Question

df <- structure(list(Cars = structure(3:1, .Label = c("Ford ", "Mazda", "Merc"), class = "factor"), Model = structure(c(2L, 1L, 3L), .Label = c("c", "e-class", "gt"), class = "factor"), Color = structure(1:3, .Label = c("Black ", "Blue", "Red"), class = "factor")), class = "data.frame", row.names = c(NA, -3L))
argument <- structure(list(NewVar = structure(3:2, .Label = c("", "SKU", "Vehicle"), class = "factor"), Input = structure(2:3, .Label = c("", "Cars", "Model"), class = "factor")), row.names = 1:2, class = "data.frame")
after <- structure(list(Cars = structure(3:1, .Label = c("Ford ", "Mazda", "Merc"), class = "factor"), Model = structure(c(2L, 1L, 3L), .Label = c("c", "e-class", "gt"), class = "factor"), Color = structure(1:3, .Label = c("Black ", "Blue", "Red"), class = "factor"), Vehicle = structure(3:1, .Label = c("Ford ", "Mazda", "Merc"), class = "factor"), SKU = structure(c(2L, 1L, 3L), .Label = c("c", "e-class", "gt"), class = "factor")), class = "data.frame", row.names = c(NA, -3L))
df
#>    Cars   Model  Color
#> 1  Merc e-class Black 
#> 2 Mazda       c   Blue
#> 3 Ford       gt    Red
argument
#>    NewVar Input
#> 1 Vehicle  Cars
#> 2     SKU Model
after
#>    Cars   Model  Color Vehicle     SKU
#> 1  Merc e-class Black     Merc e-class
#> 2 Mazda       c   Blue   Mazda       c
#> 3 Ford       gt    Red   Ford       gt

我正在尝试for遍历各列，并根据“参数” data.frame创建一个副本。这是我尝试过的方法，但实际上没有用。我有df，自变量和dputs之后。本质上，该参数具有一个标题为“ new var”和“ input”的列。在论点上，我在“ new var”中有“ Vehicle”，在“ Cars”中有“ input”。这意味着在原始df中，复制“汽车”并将其命名为“车辆”。
List是动态的，因此我尝试根据参数的长度将其保留在for循环中，即，如果我也希望将颜色复制为绘画，请进行更改。

for (i in 1:length(argument$Input) ){
  df %>%   
  mutate( paste0(argument[i,1]) =  !!as.name(argument[i,2])
     }

请注意，argument[i,1]指的是input，因此“变量”是新的列名。在dplyr中，这会影响df，因此我认为!!as.name将拉出已经存在的变量，即“汽车”已经在df之前。

Answer 1

我们可以通过'argument'的'Input'列选择感兴趣的列，然后使用rename_at更改列名。 “参数”数据集列为factor类，在进行更改时将其转换为character。

library(dplyr)
df %>%
    mutate_at(vars(as.character(argument$Input)), list(new= ~ .)) %>% 
    rename_at(vars(matches('new')), ~ as.character(argument[[1]]))
#     Cars   Model  Color Vehicle     SKU
#1  Merc e-class Black     Merc e-class
#2 Mazda       c   Blue   Mazda       c
#3 Ford       gt    Red   Ford       gt

如果我们将OP的for循环与:=配合使用，请确保对!!的{{1}}上的名称字符串求值（lhs）< / p>

:=

Answer 2

这是基本的R选项，当您已有字符列名称时，有时可以更轻松地使用它。注意字符的强制性，因为您提供的数据是作为因素给出的。

library(tidyverse)
df <- structure(list(Cars = structure(3:1, .Label = c("Ford ", "Mazda", "Merc"), class = "factor"), Model = structure(c(2L, 1L, 3L), .Label = c("c", "e-class", "gt"), class = "factor"), Color = structure(1:3, .Label = c("Black ", "Blue", "Red"), class = "factor")), class = "data.frame", row.names = c(NA, -3L))
argument <- structure(list(NewVar = structure(3:2, .Label = c("", "SKU", "Vehicle"), class = "factor"), Input = structure(2:3, .Label = c("", "Cars", "Model"), class = "factor")), row.names = 1:2, class = "data.frame")
after <- structure(list(Cars = structure(3:1, .Label = c("Ford ", "Mazda", "Merc"), class = "factor"), Model = structure(c(2L, 1L, 3L), .Label = c("c", "e-class", "gt"), class = "factor"), Color = structure(1:3, .Label = c("Black ", "Blue", "Red"), class = "factor"), Vehicle = structure(3:1, .Label = c("Ford ", "Mazda", "Merc"), class = "factor"), SKU = structure(c(2L, 1L, 3L), .Label = c("c", "e-class", "gt"), class = "factor")), class = "data.frame", row.names = c(NA, -3L))

add_column <- function(df, new_col, existing_col) {
  new_col <- as.character(new_col)
  existing_col <- as.character(existing_col)
  df[[new_col]] <- df[[existing_col]]
  return(df)
}

for (i in 1:nrow(argument)) {
  df <- add_column(df, argument$NewVar[i], argument$Input[i])
}

df
#>    Cars   Model  Color Vehicle     SKU
#> 1  Merc e-class Black     Merc e-class
#> 2 Mazda       c   Blue   Mazda       c
#> 3 Ford       gt    Red   Ford       gt

^{由reprex package（v0.3.0）于2019-07-10创建}

在一个df中使用名称创建另一个数据框的相同列副本

2 个答案: