动态将数据框列传递给“聚合”时如何保留列名 [英] How to preserve column names when dynamically passing data frame columns to `aggregate`

问题描述

具有如下数据框

df1 <- data.frame(a=seq(1.1,9.9,1.1), b=seq(0.1,0.9,0.1),
                  c=rev(seq(10.1, 99.9, 11.1)))

我想通过汇总列 b 和 c a

I want to aggregate cols b and c by a

所以我会做这样的事情

aggregate(cbind(b,c) ~ a, data = df1, mean)

这样就可以完成。但是我想不使用函数中的硬编码列名来概括。

This would get it done. However I want to generalize without hard coded column names like in a function.

myAggFunction <- function (df, col_main, col_1, col_2){
    return (aggregate(cbind(df[,col1], df[,col2]) ~ df[,col_main], df, mean))
    }
myAggFunction(df, 1, 2, 3)

我遇到的问题是返回的数据框的列名如下所示

The issue I have is that the col names of the returned data frame is as below

 df2[, 1]  V1   V2

如何在返回的数据框中获取原始数据框中的列名？

How do I get the column names in the original data frame in the returned data frame?

推荐答案

我将假设一个一般情况，您有多个LHS（左侧）和多个RHS（右侧）。

I will be assuming a general case, where you have multiple LHS (left hand sides) as well as multiple RHS (right hand sides).

使用数据框方法

## S3 method for class 'data.frame'
aggregate(x, by, FUN, ..., simplify = TRUE, drop = TRUE)

如果将对象作为主题列表，您将保留名称。因此，不要使用 [，] 来访问数据框，而要使用 [] 来访问数据框。您可以将函数构造为：

If you pass object as a named list, you get names preserved. So do not access your data frame with [, ], but with []. You may construct your function as:

## `LHS` and `RHS` are vectors of column names or numbers giving column positions
fun1 <- function (df, LHS, RHS){
  ## call `aggregate.data.frame`
  aggregate.data.frame(df[LHS], df[RHS], mean)
  }

仍然使用公式方法吗？

## S3 method for class 'formula'
aggregate(formula, data, FUN, ...,
          subset, na.action = na.omit)

这有点乏味，但是我们想要通过以下方式构造一个不错的公式：

It is slightly tedious, but we want to construct a nice formula via:

as.formula( paste(paste0("cbind(", toString(LHS), ")"),
                  paste(RHS, collapse = " + "), sep = " ~ ") )

例如：

LHS <- c("y1", "y2", "y3")
RHS <- c("x1", "x2")
as.formula( paste(paste0("cbind(", toString(LHS), ")"),
                  paste(RHS, collapse = " + "), sep = "~") )
# cbind(y1, y2, y3) ~ x1 + x2

如果将此公式提供给汇总，您将得到保留的体面的列名。

If you feed this formula to aggregate, you will get decent column names preserved.

因此，应这样构造函数：

So construct your function as such:

fun2 <- function (df, LHS, RHS){
  ## ideally, `LHS` and `RHS` should readily be vector of column names
  ## but specifying vector of numeric positions are allowed
  if (is.numeric(LHS)) LHS <- names(df)[LHS]
  if (is.numeric(RHS)) RHS <- names(df)[RHS]
  ## make a formula 
  form <- as.formula( paste(paste0("cbind(", toString(LHS), ")"),
                      paste(RHS, collapse = " + "), sep = "~") )
  ## call `aggregate.formula`
  stats:::aggregate.formula(form, df, mean)
  }

---

备注

aggregate.data .frame 是最好的。 aggregate.formula 是一个包装器，将在内部调用 model.frame 首先构建一个数据帧。

aggregate.data.frame is the best. aggregate.formula is a wrapper and will call model.frame inside to construct a data frame first.

我可以选择公式方法，因为构造公式的方式对于 lm 等有用。

I give "formula" method as an option, because the way I construct a formula is useful for lm, etc.

简单，可重现的示例

set.seed(0)
dat <- data.frame(y1 = rnorm(10), y2 = rnorm(10),
                  x1 = gl(2,5, labels = letters[1:2]))

## "data.frame" method with `fun1`
fun1(dat, 1:2, 3)
#  x1          y1         y2
#1  a  0.79071819 -0.3543499
#2  b -0.07287026 -0.3706127

## "formula" method with `fun2`
fun2(dat, 1:2, 3)
#  x1          y1         y2
#1  a  0.79071819 -0.3543499
#2  b -0.07287026 -0.3706127

fun2(dat, c("y1", "y2"), "x1")
#  x1          y1         y2
#1  a  0.79071819 -0.3543499
#2  b -0.07287026 -0.3706127

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