什么是最有效的(即高效/适当的)的方式来清理含有需要被倒塌多层次的因素? 也就是说,如何将两个或多个因子水平合二为一。
这里就是两个层次“是”和“Y”应该被倒塌的例子为“是”,“否”和“N”晕倒“否”:
## 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
一种选择是当然的清洁使用前手串sub
和朋友。
另一种方法,是让重复的标签,然后把它们
## 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)
然而, 有没有更有效的方法是什么?
虽然我知道, levels
和labels
的参数应该是向量,我列表和命名名单,并命名为载体实验,看看会发生什么不用说,有下列情形的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"))
Answer 1:
使用levels
的功能,并通过它命名列表,名称是水平的需要的名称和要素是应该改名为当前的名字。
x <- c("Y", "Y", "Yes", "N", "No", "H")
x <- factor(x)
levels(x) <- list(Yes=c("Y", "Yes"), No=c("N", "No"))
x
## [1] Yes Yes Yes No No <NA>
## Levels: Yes No
随着中提到levels
文件; 还看到有例子。
值:对于“因子”的方法,字符串的与长度的矢量的至少的“X”,或命名列表指定如何重命名层次等级的数目。
这也可以在一行中进行,如马立克在这里所做的: https://stackoverflow.com/a/10432263/210673 ; 该levels<-
魔法这里要解释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
Answer 2:
因为这个问题是名为清理因子水平(压扁多层次/标签),则forcats
包应该在这里提及为好,为完整起见。 forcats
在2016年八月出现在CRAN。
有清理因子水平提供了一些便利的功能:
x <- c("Y", "Y", "Yes", "N", "No", "H")
library(forcats)
崩溃因子水平成手动定义的组
fct_collapse(x, Yes = c("Y", "Yes"), No = c("N", "No"), NULL = "H")
#[1] Yes Yes Yes No No <NA>
#Levels: No Yes
手动更改因子水平
fct_recode(x, Yes = "Y", Yes = "Yes", No = "N", No = "No", NULL = "H")
#[1] Yes Yes Yes No No <NA>
#Levels: No Yes
自动重新标记因子水平,必要时崩溃
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
需要注意的是fct_relabel()
与因子水平的作品,所以它需要一个因素,因为第一个参数。 其他两个函数, fct_collapse()
和fct_recode()
同时接受的字符向量其是未公开的特性。
通过重新排序首次亮相因子水平
由OP给出的预期输出
[1] Yes Yes Yes No No <NA>
Levels: Yes No
在这里,他们出现在水平排序x
是不同于缺省( ?factor
: 一个因素的水平是默认排序 )。
以与预期的输出线,这可以通过使用可实现fct_inorder()
折叠之前的水平:
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")
都返回相同的顺序与水平的预期输出,现在。
Answer 3:
也许一个名为向量作为密钥可能是有用的:
> factor(unname(c(Y = "Yes", Yes = "Yes", N = "No", No = "No", H = NA)[x]))
[1] Yes Yes Yes No No <NA>
Levels: No Yes
这看起来非常相似,你的最后一次尝试......但是这一个工程:-)
Answer 4:
另一种方法是,使含有的映射的表:
# 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
我喜欢这种方式,因为它留下一个容易检查对象总结地图背后; 和data.table代码看起来就像任何其他在语法加入。
当然,如果你不希望像一个对象fmap
总结了变化,它可以是一个“一班轮”:
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
Answer 5:
我不知道你的真实使用情况,但将strtrim
在这里任何使用...
factor( strtrim( x , 1 ) , levels = c("Y" , "N" ) , labels = c("Yes" , "No" ) )
#[1] Yes Yes Yes No No <NA>
#Levels: Yes No
Answer 6:
类似@阿龙的方法,但稍微简单的将是:
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
Answer 7:
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).
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
Answer 8:
您可以使用下面的功能组合/折叠多重因素:
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
}
例:
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)))
检查结构
str(x)
# Factor w/ 6 levels "N","No","y","Y",..: 4 4 4 4 4 4 4 4 4 4 ...
使用功能:
x_new <- combofactor(c("Y","N","y","yes"),c("Yes","No","Yes","Yes"),x)
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 ...
Answer 9:
首先,我们注意到,在这种特殊情况下,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
在更一般的情况下,我会跟去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
稍微改变,如果出发点是一个因素:
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
文章来源: Cleaning up factor levels (collapsing multiple levels/labels)