Question

我有一个像

这样的功能

import random
import multiprocessing


def list_append(count, id, out_queue):
    a = [random.random() for i in range(count)]
    out_queue.put((id, a))

if __name__ == "__main__":
    size = 10000   # Number of random numbers to add
    procs = 2   # Number of processes to create
    jobs = []
    q = multiprocessing.Queue()
    for i in range(0, procs):
        process = multiprocessing.Process(target=list_append,
                                          args=(size, i, q))
        process.start()
        jobs.append(process)
    result += [q.get() for _ in range(procs)]
    for j in jobs:
        j.join()
    print("List processing complete.")

如果我然后用

调用它

dbh2vol <- function(dbh,hgt,...){

  if (missing(hgt)){
    hgt = 2
    cat("hgt is missing. Set to 2")
  }
  vol  = dbh * hgt
  return(vol)
}

一切都很好。如果我有像

这样的东西

Vol = dbh2vol(dbh=10,hgt=4)
#40

正如所料。但如果那时我做了

Vol = dbh2vol(dbh=10,param=4)
#hgt is missing. Set to 2[1] 20

Vol = dbh2vol(dbh=10,h=4) #40被解释为h=4。为什么以及如何避免这种行为，即参数如何与其名称完全匹配？

Answer 1

你观察到的是命名函数参数的部分匹配，据我所知你不能禁用它。请参阅r语言定义以获取更多信息https://cran.r-project.org/doc/manuals/R-lang.html#Argument-matching

也许您可以将hgt参数放入三个点并检查一下？当然你放松了位置匹配。

dbh2vol <- function(dbh, ...){
  hgt <- list(...)$hgt
  if (is.null(hgt)) {
    hgt <-  sqrt(dbh)
    cat("hgt has been calculated with some_other_function")
  }
  vol = dbh * hgt
  return(vol)
}

Answer 2

您可以尝试这样做，而不是将dbh和hgt作为参数传递：

pars <- list(
    dbh = 10,
    h = 5
)

dbh2vol <- function(pars){
  with(pars,{
    if (!exists("hgt")){
        hgt <- 2
        cat("hgt is missing. Set to 2")
    }
    vol <- dbh * hgt
    return(vol)  
  })
}

dbh2vol(pars)
## hgt is missing. Set to 2[1] 20

这使您可以轻松地将参数传递给其他函数。

函数输入中的R缺少参数与名称不匹配

2 个答案: