如何在R中的切割函数中放置多个断点?
我有3列,我想根据每个列的3个不同的分隔符(如示例中所示)设置标签,但我不知道如何设置,因为我可以为同一分隔符的多个列设置标签,但不能为不同的分隔符(br1、br2、br3)设置标签如何在R中的切割函数中放置多个断点?,r,function,multiple-columns,cut,R,Function,Multiple Columns,Cut,我有3列,我想根据每个列的3个不同的分隔符(如示例中所示)设置标签,但我不知道如何设置,因为我可以为同一分隔符的多个列设置标签,但不能为不同的分隔符(br1、br2、br3)设置标签 var您可以使用Map: Map(function(x, y, labels=1:9) cut(x, y, labels = labels), df[, 2:4], list(br1, br2, br3)) 输出是df列的列表。可以使用as.data.frame将其转换为数据帧。您还可以将其他参数添加到cut(例
var您可以使用Map
:
Map(function(x, y, labels=1:9) cut(x, y, labels = labels), df[, 2:4], list(br1, br2, br3))
输出是df
列的列表。可以使用as.data.frame
将其转换为数据帧。您还可以将其他参数添加到cut
(例如,include_lowest
)。提供的间隔之外的值为NA
s
# OUTPUT
$x1
[1] 6 8 <NA> 8 9 4 5 <NA> 6 8
Levels: 1 2 3 4 5 6 7 8 9
$x2
[1] 8 6 6 5 3 7 8 6 9 8
Levels: 1 2 3 4 5 6 7 8 9
$x3
[1] 8 9 <NA> 9 9 9 <NA> 9 8 8
Levels: 1 2 3 4 5 6 7 8 9
#输出
$x1
[1] 6 8 8 9 4 5 6 8
级别:123456789
$x2
[1] 8 6 6 5 3 7 8 6 9 8
级别:123456789
$x3
[1] 8 9 9 9 9 9 8 8
级别:123456789
数据
set.seed(123)
var <- 1:10
x1 <- rnorm(10, mean=100, sd=25)
x2 <- rnorm(10, mean=100, sd=25)
x3 <- rnorm(10, mean=100, sd=25)
df <- data.frame(var,x1,x2,x3)
#With 1 break for all the columns
br1 <-c(50,60,70,80,90,100,110,120,130,140)
br2 <-c(30,40,45,55,61,70,98,105,115,138)
br3<-c(20,25,30,35,38,42,45,70,95,132)
set.seed(123)
var考虑使用get()
将cut
值指定为带有for
循环的新列,以使用列后缀中的变量名称:
set.seed(1082019)
#...
for(b in c("br1", "br2", "br3"))
df[paste0(names(df)[2:4], "_", b)] <- lapply(df[, 2:4], cut, br=get(b), labels=c(1:9))
df
# var x1 x2 x3 x1_br1 x2_br1 x3_br1 x1_br2 x2_br2 x3_br2 x1_br3 x2_br3 x3_br3
# 1 1 121.95508 98.40327 139.31413 8 5 9 9 7 <NA> 9 9 <NA>
# 2 2 105.28775 116.99844 83.12366 6 7 4 8 9 6 9 9 8
# 3 3 80.17226 118.92694 104.57693 4 7 6 6 9 7 8 9 9
# 4 4 146.94335 90.50056 58.35752 <NA> 5 1 <NA> 6 4 <NA> 8 7
# 5 5 98.15953 23.58072 86.67441 5 <NA> 4 7 <NA> 6 9 1 8
# 6 6 137.52613 74.83507 95.49531 9 3 5 9 6 6 <NA> 8 9
# 7 7 51.41213 141.01571 68.36462 1 <NA> 2 3 <NA> 5 7 <NA> 7
# 8 8 74.05926 134.66125 93.40060 3 9 5 6 9 6 8 <NA> 8
# 9 9 63.16221 52.25081 76.96090 2 1 3 5 3 6 7 7 8
# 10 10 123.96491 73.03856 138.41414 8 3 9 9 6 <NA> 9 8 <NA>
set.seed(1082019)
#...
对于(c中的b(“br1”、“br2”、“br3”))
df[paste0(names(df)[2:4],“u”,b)]我的答案有帮助吗?@slava kohut你的答案帮助了我,现在我很容易继续我的工作。谢谢。考虑一下投票或接受Tististo,Muthas Gracias!
set.seed(123)
var <- 1:10
x1 <- rnorm(10, mean=100, sd=25)
x2 <- rnorm(10, mean=100, sd=25)
x3 <- rnorm(10, mean=100, sd=25)
df <- data.frame(var,x1,x2,x3)
#With 1 break for all the columns
br1 <-c(50,60,70,80,90,100,110,120,130,140)
br2 <-c(30,40,45,55,61,70,98,105,115,138)
br3<-c(20,25,30,35,38,42,45,70,95,132)
set.seed(1082019)
#...
for(b in c("br1", "br2", "br3"))
df[paste0(names(df)[2:4], "_", b)] <- lapply(df[, 2:4], cut, br=get(b), labels=c(1:9))
df
# var x1 x2 x3 x1_br1 x2_br1 x3_br1 x1_br2 x2_br2 x3_br2 x1_br3 x2_br3 x3_br3
# 1 1 121.95508 98.40327 139.31413 8 5 9 9 7 <NA> 9 9 <NA>
# 2 2 105.28775 116.99844 83.12366 6 7 4 8 9 6 9 9 8
# 3 3 80.17226 118.92694 104.57693 4 7 6 6 9 7 8 9 9
# 4 4 146.94335 90.50056 58.35752 <NA> 5 1 <NA> 6 4 <NA> 8 7
# 5 5 98.15953 23.58072 86.67441 5 <NA> 4 7 <NA> 6 9 1 8
# 6 6 137.52613 74.83507 95.49531 9 3 5 9 6 6 <NA> 8 9
# 7 7 51.41213 141.01571 68.36462 1 <NA> 2 3 <NA> 5 7 <NA> 7
# 8 8 74.05926 134.66125 93.40060 3 9 5 6 9 6 8 <NA> 8
# 9 9 63.16221 52.25081 76.96090 2 1 3 5 3 6 7 7 8
# 10 10 123.96491 73.03856 138.41414 8 3 9 9 6 <NA> 9 8 <NA>