Question

我正在寻找更有效的代码版本，它将取代数据框中的因素。

这是我的数据集：

structure(list(Rio.Olympics.Sports.Participating.Team = structure(c(2L, 
2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 1L, 1L, 1L, 1L, 1L, 
1L, 1L, 1L, 1L, 1L, 1L, 1L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 
4L, 4L, 4L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 3L, 
3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L), .Label = c("American Gymnastics", 
"American Swimmers", "Boxing", "European Gymnastics", "Running", 
"Free-style swimming", "Breaststroke Swimming", "Diving", "Athletics", 
"Soccer"), class = "factor"), Calendar.Quarter = structure(c(16071, 
16161, 16252, 16344, 16436, 16526, 16617, 16709, 16801, 16892, 
16983, 17075, 16071, 16161, 16252, 16344, 16436, 16526, 16617, 
16709, 16801, 16892, 16983, 17075, 16071, 16161, 16252, 16344, 
16436, 16526, 16617, 16709, 16801, 16892, 16983, 17075, 16071, 
16161, 16252, 16344, 16436, 16526, 16617, 16709, 16801, 16892, 
16983, 17075, 16071, 16161, 16252, 16344, 16436, 16526, 16617, 
16709, 16801, 16892, 16983, 17075), class = "Date"), Randomized.Viewers = c(49, 
45, 51, 55, 47, 48, 54, 57, 53, 50, 52, 58, 32, 29, 33, 40, 34, 
36, 31, 39, 37, 30, 35, 41, 5, 1, 25, 46, 38, 4, 56, 27, 21, 
43, 42, 44, 2, 59, 3, 10, 60, 7, 14, 24, 13, 16, 17, 28, 15, 
6, 19, 23, 11, 12, 20, 22, 9, 8, 18, 26)), .Names = c("Rio.Olympics.Sports.Participating.Team", 
"Calendar.Quarter", "Randomized.Viewers"), row.names = c(NA, 
-60L), class = "data.frame")

现在，我想更改因子标签。这是我做的：

Old_labels <- c("American Swimmers", "American Gymnastics", 
               "European Gymnastics", "Running", "Boxing")
New_labels <- c("Jupitean Swimmers", "Saturnish Gymastics", 
               "Plutoish Gymnastics", "Walking", "Fighting")
Apply_lables <- data.frame(Old_labels, New_labels)
colnames(Apply_lables)[1] <- "Old_labels"

最后，这段代码可以解决问题：

p1 <- p
p1$Rio.Olympics.Sports.Participating.Team <- 
  Apply_lables[match(p$Rio.Olympics.Sports.Participating.Team, 
                     Apply_lables$Old_labels), "New_labels"]

以下是修改后的数据框：

 Rio.Olympics.Sports.Participating.Team Calendar.Quarter Randomized.Viewers
1                      Jupitean Swimmers       2014-01-01                 49
2                      Jupitean Swimmers       2014-04-01                 45
3                      Jupitean Swimmers       2014-07-01                 51
4                      Jupitean Swimmers       2014-10-01                 55
5                      Jupitean Swimmers       2015-01-01                 47
6                      Jupitean Swimmers       2015-04-01                 48

问题：作为R的初学者，我在几个小时内挣扎了很多。虽然我设法得到了我想要的东西，但有没有更好的方法（即更少的代码行和更快的实现）来改变基于查找表的因素？我的原始数据集大约有1M行，上面的代码需要花费很多时间才能运行。

我在SO上研究了这个主题，但我不认为这在任何地方都有过。虽然有一些帖子谈到使用match（）来使用查找表来更改行。

Answer 1

如果您想要从特定列表中替换Old_labels，我认为您可以将levels()置为所需因子的子集，并在那里推送新标签：

levels(dta$Rio.Olympics.Sports.Participating.Team)[
    dta$Rio.Olympics.Sports.Participating.Team %in% Old_labels] <-
    New_labels

使用查找表替换因子值

1 个答案: