Question

我是R的新手，我编写了一个代码，我相信可以使用for循环来缩短它。问题是我不知道如何编写循环。

我有一个带有“ TestGrade”列的数据框，其值类似于“ Grade 1”或“ Kindergarten”。我试图将该列更改为仅一个数字值。例如，将“幼儿园”更改为0，将“等级1”更改为1。我将在示例数据帧的下方提供代码，以及如何无循环解决问题。

任何指导将不胜感激！

##Sample Data
FirstInitial <- c("A", "D", "M", "C", "J", "S", "K", "L", "M", "K", "G", "B", "F")
LastInitial <- c("S", "M", "T", "M", "A", "B", "H", "M", "S", "W", "L", "Z", "P")
TestGrade <- c('Kindergarten', 'Grade 1','Grade 2', 'Grade 3','Grade 4', 'Grade 5', 'Grade 6','Grade 7','Grade 8', 'Grade 9', 'Grade 10', 'Grade 11','Grade 12')

df <- data.frame(FirstInitial, LastInitial, TestGrade)

##The codes current function
if(any(df$TestGrade == 'Kindergarten')){
  df$TestGrade <- gsub('Kindergarten', '0', df$TestGrade)
}
if(any(df$TestGrade == 'Grade 1')){
  df$TestGrade <- gsub('Grade 1', '1', df$TestGrade)
}
if(any(df$TestGrade == 'Grade 2')){
  df$TestGrade <- gsub('Grade 2', '2', df$TestGrade)
}
if(any(df$TestGrade == 'Grade 3')){
  df$TestGrade <- gsub('Grade 3', '3', df$TestGrade)
}
if(any(df$TestGrade == 'Grade 4')){
  df$TestGrade <- gsub('Grade 4', '4', df$TestGrade)
}
if(any(df$TestGrade == 'Grade 5')){
  df$TestGrade <- gsub('Grade 5', '5', df$TestGrade)
}

if(any(df$TestGrade == 'Grade 6')){
  df$TestGrade <- gsub('Grade 6', '6', df$TestGrade)
}
if(any(df$TestGrade == 'Grade 7')){
  df$TestGrade <- gsub('Grade 7', '7', df$TestGrade)
}
if(any(df$TestGrade == 'Grade 8')){
  df$TestGrade <- gsub('Grade 8', '8', df$TestGrade)
}
if(any(df$TestGrade == 'Grade 9')){
  df$TestGrade <- gsub('Grade 9', '9', df$TestGrade)
}
if(any(df$TestGrade == 'Grade 10')){
  df$TestGrade <- gsub('Grade 10', '10', df$TestGrade)
}
if(any(df$TestGrade == 'Grade 11')){
  df$TestGrade <- gsub('Grade 11', '11', df$TestGrade)
}
if(any(df$TestGrade == 'Grade 12')){
  df$TestGrade <- gsub('Grade 12', '12', df$TestGrade)
}

Answer 1

我们可以使用ifelse，为“幼儿园”分配0，并从其他人中删除“成绩”

as.numeric(ifelse(df$TestGrade == "Kindergarten", 0, 
          sub("Grade ", "", df$TestGrade)))

#[1]  0  1  2  3  4  5  6  7  8  9 10 11 12

Answer 2

我们可以使用case_when

library(dplyr)
library(readr)
df %>%
  mutate(TestGrade = case_when(as.character(TestGrade) == "Kindergarten"~ 0,
                               TRUE ~ parse_number(TestGrade)))

#   FirstInitial LastInitial TestGrade
#1             A           S         0
#2             D           M         1
#3             M           T         2
#4             C           M         3
#5             J           A         4
#6             S           B         5
#7             K           H         6
#8             L           M         7
#9             M           S         8
#10            K           W         9
#11            G           L        10
#12            B           Z        11
#13            F           P        12

Answer 3

首先简化：您不需要任何if(any(...))。 gsub很聪明，就像是查找/替换。命令gsub('Grade 9', '9', df$TestGrade)将'Grade 9'替换为'9'，并且不会触及其他任何内容。因此，删除所有if语句，我们得到：

df$TestGrade <- gsub('Kindergarten', '0', df$TestGrade)
df$TestGrade <- gsub('Grade 1', '1', df$TestGrade)
df$TestGrade <- gsub('Grade 2', '2', df$TestGrade)
df$TestGrade <- gsub('Grade 3', '3', df$TestGrade)
df$TestGrade <- gsub('Grade 4', '4', df$TestGrade)
df$TestGrade <- gsub('Grade 5', '5', df$TestGrade)
df$TestGrade <- gsub('Grade 6', '6', df$TestGrade)
df$TestGrade <- gsub('Grade 7', '7', df$TestGrade)
df$TestGrade <- gsub('Grade 8', '8', df$TestGrade)
df$TestGrade <- gsub('Grade 9', '9', df$TestGrade)
df$TestGrade <- gsub('Grade 10', '10', df$TestGrade)
df$TestGrade <- gsub('Grade 11', '11', df$TestGrade)
df$TestGrade <- gsub('Grade 12', '12', df$TestGrade)

下一个改进，我们可以做一个循环。这与上面的代码完全等效，只需要较少的键入即可。

pattern = c("Kindergarten", paste("Grade", 1:12))
replacement = as.character(0:12)

for (i in seq_along(pattern)) {
  df$TestGrade <- gsub(pattern[i], replacement[i], df$TestGrade)
}

更好的是，我们可以变得更聪明，使幼儿园成为特殊情况，并从其他所有内容中删除"Grade "，如Juian和Ronak的答案。另一个变化是：

df$TestGrade = as.character(df$TestGrade) # needed only if it is a factor
df$TestGrade[df$TestGrade == "Kindergarten"] = 0
df$TestGrade = sub("Grade ", "", df$TestGrade)
df$TestGrade = as.numeric(df$TestGrade) # if needed

如果我们真的想花哨的话，可以在fixed = TRUE内设置sub()。这表明sub仅需要完全匹配，而并非试图使用正则表达式。这将使代码运行得更快，但是除非您有很多数据，否则您将不会发现任何区别。如果您有100,000+行，则此方法将非常快：

# optimized
df$TestGrade = as.character(df$TestGrade) # needed only if it is a factor
df$TestGrade[df$TestGrade == "Kindergarten"] = 0
df$TestGrade = as.integer(sub("Grade ", "", df$TestGrade, fixed = TRUE))

Answer 4

无需使用两行代码进行for循环即可完成此操作。我还建议您在运行这些行之前在stringsAsFactors = F命令中添加data.frame

df$TestGrade[df$TestGrade == "Kindergarten"] = 0
df$TestGrade <- gsub("Grade ", "", df$TestGrade)

> df
   FirstInitial LastInitial TestGrade
1             A           S         0
2             D           M         1
3             M           T         2
4             C           M         3
5             J           A         4
6             S           B         5
7             K           H         6
8             L           M         7
9             M           S         8
10            K           W         9
11            G           L        10
12            B           Z        11
13            F           P        12

Answer 5

您可以编写一个关键点并将等级设置为一个因子。即使成绩格式发生变化，这也将起作用。

key <- c('Kindergarten',
         'Grade 1',
         'Grade 2',
         'Grade 3',
         'Grade 4',
         'Grade 5',
         'Grade 6',
         'Grade 7',
         'Grade 8',
         'Grade 9',
         'Grade 10',
         'Grade 11',
         'Grade 12')
dat <- c('Grade 3', 'Grade 5', 'Grade 2')
dat <- factor(dat, levels = key)
dat <- as.numeric(dat) - 1
dat

我们在末尾减去1，因为因子从1开始并且您希望幼儿园设置为0。

Answer 6

这在这里解决了您的问题：

df$TestGrade <- sapply(df$TestGrade,function(el)
  {
  if(el == "Kindergarten") return(0)
  else return(as.numeric(sub("Grade ","",el)))
}

使用for循环更改数据框中的列

6 个答案: