用subset()代替data.frames的列表 [英] A replacement for `subset()` for a list of data.frames

问题描述

函数foo1可以使用一个或多个请求的变量(例如，by = ESL == 1或by == ESL == 1 & type == 4)对数据帧列表进行子集化(使用subset()).

Function foo1 can subset (using subset()) a list of data.frames by one or more requested variables (e.g., by = ESL == 1 or by == ESL == 1 & type == 4).

但是，我知道 在R中使用subset() 的危险因此，我想知道在下面的foo1中，我可以用什么代替subset()来获得相同的输出?

However, I'm aware of the danger of using subset() in R. Thus, I wonder in foo1 below, what I can use instead of subset() to get the same output?

foo1 <- function(data, by){

  s <- substitute(by)
  L <- split(data, data$study.name) ; L[[1]] <- NULL

  lapply(L, function(x) do.call("subset", list(x, s))) ## What to use instead of `subset`
                                                       ## to get the same output?
}

# EXAMPLE OF USE:
D <- read.csv("https://raw.githubusercontent.com/izeh/i/master/k.csv", header=TRUE) # DATA
foo1(D, ESL == 1) 

推荐答案

您可以使用该语言进行计算.基于我对在$登录R后使用替代项"的答案:

You can compute on the language. Building on my answer to "Working with substitute after $ sign in R":

foo1 <- function(data, by){

  s <- substitute(by)
  L <- split(data, data$study.name) ; L[[1]] <- NULL

  E <- quote(x$a)
  E[[3]] <- s[[2]]
  s[[2]] <- E

  eval(bquote(lapply(L, function(x) x[.(s),])))
}

foo1(D, ESL == 1) 

对于任意子集表达式，这变得更加复杂.您需要一个递归函数，该函数爬网解析树并将调用插入到$的正确位置.

This gets more complex for arbitrary subset expressions. You'd need a recursive function that crawls the parse tree and inserts the calls to $ at the right places.

就我个人而言，我只使用package data.table，因为它不需要$，因此可以更轻松地使用它，即，您只需更改s就可以执行eval(bquote(lapply(L, function(x) setDT(x)[.(s),]))).太太，我根本不会这样做.子集化之前，确实没有任何理由进行拆分.

Personally, I'd just use package data.table where this is easier because you don't need $, i.e., you can just do eval(bquote(lapply(L, function(x) setDT(x)[.(s),]))) without changing s. OTOH, I wouldn't do this at all. There is really no reason to split before subsetting.

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