets up an empty list for results
for(i in seq(nrow(df))){
  res[[i]] <- do.call(hist.fn, c(n=n, mu=mu, isim=isim, df[i,]) ) 
}
res # show results

Answer 2

sd <- 1:3
n.sub <- 4:6
iboot <- 7:9

funct1<-function(x,y,z) print(x+y+z)

for (i in 1:length(sd)){
  funct1(sd[i],n.sub[i],iboot[i])
}

只是一个例子。用循环做。

Answer 3

也许是这样的？

    n = 10000         
    mu = 0                          
    sd = 1                          
    n.sub = 100                       
    iboot = 100
    isim = 1000 
    sd <- c(1,10,100,1000)
    n.sub <- c(4,10,100,1000)
    iboot <- c(100,1000,10000)
    # hist.fn parameters: n,mu,sd,n.sub,iboot
    params <- expand.grid(n = n, mu = mu, sd = sd, 
              n.sub = n.sub, iboot = iboot)
    apply(params, 1, FUN = function(x) do.call(hist.fn, as.list(x) ) )

你可能想把这些：

    var.values <- NULL         
var.values.pop <- NULL

在 hist.fn 内部，因为在函数外部为变量赋值并不像您想象的那样有效。

来自列表的函数输入

Function inputs from a list

simulation

r

function

list

dataframe