R在循环中重新编码变量
再见 下面是一个可复制的示例R在循环中重新编码变量,r,dplyr,paste,recode,R,Dplyr,Paste,Recode,再见 下面是一个可复制的示例 df <- data.frame("STUDENT"=c(1,2,3,4,5), "TEST1"=c(6,88,17,5,18), "TEST2"=c(34,NA,87,88,82), "TEST3"=c(87,62,13,8,71), "TEST1NEW"=c(0,1,0,0,0), "TE
df <- data.frame("STUDENT"=c(1,2,3,4,5),
"TEST1"=c(6,88,17,5,18),
"TEST2"=c(34,NA,87,88,82),
"TEST3"=c(87,62,13,8,71),
"TEST1NEW"=c(0,1,0,0,0),
"TEST2NEW"=c(0,NA,1,1,1),
"TEST3NEW"=c(1,1,0,0,1)
df你可以
df[, paste0("TEST", 1:3, "_NEW")] <- as.integer(df[,-1] >= 50)
df
# STUDENT TEST1 TEST2 TEST3 TEST1_NEW TEST2_NEW TEST3_NEW
#1 1 6 34 87 0 0 1
#2 2 88 NA 62 1 NA 1
#3 3 17 87 13 0 1 0
#4 4 5 88 8 0 1 0
#5 5 18 82 71 0 1 1
这真的很好@markus。现在,如果我想操纵重新编码并获得更具体的结果,比如说如果我想要三个类别而不是0和1,并且说我想要4、7和8,我怎么能做到这一点as.integer(df[,-1]=20&=65,8)
是我能想到的最好的方法,如何处理NA值?因此,在TEST1
中缺失会在TEST1NEW
中产生NA,而不是0Ciao@Salman。样本输出在df中。非常感谢
df[, paste0("TEST", 1:3, "_NEW")] <- as.integer(df[,-1] >= 50)
df
# STUDENT TEST1 TEST2 TEST3 TEST1_NEW TEST2_NEW TEST3_NEW
#1 1 6 34 87 0 0 1
#2 2 88 NA 62 1 NA 1
#3 3 17 87 13 0 1 0
#4 4 5 88 8 0 1 0
#5 5 18 82 71 0 1 1
df <- data.frame(
"STUDENT" = c(1, 2, 3, 4, 5),
"TEST1" = c(6, 88, 17, 5, 18),
"TEST2" = c(34, NA, 87, 88, 82),
"TEST3" = c(87, 62, 13, 8, 71)
)
library(dplyr)
df[, paste0("TEST", 1:3, "_NEW")] <- case_when(df[,-1] < 20 ~ 4L,
df[,-1] >= 65 ~ 8L,
is.na(df[,-1]) ~ NA_integer_,
TRUE ~ 7L)