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 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 = ">") )

Returns:

[[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"

这是我的数据中所有可能的 2 个字符序列（按顺序拆分）

我知道想要得到 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 return 所有组合，包括不直接跟随的组合。

我也不希望它 return 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

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 2

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"

考虑到这个问题，将这些数据视为列表是有意义的用分隔符 >:

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来自单个向量的长度。一旦我们可以做到这一点，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要简单

有了这个目标，我们就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列表：

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对数据应用函数将分隔的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，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) 创建于 2018-05-21。

Answer 3

您还可以将第二个 sapply 中的 paste-命令调整为：

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"

R：计算字符串列表中的所有组合（特定顺序）

R: Count all combinations in a list of strings (Specific Order)

combinations

r

count

strsplit

sapply