lapply之后从列表返回到data.frame [英] returning from list to data.frame after lapply

问题描述

我有一个关于lapply的非常简单的问题.我正在从STATA过渡到R，我认为有一些非常基本的概念，我不会在R中循环.但是我整个下午都在阅读有关它的内容，无法找到一种合理的方法来完成这一非常简单的事情

I have a very simply question about lapply. I am transitioning from STATA to R and I think there is some very basic concept that I am not getting about looping in R. But I have been reading about it all afternoon and can't figure out a reasonable way to do this very simple thing.

我有三个数据帧df1，df2和df3，它们都具有相同的列名，顺序相同等.

I have three data frames df1, df2, and df3 that all have the same column names, in the same order, etc.

我想一次重命名它们的列.

I want to rename their columns all at once.

我将数据帧放在列表中:

I put the data frames in a list:

dflist <- list(df1, df2, df3)

我希望新名称是什么

varlist <- c("newname1", "newname2", "newname3")

编写一个用varlist中的名称替换名称的函数，然后将其套用在数据帧上

Write a function that replaces names with those in varlist, and lapply it over the data frames

ChangeNames <- function(x) {
  names(x) <- varlist 
  return(x)
}

dflist <- lapply(dflist, ChangeNames)

据我所知，R更改了我放入列表中的数据帧副本的名称，但未更改原始数据帧本身.我希望重命名数据框本身，而不是重命名列表中的元素(被困在列表中).

So, as far as I understand, R has changed the names of the copies of the data frames that I put in the list, but not the original data frames themselves. I want the data frames themselves to be renamed, not the elements of the list (which are trapped in a list).

现在，我可以走了

df1 <- as.data.frame(dflist[1])
df2 <- as.data.frame(dflist[2])
df2 <- as.data.frame(dflist[3])

但这似乎很奇怪.您需要一个循环来取回循环的元素吗?

But that seems weird. You need a loop to get back the elements of a loop?

基本上:将一些数据帧放入列表中并通过lapply在其上运行函数后，如何将它们从列表中移出而又不从第一个平方开始呢?

Basically: once you've put some data frames in a list and run your function on them via lapply, how do you get them back out of the list, without starting back at square one?

推荐答案

如果只想更改名称，在R中就不太难了.请记住，赋值运算符<-，可以依次应用.因此:

If you just want to change the names, that isn't too hard in R. Bear in mind that the assignment operator, <-, can be applied in sequence. Hence:

names(df1) <- names(df2) <- names(df3) <- c("newname1", "newname2", "newname3")

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