我遇到了一个小问题，但没有找到合适的搜索词。我有来自“A”-“N”的字母，并且想根据它们在字母表中的位置将这些大于“G”的字母替换为“A”-“G”。使用gsub它似乎很麻烦。还是有任何正则表达式可以做得更聪明？
k <- rep(LETTERS[1:14],2)
gsub(pattern="H", replace="A", x=k)
gsub(pattern="I", replace="B", x=k)
gsub(pattern="J", replace="C", x=k)
gsub(pattern="K", replace="D", x=k)
# etc.
是不是有某种方法可以将字符转换为整数，然后简单地在整数值内计算，然后再回滚？或者有任何字母的倒数吗？
as.numeric()并as.integer()返回NA。
这将 HN 转换为 AG：
chartr("HIJKLMN", "ABCDEFG", k)
每当我看到这样的问题时，我的第一个想法是match：
AG <- LETTERS[1:7]
HN <- LETTERS[8:14]
k <- rep(LETTERS[1:14],2)
n <- AG[match(k, HN)]
ifelse(is.na(n), k, n)
# [1] "A" "B" "C" "D" "E" "F" "G" "A" "B" "C" "D" "E" "F" "G" "A" "B" "C" "D" "E"
#[20] "F" "G" "A" "B" "C" "D" "E" "F" "G"
我会LETTERS以同样的方式构造一个反函数：
invLETTERS <- function(x) match(x, LETTERS[1:26])
invLETTERS(k)
# [1] 1 2 3 4 5 6 7 8 9 10 11 12 13 14 1 2 3 4 5 6 7 8 9 10 11
#[26] 12 13 14
这是一个干净直接的解决方案：
k <- rep(LETTERS[1:14],2)
# (1) Create a lookup vector whose elements can be indexed into
# by their names and will return their associated values
subs <- setNames(rep(LETTERS[1:7], 2), LETTERS[1:14])
subs
# A B C D E F G H I J K L M N
# "A" "B" "C" "D" "E" "F" "G" "A" "B" "C" "D" "E" "F" "G"
# (2) Use it.
unname(subs[k])
# [1] "A" "B" "C" "D" "E" "F" "G" "A" "B" "C" "D" "E" "F" "G"
# [15] "A" "B" "C" "D" "E" "F" "G" "A" "B" "C" "D" "E" "F" "G"
我确信有一种方法可以使它更紧凑，但这可能是您在第二个非正则表达式想法中考虑的那种事情：
k <- factor(k)
> k1 <- as.integer(k) %% 7
> k1[k1 == 0] <- 7
> LETTERS[k1]
[1] "A" "B" "C" "D" "E" "F" "G" "A" "B" "C" "D" "E" "F" "G" "A" "B" "C" "D" "E" "F" "G" "A"
[23] "B" "C" "D" "E" "F" "G"
可能有一种巧妙的方法来回避 0 索引问题，但我现在感觉不是很聪明。
编辑
评论中的好建议。首先，处理 0 形式的模运算：
k1 <- ((as.integer(k)-1) %%7) + 1
并结合match它变成一个单行：
k1 <- LETTERS[((match(k, LETTERS)-1) %% 7) + 1]
如果您的问题仅与 AN 有关：
set.seed(1)
k = sample(LETTERS[1:14], 42, replace=TRUE)
temp = match(k, LETTERS)
# > table(k)
# k
# A B C D E F G I J K L M N
# 2 2 5 2 1 6 3 3 5 4 3 3 3
k[which(temp > 7)] = LETTERS[temp[temp > 7] -7]
# > table(k)
# k
# A B C D E F G
# 2 5 10 6 4 9 6