r - 将多个级别编码为 2 个因子标签

Question

我有一个包含一些列的数据框:

• 我想转化为一个因素，
• 其中不同级别编码为-2, -1, 0, 1, 2, 3, 4

• 我希望按照以下约定将级别标记为 0 或 1:

  -2 = 1
  -1 = 1
   0 = 0
   1 = 1
   2 = 1
   3 = 1
   4 = 0
  

我有以下代码:

#Convert to factor
dat[idx] <- lapply(dat[idx], factor, levels = -2:4, labels = c(1, 1, 0, 1, 1, 1, 0))

#Drop unused factor levels
dat <- droplevels(dat)

这有效，但它给了我以下警告:

In `levels<-`(`*tmp*`, value = if (nl == nL) as.character(labels) else paste0(labels,  :
duplicated levels in factors are deprecated

我尝试了以下代码(根据 Ananda Mahto 的建议)但没有成功:

levels(dat[idx]) <- list(`0` = c(0, 4), `1` = c(-2, -1, 1, 2, 3))

我认为必须有更好的方法来做到这一点，有什么建议吗？

我的数据是这样的:

structure(list(Timestamp = structure(c(1380945601, 1380945603, 
1380945605, 1380945607, 1380945609, 1380945611, 1380945613, 1380945615, 
1380945617, 1380945619), class = c("POSIXct", "POSIXt"), tzone = ""), 
FCB2C01 = c(0, 0, 0, 0, 0, 0, 0, 0, 0, 0), RCB2C01 = c(0, 
0, 0, 0, 0, 0, 0, 0, 0, 0), FCB2C02 = c(1, 1, 1, 1, 1, 1, 
1, 1, 1, 1), RCB2C02 = c(0, 0, 0, 0, 0, 0, 0, 0, 0, 0), FCB2C03 = c(0, 
0, 0, 0, 0, 0, 0, 0, 0, 0), RCB2C03 = c(0, 0, 0, 0, 0, 0, 
0, 0, 0, 0), FCB2C04 = c(1, 1, 1, 1, 1, 1, 1, 1, 1, 1), RCB2C04 = c(0, 
0, 0, 0, 0, 0, 0, 0, 0, 0), FCB2C05 = c(1, 1, 1, 1, 1, 1, 
1, 1, 1, 1), RCB2C05 = c(0, 0, 0, 0, 0, 0, 0, 0, 0, 0), FCB2C06 = c(1, 
1, 1, 1, 1, 1, 1, 1, 1, 1), RCB2C06 = c(0, 0, 0, 0, 0, 0, 
0, 0, 0, 0), FCB2C07 = c(1, 1, 1, 1, 1, 1, 1, 1, 1, 1), RCB2C07 = c(0, 
0, 0, 0, 0, 0, 0, 0, 0, 0), FCB2C08 = c(1, 1, 1, 1, 1, 1, 
1, 1, 1, 1), RCB2C08 = c(0, 0, 0, 0, 0, 0, 0, 0, 0, 0), FCB2C09 = c(1, 
1, 1, 1, 1, 1, 1, 1, 1, 1), RCB2C09 = c(0, 0, 0, 0, 0, 0, 
0, 0, 0, 0), FCB2C10 = c(1, 1, 1, 1, 1, 1, 1, 1, 1, 1), RCB2C10 = c(0, 
0, 0, 0, 0, 0, 0, 0, 0, 0)), .Names = c("Timestamp", "FCB2C01", 
"RCB2C01", "FCB2C02", "RCB2C02", "FCB2C03", "RCB2C03", "FCB2C04", 
"RCB2C04", "FCB2C05", "RCB2C05", "FCB2C06", "RCB2C06", "FCB2C07", 
"RCB2C07", "FCB2C08", "RCB2C08", "FCB2C09", "RCB2C09", "FCB2C10", 
"RCB2C10"), row.names = c(NA, 10L), class = "data.frame")

和列索引:

    idx <- seq(2,21,2)

Answer 1

如果我正确理解你想要做什么，“正确”的方法是使用 levels 函数来指定你的级别。比较以下内容:

set.seed(1)
x <- sample(-2:4, 10, replace = TRUE)

YourApproach <- factor(x, levels = -2:4, labels = c(1, 1, 0, 1, 1, 1, 0))
# Warning message:
# In `levels<-`(`*tmp*`, value = if (nl == nL) as.character(labels) else paste0(labels,  :
#   duplicated levels in factors are deprecated
YourApproach
#  [1] 1 0 1 0 1 0 0 1 1 1
# Levels: 1 1 0 1 1 1 0

xFac <- factor(x, levels = -2:4)
levels(xFac) <- list(`0` = c(0, 4), `1` = c(-2, -1, 1, 2, 3))
xFac
#  [1] 1 0 1 0 1 0 0 1 1 1
# Levels: 0 1

请注意每个“级别”的差异。这也意味着底层数字表示将有所不同:

> as.numeric(YourApproach)
 [1] 2 3 5 7 2 7 7 5 5 1
> as.numeric(xFac)
 [1] 2 1 2 1 2 1 1 2 2 2