对长度大于1的列表组件使用“a”

Question

我有一个类似于以下内容的清单：

> x <- list(
    j = "first",
    k = "second",
    l = "third",
    m = c("first", "second"),
    n = c("first", "second", "third"),
    o = c("first", "third"),
    p = c("second", "third"),
    q = "first",
    r = "third")

我需要根据字符串组件的值创建它们的索引。对于包含单个字符串元素的组件，我可以很容易地使用which完成此操作：

> which(x == "third")
l r
3 9

甚至对于包含单个字符串元素的多个组件：

> which(x == "first" | x == "second" | x == "third")
j k l q r
1 2 3 8 9

返回的值显示列表中组件的数量及其名称。但是，有时我需要获得包含多个元素(如m (c("first", "second"))或n (c("first", "second", "third")) )的字符向量的组件索引。也就是说，组件中字符向量的长度将大于1。我认为以下内容可以工作：

which(x == c("first", "second"))

我错了，结果是：

j k
1 2
Warning message:
In x == c("first", "second") :
  longer object length is not a multiple of shorter object length

当我尝试一个以上的条件时，情况也是一样的：

> which(x == "first" | x == c("first", "second"))
j k q
1 2 8
Warning message:
In x == c("first", "second") :
  longer object length is not a multiple of shorter object length

which(x == c("first", "second"))的期望输出是：

m
4

which(x == "first" | x == c("first", "second"))的期望输出是：

j m q
1 4 8

这是怎么做到的？不用用which，你.

Answer 1

通过使用"list" == "character"，"list“被转换为”as.character(x)“，我认为这是不需要的。您可以使用match比较“list”的相应元素：

ff = function(x, table) which(setNames(table %in% x, names(table)))
ff(list("third"), x)
#l r 
#3 9 
ff(list("first", "second", "third"), x) # your "|"
#j k l q r 
#1 2 3 8 9 
ff(list(c("first", "second")), x)
#m 
#4 
ff(list("first", c("first", "second")), x) # your "|"
#j m q 
#1 4 8