What is the most effective (ie efficient / appropriate) way to clean up a factor containing multiple levels that need to be collapsed? That is, how to combine two or more factor levels into one.

清理包含需要折叠的多个级别的因素的最有效(即有效/适当)方法是什么？也就是说，如何将两个或两个以上的因素水平组合成一个水平。

Here's an example where the two levels "Yes" and "Y" should be collapsed to "Yes", and "No" and "N" collapsed to "No":

下面是一个示例，其中两个级别“Yes”和“Y”应折叠为“Yes”，而“No”和“N”应折叠为“No”：

## Given: 
x <- c("Y", "Y", "Yes", "N", "No", "H")   # The 'H' should be treated as NA

## expectedOutput
[1] Yes  Yes  Yes  No   No   <NA>
Levels: Yes No  # <~~ NOTICE ONLY **TWO** LEVELS

---

One option is of course to clean the strings before hand using sub and friends.

当然，一种选择是事先使用SUB和FRIENDS清理字符串。

Another method, is to allow duplicate label, then drop them

另一种方法是允许重复标注，然后删除它们

## Duplicate levels ==> "Warning: deprecated"
x.f <- factor(x, levels=c("Y", "Yes", "No", "N"), labels=c("Yes", "Yes", "No", "No"))

## the above line can be wrapped in either of the next two lines
factor(x.f)      
droplevels(x.f) 

However, is there a more effective way?

然而，有没有更有效的方法呢？

---

While I know that the levels and labels arguments should be vectors, I experimented with lists and named lists and named vectors to see what happens
Needless to say, none of the following got me any closer to my goal.

虽然我知道级别和标签参数应该是向量，但我尝试了列表、命名列表和命名向量，看看会发生什么，不用说，以下这些都不会让我更接近我的目标。

  factor(x, levels=list(c("Yes", "Y"), c("No", "N")), labels=c("Yes", "No"))
  factor(x, levels=c("Yes", "No"), labels=list(c("Yes", "Y"), c("No", "N")))

  factor(x, levels=c("Y", "Yes", "No", "N"), labels=c(Y="Yes", Yes="Yes", No="No", N="No"))
  factor(x, levels=c("Y", "Yes", "No", "N"), labels=c(Yes="Y", Yes="Yes", No="No", No="N"))
  factor(x, levels=c("Yes", "No"), labels=c(Y="Yes", Yes="Yes", No="No", N="No"))

优秀答案推荐

UPDATE 2: See Uwe's answer which shows the new "tidyverse" way of doing this, which is quickly becoming the standard.

更新2：请看Uwe的答案，它展示了一种新的“整洁”方式，这种方式正在迅速成为标准。

UPDATE 1: Duplicated labels (but not levels!) are now indeed allowed (per my comment above); see Tim's answer.

更新1：标签重复(但不是级别！)现在确实被允许了(根据我上面的评论)；请参见Tim的回答。

ORIGINAL ANSWER, BUT STILL USEFUL AND OF INTEREST:
There is a little known option to pass a named list to the levels function, for exactly this purpose. The names of the list should be the desired names of the levels and the elements should be the current names that should be renamed. Some (including the OP, see Ricardo's comment to Tim's answer) prefer this for ease of reading.

最初的答案，但仍然有用和有趣：有一个鲜为人知的选项，将命名列表传递给级别函数，就是为了这个目的。列表的名称应该是所需的级别名称，元素应该是应该重命名的当前名称。有些人(包括OP，参见李嘉图对Tim回答的评论)为了便于阅读而倾向于这样做。

x <- c("Y", "Y", "Yes", "N", "No", "H", NA)
x <- factor(x)
levels(x) <- list("Yes"=c("Y", "Yes"), "No"=c("N", "No"))
x
## [1] Yes  Yes  Yes  No   No   <NA>  <NA>
## Levels: Yes No

As mentioned in the levels documentation; also see the examples there.

如级别文档中所述；另请参阅那里的示例。

value: For the 'factor' method, a
vector of character strings with length at least the number
of levels of 'x', or a named list specifying how to rename
the levels.

This can also be done in one line, as Marek does here: https://stackoverflow.com/a/10432263/210673; the levels<- sorcery is explained here https://stackoverflow.com/a/10491881/210673.

这也可以用一行来完成，就像Marek在这里所做的：https://stackoverflow.com/a/10432263/210673；the Level<-巫术在这里被解释为https://stackoverflow.com/a/10491881/210673.

> `levels<-`(factor(x), list(Yes=c("Y", "Yes"), No=c("N", "No")))
[1] Yes  Yes  Yes  No   No   <NA>
Levels: Yes No

As the question is titled Cleaning up factor levels (collapsing multiple levels/labels), the forcats package should be mentioned here as well, for the sake of completeness. forcats appeared on CRAN in August 2016.

由于问题的标题是清理因素级别(折叠多个级别/标签)，为了完整起见，这里也应该提到forcats包。2016年8月，CRAN上出现了作用力。

There are several convenience functions available for cleaning up factor levels:

有几个方便的功能可用于清理因子水平：

x <- c("Y", "Y", "Yes", "N", "No", "H") 

library(forcats)

Collapse factor levels into manually defined groups

fct_collapse(x, Yes = c("Y", "Yes"), No = c("N", "No"), NULL = "H")
#[1] Yes  Yes  Yes  No   No   <NA>
#Levels: No Yes

Change factor levels by hand

fct_recode(x, Yes = "Y", Yes = "Yes", No = "N", No = "No", NULL = "H")
#[1] Yes  Yes  Yes  No   No   <NA>
#Levels: No Yes

Automatically relabel factor levels, collapse as necessary

fun <- function(z) {
  z[z == "Y"] <- "Yes"
  z[z == "N"] <- "No"
  z[!(z %in% c("Yes", "No"))] <- NA
  z
}
fct_relabel(factor(x), fun)
#[1] Yes  Yes  Yes  No   No   <NA>
#Levels: No Yes

Note that fct_relabel() works with factor levels, so it expects a factor as first argument. The two other functions, fct_collapse() and fct_recode(), accept also a character vector which is an undocumented feature.

请注意，fct_relabel()处理因子级别，因此它需要一个因子作为第一个参数。另外两个函数FCT_CLUSLE()和FCT_RECODE()也接受字符向量，这是一个未记录的功能。

Reorder factor levels by first appearance

The expected output given by the OP is

OP给出的预期输出为

[1] Yes  Yes  Yes  No   No   <NA>
Levels: Yes No

Here the levels are ordered as they appear in x which is different from the default (?factor: The levels of a factor are by default sorted).

在这里，级别是按照它们在x中出现的顺序排列的，这不同于默认的(？系数：系数的级别在默认情况下是排序的)。

To be in line with the expected output, this can be achieved by using fct_inorder() before collapsing the levels:

为了与预期的输出一致，可以在折叠级别之前使用fct_inorder()来实现：

fct_collapse(fct_inorder(x), Yes = c("Y", "Yes"), No = c("N", "No"), NULL = "H")
fct_recode(fct_inorder(x), Yes = "Y", Yes = "Yes", No = "N", No = "No", NULL = "H")

Both return the expected output with levels in the same order, now.

现在，两者都以相同的顺序返回具有相同级别的预期输出。

Since R 3.5.0 (2018-04-23) you can do this in one clear and simple line:

从R 3.5.0(2018-04-23)开始，您可以用一句简单明了的话来实现这一点：

x = c("Y", "Y", "Yes", "N", "No", "H") # The 'H' should be treated as NA

tmp = factor(x, levels= c("Y", "Yes", "N", "No"), labels= c("Yes", "Yes", "No", "No"))
tmp
# [1] Yes  Yes  Yes  No   No   <NA>
# Levels: Yes No

1 line, maps multiple values to the same level, sets NA for missing levels" – h/t @Aaron

1行，将多个值映射到同一级别，为缺少的级别设置NA“-h/t@aron

Perhaps a named vector as a key might be of use:

也许作为键的命名向量可能会有用：

> factor(unname(c(Y = "Yes", Yes = "Yes", N = "No", No = "No", H = NA)[x]))
[1] Yes  Yes  Yes  No   No   <NA>
Levels: No Yes

This looks very similar to your last attempt... but this one works :-)

这看起来和你上一次的尝试非常相似。但这一款管用：-)

Another way is to make a table containing the mapping:

另一种方法是制作一个包含映射的表：

# stacking the list from Aaron's answer
fmap = stack(list(Yes = c("Y", "Yes"), No = c("N", "No")))

fmap$ind[ match(x, fmap$values) ]
# [1] Yes  Yes  Yes  No   No   <NA>
# Levels: No Yes

# or...

library(data.table)
setDT(fmap)[x, on=.(values), ind ]
# [1] Yes  Yes  Yes  No   No   <NA>
# Levels: No Yes

I prefer this way, since it leaves behind an easily inspected object summarizing the map; and the data.table code looks just like any other join in that syntax.

我更喜欢这种方式，因为它留下了一个易于检查的对象来总结映射；而且data.table代码看起来就像该语法中的任何其他连接。

---

Of course, if you don't want an object like fmap summarizing the change, it can be a "one-liner":

当然，如果您不希望像FMAP这样的对象总结更改，它可以是“一行程序”：

library(data.table)
setDT(stack(list(Yes = c("Y", "Yes"), No = c("N", "No"))))[x, on=.(values), ind ]
# [1] Yes  Yes  Yes  No   No   <NA>
# Levels: No Yes

First let's note that in this specific case we can use partial matching:

首先，让我们注意，在这个特定的情况下，我们可以使用部分匹配：

x <- c("Y", "Y", "Yes", "N", "No", "H")
y <- c("Yes","No")
x <- factor(y[pmatch(x,y,duplicates.ok = TRUE)])
# [1] Yes  Yes  Yes  No   No   <NA>
# Levels: No Yes

In a more general case I'd go with dplyr::recode:

在更一般的情况下，我会使用dplyr：：Recode：

library(dplyr)
x <- c("Y", "Y", "Yes", "N", "No", "H")
y <- c(Y="Yes",N="No")
x <- recode(x,!!!y)
x <- factor(x,y)
# [1] Yes  Yes  Yes  No   No   <NA>
# Levels: Yes No

Slightly altered if the starting point is a factor:

如果起始点是一个因素，则略有更改：

x <- factor(c("Y", "Y", "Yes", "N", "No", "H"))
y <- c(Y="Yes",N="No")
x <- recode_factor(x,!!!y)
x <- factor(x,y)
# [1] Yes  Yes  Yes  No   No   <NA>
# Levels: Yes No

I add this answer to demonstrate the accepted answer working on a specific factor in a dataframe, since this was not initially obvious to me (though it probably should have been).

我添加这个答案是为了演示可接受的答案在数据帧中的特定因素上起作用，因为这最初对我来说并不明显(尽管它可能应该是显而易见的)。

levels(df$var1)
# "0" "1" "Z"
summary(df$var1)
#    0    1    Z 
# 7012 2507    8 
levels(df$var1) <- list("0"=c("Z", "0"), "1"=c("1"))
levels(df$var1)
# "0" "1"
summary(df$var1)
#    0    1 
# 7020 2507

I don't know your real use-case, but would strtrim be of any use here...

我不知道你的实际用例，但strtrim在这里有什么用处吗？

factor( strtrim( x , 1 ) , levels = c("Y" , "N" ) , labels = c("Yes" , "No" ) )
#[1] Yes  Yes  Yes  No   No   <NA>
#Levels: Yes No

Similar to @Aaron's approach, but slightly simpler would be:

类似于@Aaron的方法，但稍微简单一点：

x <- c("Y", "Y", "Yes", "N", "No", "H")
x <- factor(x)
# levels(x)  
# [1] "H"   "N"   "No"  "Y"   "Yes"
# NB: the offending levels are 1, 2, & 4
levels(x)[c(1,2,4)] <- c(NA, "No", "Yes")
x
# [1] Yes  Yes  Yes  No   No   <NA>
# Levels: No Yes

You may use the below function for combining/collapsing multiple factors:

您可以使用以下功能组合/折叠多个因素：

combofactor <- function(pattern_vector,
         replacement_vector,
         data) {
 levels <- levels(data)
 for (i in 1:length(pattern_vector))
      levels[which(pattern_vector[i] == levels)] <-
        replacement_vector[i]
 levels(data) <- levels
  data
}

Example:

示例：

Initialize x

初始化x

x <- factor(c(rep("Y",20),rep("N",20),rep("y",20),
rep("yes",20),rep("Yes",20),rep("No",20)))

Check the structure

检查结构

str(x)
# Factor w/ 6 levels "N","No","y","Y",..: 4 4 4 4 4 4 4 4 4 4 ...

Use the function:

使用函数：

x_new <- combofactor(c("Y","N","y","yes"),c("Yes","No","Yes","Yes"),x)

Recheck the structure:

重新检查结构：

str(x_new)
# Factor w/ 2 levels "No","Yes": 2 2 2 2 2 2 2 2 2 2 ...