Question

我有一个因子变量，它由两个_分隔的子字符串组成，例如string1_string2。我想分别设置前缀（“ string1”）和后缀（“ string2”）的因子水平，然后为串联字符串定义整体因子水平集。此外，第一个子字符串与第二个子字符串中的级别优先级可能会有所不同。

我想要实现的一个小例子：

# reproducible data

x <- factor(c("DBO_A", "PH_A", "COND_A", "DBO_B", "PH_B", "COND_B", "DBO_C", "PH_C", "COND_C"))

[1] DBO_A  PH_A   COND_A DBO_B  PH_B   COND_B DBO_C  PH_C   COND_C
Levels: COND_A COND_B COND_C DBO_A DBO_B DBO_C PH_A PH_B PH_C

如果我没有定义因子水平，它们将按字母顺序排列。现在，我想在_分隔符的左侧和右侧设置字符串的级别，例如

PH <COND <DBO 。
B <A <C （RHS）。

此外，我想指定LHS或RHS哪一方优先于另一方。根据优先级的高低，级别的总体顺序将有所不同：

（1）如果LHS级别是先例：

[1] DBO_A  PH_A   COND_A DBO_B  PH_B   COND_B DBO_C  PH_C   COND_C
Levels: PH_B PH_A PH_C COND_B COND_A COND_C DBO_B DBO_A DBO_C

（2）如果RHS级别是先例：

[1] DBO_A  PH_A   COND_A DBO_B  PH_B   COND_B DBO_C  PH_C   COND_C
Levels: PH_B COND_B DBO_B PH_A COND_A DBO_A PH_C COND_C DBO_C

现在我只是想像解决factor(x, levels = c(xx, xx, ...))这样的问题，但是我的水平比上面显示的要高，所以这看起来很荒谬。

注意：我不想更改数据的顺序，而只需更改级别的顺序。

Answer 1

使用CRAN软件包forcats，您可以合并一系列因素。下面的函数将prefix和suffix的2个向量按您希望的顺序输入。
参数sep = "_"的默认设置为问题中的分隔符。您可以根据需要传递另一个分隔符。

library(forcats)

custom_fct <- function(prefix, suffix, sep = "_"){
  lst <- lapply(prefix, function(p){
    f <- paste(p, suffix, sep = sep)
    factor(f, levels = f)
  })
  fct_c(!!!lst)
}

x <- c("PH", "COND", "DBO")
y <- c("B", "A", "C")

custom_fct(x, y)

编辑。

在OP的评论之后我才明白，解决该问题的另一种方法是将输入数据向量x强制分解为2个向量，其中一个是前缀，一个是后缀。以下函数创建了这样的向量，不需要外部程序包。

custom_fct2 <- function(x, prefix, suffix, sep = "_"){
  lst <- lapply(prefix, function(p){
    paste(p, suffix, sep = sep)
  })
  factor(x, levels = unlist(lst))
}

x <- c("DBO_A", "PH_A", "COND_A", "DBO_B",
       "PH_B", "COND_B", "DBO_C", "PH_C", "COND_C")
a <- c("PH", "COND", "DBO")
b <- c("B", "A", "C")

custom_fct2(x, a, b)
#[1] DBO_A  PH_A   COND_A DBO_B  PH_B   COND_B DBO_C  PH_C  
#[9] COND_C
#9 Levels: PH_B PH_A PH_C COND_B COND_A COND_C DBO_B ... DBO_C

Answer 2

我们可以使用base R来做到这一点。使用sub删除向量levels中的子字符串，match通过检查那些按自定义顺序的值来创建数字索引，然后重新分配levels factor通过order索引对levels向量的序列进行match

i1 <- match(sub("_.*", "", levels(x)), c("PH", "COND", "DBO"))
i2 <- match(sub(".*_", "", levels(x)), c("B", "A", "C"))
factor(x, levels = levels(x)[seq_along(levels(x))[order(i1, i2)]])

对于第二种情况，只需反转order中的索引

factor(x, levels = levels(x)[seq_along(levels(x))[order(i2, i1)]])

要重复使用，可以包装在一个函数中

f1 <- function(vec, lvls1, lvls2, flag = "former") {
   i1 <- match(sub("_.*", "", levels(vec)), lvls1)
   i2 <- match(sub(".*_", "", levels(vec)), lvls2)

   if(flag == 'former') {
     factor(vec, levels = levels(vec)[seq_along(levels(vec))[order(i1, i2)]])
   } else {
     factor(vec, levels = levels(vec)[seq_along(levels(vec))[order(i2, i1)]])

   }


}

f1(x, c("PH", "COND", "DBO"), c("B", "A", "C"))
#[1] DBO_A  PH_A   COND_A DBO_B  PH_B   COND_B DBO_C  PH_C   COND_C
#Levels: PH_B PH_A PH_C COND_B COND_A COND_C DBO_B DBO_A DBO_C


f1(x, c("PH", "COND", "DBO"), c("B", "A", "C"), flag = "latter")
#[1] DBO_A  PH_A   COND_A DBO_B  PH_B   COND_B DBO_C  PH_C   COND_C
#Levels: PH_B COND_B DBO_B PH_A COND_A DBO_A PH_C COND_C DBO_C

Answer 3

使用data.table便捷功能tstrsplit和setorderv。

为子字符串（cols <- c("V1", "V2")）创建一个（任意）列名称的向量。将向量转换为data.table（d <- data.table(x)）。将向量分为两列（(cols) := tstrsplit(x, split = "_")）。设置子字符串的因子级别（factor(V1, levels = l1)）。按第一个子字符串然后第二个子字符串，或第二个然后第一个（setorderv(d, if(prec == 1) cols else rev(cols))）排序数据。使用data.table中有序的列“ x”作为向量'x'（factor(x, levels = d$x)）的因子水平。

library(data.table)

f <- function(x, l1, l2, prec){
  cols <- c("V1", "V2")
  d <- data.table(x)
  d[ , (cols) := tstrsplit(x, split = "_")]
  d[ , `:=`(
    V1 = factor(V1, levels = l1),
    V2 = factor(V2, levels = l2))]
  setorderv(d, if(prec == 1) cols else rev(cols))
  factor(x, levels = d$x)
}

# First substring has precedence
f(x, l1 = c("PH", "COND", "DBO"), l2 = c("B", "A", "C"), prec = 1)
# [1] DBO_A  PH_A   COND_A DBO_B  PH_B   COND_B DBO_C  PH_C   COND_C
# Levels: PH_B PH_A PH_C COND_B COND_A COND_C DBO_B DBO_A DBO_C

# Second substring has precedence
f(x, l1 = c("PH", "COND", "DBO"), l2 = c("B", "A", "C"), prec = 2)
# [1] DBO_A  PH_A   COND_A DBO_B  PH_B   COND_B DBO_C  PH_C   COND_C
# Levels: PH_B COND_B DBO_B PH_A COND_A DBO_A PH_C COND_C DBO_C

一种类似的base替代方法，但是将子字符串放置在矩阵中。使用标准的正则表达式（例如，here）来抓取子字符串。转换为系数并设置水平。创建列索引（i <- c(1, 2, 1)[prec:(prec + 1)]）。订单等级为'x'（as.character(x)[order(m[ , i[1]], m[ , i[2]])])）。

f2 <- function(x, l1, l2, prec){
  m <- cbind(factor(sub("_.*", "", x), l1), factor(sub(".*_", "", x), l2))
  i <- c(1, 2, 1)[prec:(prec + 1)]
  factor(x, levels = as.character(x)[order(m[ , i[1]], m[ , i[2]])])}

f2(x, l1, l2, prec = 1)
# [1] DBO_A  PH_A   COND_A DBO_B  PH_B   COND_B DBO_C  PH_C   COND_C
# Levels: PH_B PH_A PH_C COND_B COND_A COND_C DBO_B DBO_A DBO_C

f2(x, l1, l2, prec = 2)
# [1] DBO_A  PH_A   COND_A DBO_B  PH_B   COND_B DBO_C  PH_C   COND_C
# Levels: PH_B COND_B DBO_B PH_A COND_A DBO_A PH_C COND_C DBO_C

Answer 4

类似的东西

x <- with(expand.grid(x = c("DBO", "PH", "COND"), y = c("A", "B", "C")),
          factor(paste(x, y, sep = "_"), levels = paste(x, y, sep = "_")))

您不需要写出所有可能的级别，只需写出一侧和另一侧的级别即可。

串联字符串中的自定义因子级别

4 个答案: