R：计算字符串列表中的所有组合（特定顺序）

Question

我正在尝试计算由“>”分隔的大量字符列表中的所有序列，但只计算彼此直接相邻的组合。

例如给定字符向量：

[1]Social>PaidSearch>PaidSearch>PaidSearch>PaidSearch>PaidSearch>PaidSearch>PaidSearch>PaidSearch>PaidSearch>PaidSearch>OrganicSearch>OrganicSearch>OrganicSearch
[2]Referral>Referral>Referral

我可以运行以下行来检索包含 2 个字符的所有组合：

split_fn <- sapply(p , strsplit , split = ">", perl=TRUE)

split_fn <- sapply(split_fn, function(x) paste(head(x,-1) , tail(x,-1) , sep = ">") )

返回：

[[1]]

 [1] "Social>PaidSearch"           "PaidSearch>PaidSearch"       "PaidSearch>PaidSearch"       "PaidSearch>PaidSearch"       "PaidSearch>PaidSearch"      
 [6] "PaidSearch>PaidSearch"       "PaidSearch>PaidSearch"       "PaidSearch>PaidSearch"       "PaidSearch>PaidSearch"       "PaidSearch>PaidSearch"      
[11] "PaidSearch>OrganicSearch"    "OrganicSearch>OrganicSearch" "OrganicSearch>OrganicSearch"

[[2]]

[1] "Referral>Referral" "Referral>Referral"

我的数据中所有可能的 2 个字符序列（按顺序拆分）

我知道想要 3 个字符的所有可能结果。

例如

"Social>PaidSearch>PaidSearch" "PaidSearch>PaidSearch>PaidSearch"..."Referral>Referral>Referral"

尝试使用

unlist(lapply(strsplit(p, split = ">"), function(i) combn(sort(i), 3, paste, collapse='>')))

但它会返回所有组合，包括那些不直接跟随的组合。

我也不希望它返回第一行中的最后一个值与第二行中的第一个值等的组合。

Answer 1

让我们从创建一些数据开始：

set.seed(1)

data <- lapply(1:3, function(i) sample(LETTERS[1:3], rpois(1, 6), re = T))
data <- sapply(data, paste, collapse = ">")

data
#> [1] "B>B>C>A"           "C>B>B>A>A>A>C>B>C" "C>C>B>C>C>A"

鉴于问题，将这些数据视为一个列表是有意义的 我们用分隔符>分割元素后得到的向量：

strsplit(data, ">")
#> [[1]]
#> [1] "B" "B" "C" "A"
#> 
#> [[2]]
#> [1] "C" "B" "B" "A" "A" "A" "C" "B" "C"
#> 
#> [[3]]
#> [1] "C" "C" "B" "C" "C" "A"

现在，问题的核心是找到给定的所有连续序列 单个向量的长度。一旦我们能做到这一点，申请就很简单了 我们拥有的数据列表；转换回分隔格式将 也很简单。

考虑到这个目标，我们可以创建一个函数来提取 序列；在这里，我们只是遍历每个元素并提取 给定长度的所有序列到一个列表中：

seqs <- function(x, length = 2) {
  if (length(x) < length)
    return(NULL)
  k <- length - 1
  lapply(seq_len(length(x) - k), function(i) x[i:(i + k)])
}

我们现在可以在之后的数据中应用该函数 将分隔字符拆分为向量以获得结果。我们还需要一个额外的 sapply 和 paste 来将数据转换回我们开始使用的分隔格式：

lapply(strsplit(data, ">"), function(x) {
  sapply(seqs(x, 3), paste, collapse = ">")
})
#> [[1]]
#> [1] "B>B>C" "B>C>A"
#> 
#> [[2]]
#> [1] "C>B>B" "B>B>A" "B>A>A" "A>A>A" "A>A>C" "A>C>B" "C>B>C"
#> 
#> [[3]]
#> [1] "C>C>B" "C>B>C" "B>C>C" "C>C>A"

---

进一步，为了同时得到多个长度的序列，我们可以再增加一层迭代：

lapply(strsplit(data, ">"), function(x) {
  unlist(sapply(c(2, 3), function(n) {
    sapply(seqs(x, n), paste, collapse = ">")
  }))
})
#> [[1]]
#> [1] "B>B"   "B>C"   "C>A"   "B>B>C" "B>C>A"
#> 
#> [[2]]
#>  [1] "C>B"   "B>B"   "B>A"   "A>A"   "A>A"   "A>C"   "C>B"   "B>C"  
#>  [9] "C>B>B" "B>B>A" "B>A>A" "A>A>A" "A>A>C" "A>C>B" "C>B>C"
#> 
#> [[3]]
#> [1] "C>C"   "C>B"   "B>C"   "C>C"   "C>A"   "C>C>B" "C>B>C" "B>C>C" "C>C>A"

由reprex package (v0.2.0) 于 2018 年 5 月 21 日创建。

Answer 2

使用stringr 包（或一般的正则表达式）。

library(stringr)
str_extract_all(p, "(\\w+)>(\\w+)>(\\w+)")

有重叠，但代码可以简化。

str_extract_all_overlap <- function (x) {
  extractions <- character()
  x_curr <- x
  extr <- str_extract(x_curr, "(\\w+)>(\\w+)>(\\w+)")
  i = 1
  while (!is.na(extr)) {
    extractions[i] <- extr 
    x_curr <- str_replace(x_curr, "\\w+", replacement = "")
    extr <- str_extract(x_curr, "(\\w+)>(\\w+)>(\\w+)")
    i = i + 1
  }
  return(extractions)
}

lapply(p, str_extract_all_overlap)

Answer 3

您还可以将第二个 sapply 中的 paste-command 调整为：

paste(head(x,-2), head(tail(x,-1),-1), tail(x,-2) , sep = ">")

您的完整代码现在应该如下所示：

split_fn <- sapply(p , strsplit , split = ">", USE.NAMES = FALSE)

split_fn <- sapply(split_fn, function(x) paste(head(x,-2), head(tail(x,-1),-1), tail(x,-2), sep = ">") )

结果：

> split_fn
[[1]]
 [1] "Social>PaidSearch>PaidSearch"              "PaidSearch>PaidSearch>PaidSearch"          "PaidSearch>PaidSearch>PaidSearch"         
 [4] "PaidSearch>PaidSearch>PaidSearch"          "PaidSearch>PaidSearch>PaidSearch"          "PaidSearch>PaidSearch>PaidSearch"         
 [7] "PaidSearch>PaidSearch>PaidSearch"          "PaidSearch>PaidSearch>PaidSearch"          "PaidSearch>PaidSearch>PaidSearch"         
[10] "PaidSearch>PaidSearch>OrganicSearch"       "PaidSearch>OrganicSearch>OrganicSearch"    "OrganicSearch>OrganicSearch>OrganicSearch"

[[2]]
[1] "Referral>Referral>Referral"