Question

我想循环遍历数据帧中的变量，在每个变量上调用lm（），所以我写了这个：

findvars <- function(x = samsungData, dv = 'activity', id = 'subject') {
  # Loops through the possible predictor vars, does an lm() predicting the dv
  # from each, and returns a data.frame of coefficients, one row per IV.
  r <- data.frame()
  # All varnames apart from the dependent var, and the case identifier
  ivs <- setdiff(names(x), c(dv, id))
  for (iv in ivs) {
    print(paste("trying", iv))
    m <- lm(dv ~ iv, data = x, na.rm = TRUE)
    # Take the absolute value of the coefficient, then transpose.
    c <- t(as.data.frame(sapply(m$coefficients, abs)))
    c$iv <- iv # which IV produced this row?
    r <- c(r, c)
  }
  return(r)
}

这不起作用，我相信b / c lm（）调用中的公式由函数局部变量组成，这些变量在传入的数据帧中包含命名变量的字符串（例如，＆＃34; my_dependant_var＆＃ 34;和＆＃34; this_iv＆＃34;）而不是指向实际变量对象的指针。

我尝试在eval（parse（text =））中包装该公式，但无法使其工作。

如果我对这个问题是正确的，有人可以向我解释如何让R解决这些变量的内容iv＆amp; dv进入我需要的指针？或者，如果我错了，有人可以解释还有什么事吗？

非常感谢！

以下是一些重复代码：

library(datasets)
data(USJudgeRatings)
findvars(x = USJudgeRatings, dv = 'CONT', id = 'DILG')

Answer 1

除了你的公式问题之外，你的功能中还有足够多的坏事发生，我认为有人应该引导你完成所有这些。以下是一些注释，然后是更好的版本：

  #For small examples, "growing" objects isn't a huge deal,
  # but you will regret it very, very quickly. It's a bad
  # habit. Learn to ditch it now. So don't inititalize
  # empty lists and data frames.
  r <- data.frame()

  ivs <- setdiff(names(x), c(dv, id))
  for (iv in ivs) {
    print(paste("trying", iv))
    #There is no na.rm argument to lm, only na.action
    m <- lm(dv ~ iv, data = x, na.rm = TRUE)
    #Best not to name variables c, its a common function, see two lines from now!
    # Also, use the coef() extractor functions, not $. That way, if/when
    # authors change the object structure your code won't break.
    #Finally, abs is vectorized, no need for sapply
    c <- t(as.data.frame(sapply(m$coefficients, abs)))
    #This is probably best stored in the name
    c$iv <- iv # which IV produced this row?
    #Growing objects == bad! Also, are you sure you know what happens when
    # you concatenate two data frames?
    r <- c(r, c)
  }
  return(r)
}

尝试这样的事情：

findvars <- function(x,dv,id){
  ivs <- setdiff(names(x),c(dv,id))
  #initialize result list of the appropriate length
  result <- setNames(vector("list",length(ivs)),ivs)
  for (i in seq_along(ivs)){
    result[[i]] <- abs(coef(lm(paste(dv,ivs[i],sep = "~"),data = x,na.action = na.omit)))
  }
  result
}

变量范围＆amp; R函数中的分辨率

1 个答案: