我有以下順序：
s0 <- "KDRH?THLA???RT?HLAK"
那里的通配符由 表示?。我想要做的是用這個向量中的采樣字符替換那個字符：
AADict <- c("A", "R", "N", "D", "C", "E", "Q", "G", "H",
"I", "L", "K", "M", "F", "P", "S", "T", "W", "Y", "V")
由于s0有 5 個通配符?，我將從 AADict 中取樣：
set.seed(1)
nof_wildcard <- 5
tolower(sample(AADict, nof_wildcard, TRUE))
這使[1] "d" "q" "a" "r" "l"
因此預期的結果是：
KDRH?THLA???RT?HLAK
KDRHdTHLAqarRTlHLAK
所以采樣字符的位置必須與 完全相同?，但字符的順序并不重要。例如，這個答案也是可以接受的：KDRHqTHLAdlaRTrHLAK。
我怎樣才能用 R 實作這一點？
另一個例子是：
s1 <- "FKDHKHIDVKDRHRTHLAK????RTRHLAK"
s2 <- "FKHIDVKDRHRTRHLAK??????????"
uj5u.com熱心網友回復：
一種方法是替換“？” 使用回圈“一次一個”的字符，例如
s0 <- "KDRH?THLA???RT?HLAK"
AADict <- c("A", "R", "N", "D", "C", "E", "Q", "G", "H",
"I", "L", "K", "M", "F", "P", "S", "T", "W", "Y", "V")
s0
#> [1] "KDRH?THLA???RT?HLAK"
repeat{s0 <- sub("\\?", sample(tolower(AADict), 1), s0); if(grepl("\\?", s0) == FALSE) break}
s0
#> [1] "KDRHtTHLAidwRTyHLAK"
s1 <- "FKDHKHIDVKDRHRTHLAK????RTRHLAK"
repeat{s1 <- sub("\\?", sample(tolower(AADict), 1), s1); if(grepl("\\?", s1) == FALSE) break}
s1
#> [1] "FKDHKHIDVKDRHRTHLAKrstaRTRHLAK"
s2 <- "FKHIDVKDRHRTRHLAK??????????"
repeat{s2 <- sub("\\?", sample(tolower(AADict), 1), s2); if(grepl("\\?", s2) == FALSE) break}
s2
#> [1] "FKHIDVKDRHRTRHLAKdvcfmheiqn"
另一種方法也可以允許在不替換的情況下進行采樣：
s0 <- "KDRH?THLA???RT?HLAK"
AADict <- c("A", "R", "N", "D", "C", "E", "Q", "G", "H",
"I", "L", "K", "M", "F", "P", "S", "T", "W", "Y", "V")
matches <- gregexpr("\\?", s0)
regmatches(s0, matches) <- lapply(lengths(matches), sample, x = tolower(AADict), replace = FALSE)
s0
#> [1] "KDRHdTHLAlanRTiHLAK"
由reprex 包于 2022-10-22 創建(v2.0.1)
uj5u.com熱心網友回復：
您可以將字串拆分為單個字符，這樣可以輕松替換通配符而無需回圈（這是我的第一種方法）：
replace_wc <- function(x, dict) {
x <- strsplit(x, split = "")[[1]]
ix <- grepl("\\?", x)
x[ix] <- sample(dict, sum(ix), replace = TRUE)
return(paste0(x, collapse = ""))
}
s0 <- "KDRH?THLA???RT?HLAK"
AADict <- c(
"A", "R", "N", "D", "C", "E", "Q", "G", "H",
"I", "L", "K", "M", "F", "P", "S", "T", "W", "Y", "V"
)
set.seed(1)
replace_wc(s0, tolower(AADict))
#> [1] "KDRHdTHLAqarRTlHLAK"
uj5u.com熱心網友回復：
這是一個向量化函式，用于替換"?"字串向量中的字符。
fun <- function(x, dict = AADict) {
dict <- tolower(dict)
inx <- gregexpr("\\?", x)
sapply(seq_along(x), \(j) {
for(i in inx[[j]]) {
substr(x[j], i, i) <- sample(dict, 1L)
}
x[j]
})
}
AADict <- c("A", "R", "N", "D", "C", "E", "Q", "G", "H",
"I", "L", "K", "M", "F", "P", "S", "T", "W", "Y", "V")
s0 <- "KDRH?THLA???RT?HLAK"
s1 <- "FKDHKHIDVKDRHRTHLAK????RTRHLAK"
s2 <- "FKHIDVKDRHRTRHLAK??????????"
fun(s0)
#> [1] "KDRHsTHLAwppRTwHLAK"
fun(s1)
#> [1] "FKDHKHIDVKDRHRTHLAKyfqfRTRHLAK"
fun(s2)
#> [1] "FKHIDVKDRHRTRHLAKnsfehqwmkv"
fun(c(s0, s1, s2))
#> [1] "KDRHiTHLAdssRTgHLAK" "FKDHKHIDVKDRHRTHLAKcdivRTRHLAK"
#> [3] "FKHIDVKDRHRTRHLAKfrpafwpnif"
使用reprex v2.0.2創建于 2022-10-22
轉載請註明出處，本文鏈接：https://www.uj5u.com/net/520261.html
標籤：r细绳
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