在 R 中创建一个函数，将字符串转换为整数，覆盖整个数据帧

Question

我需要在 R 中创建一个函数，根据先前确定的“翻译代码”将数据帧的所有条目（字符串）转换为整数。

输入数据示例：

Question 1          Question 2    Question 3

Strongly Agree      Agree         Disagree

Strongly Disagree   Neutral       Don't Know

我将使用的数据集将包含 1000 多行和 50 列。每个答案都需要转换为整数值。翻译公式为：

非常不同意 = 1，不同意 = 2，中立 = 3，同意 = 4，非常同意 = 5，不知道 = 0。

所以这个样本数据的函数输出将是

Question 1  Question 2  Question 3

5           4           2

1           3           0

我的功能尝试：

transform <- function(x)

{
  for (i in x[i, ]

  {
  if (i == 'Discordo fortemente')  {i == 1}
  if (i == 'Discordo')  {i == 2}
  if (i == 'Não concordo nem discordo') {i == 3}
  if (i == 'Concordo')  {i == 4}
  if (i == 'Concordo fortemente')  {i == 5}
  if (i == 'Não sei dizer')  {i == 0}
  }

}

上面的语言是葡萄牙语。显然代码不起作用，我已经把头撞在墙上将近 2 个小时了。欢迎任何对我的问题的解决方案，尽管我的想法是构建一个适用于一列的函数，然后将它与 lapply 一起使用。

Answer 1

我建议使用case_when 函数。例如

library(dplyr)
x %>& 
 mutate_all(~case_when(.x == 'Discordo fortemente' ~ 1,
                       .x == 'Discordo' ~ 2, 
                       .x == 'Não concordo nem discordo' ~ 3, 
                       .x == 'Concordo' ~ 4, 
                       .x == 'Concordo fortemente' ~ 5, 
                       .x == 'Não sei dizer' ~ 0))

这里，x 是您的数据。此代码修改所有列。 如果您有其他不想转换的列，可以使用 mutate_at 代替 mutate_all 函数。

如果你想让你的代码工作，你必须修改如下：

transform <- function(x) {

  y <- seq_along(x)

  for (i in 1:length(x)) {
    if (x[i] == 'Discordo fortemente')  {y[i] = 1}
    if (x[i] == 'Discordo')  {y[i] = 2}
    if (x[i] == 'Não concordo nem discordo') {y[i] = 3}
    if (x[i] == 'Concordo')  {y[i] = 4}
    if (x[i] == 'Concordo fortemente')  {y[i] = 5}
    if (x[i] == 'Não sei dizer')  {y[i]= 0}
}

  return(y)
}

transform(c("Discordo", 'Concordo fortemente', 'Não sei dizer'))
[1] 2 5 0

Answer 2

为什么不这样：

library(dplyr)
transform_fct <- function(var) {
  case_when(
    var == "Strongly disagree" ~  1,
    var == "Disagree" ~ 2,
    var == "Neutral" ~ 3,
    var == "Agree" ~ 4,
    var == "Strongly agree" ~ 5,
    var == "Don't know" ~ 0
  )
}
x <- x %>%
  mutate_all(transform_fct)

Answer 3

for (i in colnames(x)) {
  x[,i] <- sapply(x[,i], function(j) switch(j,
                   "Discordo fortemente" = 1,
                   "Discordo" = 2,
                   "Não concordo nem discordo" = 3,
                   "Concordo" = 4,
                   "Concordo fortemente" = 5,
                   0))
}

如果您不想学习 dplyr，这种方法使用 base R，但总体上会变得笨拙。

Answer 4

如果你有一致的情况，你可以这样做：

mapping <- c(`Strongly disagree` = 1, Disagree = 2, Neutral = 3, Agree = 4,
  `Strongly agree` = 5, `Don't know` = 0.)

df[] <- lapply(df, function(x) mapping[x])

或

df[] <- mapping[unlist(df)]

因为你没有，你可以做到：

mapping <- setNames(mapping,toupper(names(mapping)))
df[] <- lapply(df, function(x) mapping[toupper(x)])
df
#   Question.1 Question.2 Question.3
# 1          5          4          2
# 2          1          3          0

或

df[] <- mapping[toupper(unlist(df))] # (same output)

数据

df <- read.table(header=TRUE,stringsAsFactors=FALSE,text="
'Question 1'          'Question 2'    'Question 3'
'Strongly Agree'      Agree         Disagree
'Strongly Disagree'   Neutral       'Don\\'t Know'")