如何將字符串拆分爲給定長度的子字符串？

Question

我有一個字符串，例如：

"aabbccccdd"

我想要打破這個串入長度爲2的子串的矢量：

"aa" "bb" "cc" "cc" "dd"

Answer 1

這裏是單程

substring("aabbccccdd", seq(1, 9, 2), seq(2, 10, 2)) 
#[1] "aa" "bb" "cc" "cc" "dd" 

或更一般地

text <- "aabbccccdd" 
substring(text, seq(1, nchar(text)-1, 2), seq(2, nchar(text), 2)) 
#[1] "aa" "bb" "cc" "cc" "dd" 

編輯：這是很多，要快得多

sst <- strsplit(text, "")[[1]] 
out <- paste0(sst[c(TRUE, FALSE)], sst[c(FALSE, TRUE)]) 

它首先拆分字符串轉換成字符。然後，將偶數元素和奇怪元素粘貼在一起。

計時

text <- paste(rep(paste0(letters, letters), 1000), collapse="") 
g1 <- function(text) { 
    substring(text, seq(1, nchar(text)-1, 2), seq(2, nchar(text), 2)) 
} 
g2 <- function(text) { 
    sst <- strsplit(text, "")[[1]] 
    paste0(sst[c(TRUE, FALSE)], sst[c(FALSE, TRUE)]) 
} 
identical(g1(text), g2(text)) 
#[1] TRUE 
library(rbenchmark) 
benchmark(g1=g1(text), g2=g2(text)) 
# test replications elapsed relative user.self sys.self user.child sys.child 
#1 g1   100 95.451 79.87531 95.438  0   0   0 
#2 g2   100 1.195 1.00000  1.196  0   0   0 

Answer 2

string <- "aabbccccdd" 
# total length of string 
num.chars <- nchar(string) 

# the indices where each substr will start 
starts <- seq(1,num.chars, by=2) 

# chop it up 
sapply(starts, function(ii) { 
    substr(string, ii, ii+1) 
}) 

其中給出

[1] "aa" "bb" "cc" "cc" "dd" 

Answer 3

人們可以使用一個矩陣來組中的字符：

s2 <- function(x) { 
    m <- matrix(strsplit(x, '')[[1]], nrow=2) 
    apply(m, 2, paste, collapse='') 
} 

s2('aabbccddeeff') 
## [1] "aa" "bb" "cc" "dd" "ee" "ff" 

不幸的是，第是奇數字符串長度的輸入休息，給予警告：

s2('abc') 
## [1] "ab" "ca" 
## Warning message: 
## In matrix(strsplit(x, "")[[1]], nrow = 2) : 
## data length [3] is not a sub-multiple or multiple of the number of rows [2] 

更不幸的是，g1和g2從@GSee不返回不正確的結果爲奇數字符串長度的輸入：

g1('abc') 
## [1] "ab" 

g2('abc') 
## [1] "ab" "cb" 

這裏是s2精神中的功能，對每個組中的字符數取參數，並在必要時留下最後一個條目：

s <- function(x, n) { 
    sst <- strsplit(x, '')[[1]] 
    m <- matrix('', nrow=n, ncol=(length(sst)+n-1)%/%n) 
    m[seq_along(sst)] <- sst 
    apply(m, 2, paste, collapse='') 
} 

s('hello world', 2) 
## [1] "he" "ll" "o " "wo" "rl" "d" 
s('hello world', 3) 
## [1] "hel" "lo " "wor" "ld" 

（它確實比g2慢，但通過的7約爲係數比g1更快）

Answer 4

醜但工程

sequenceString <- "ATGAATAAAG" 

J=3#maximum sequence length in file 
sequenceSmallVecStart <- 
    substring(sequenceString, seq(1, nchar(sequenceString)-J+1, J), 
    seq(J,nchar(sequenceString), J)) 
sequenceSmallVecEnd <- 
    substring(sequenceString, max(seq(J, nchar(sequenceString), J))+1) 
sequenceSmallVec <- 
    c(sequenceSmallVecStart,sequenceSmallVecEnd) 
cat(sequenceSmallVec,sep = "\n") 

給出 ATG AAT AAA ģ

Answer 5

有兩種容易的可能性：

s <- "aabbccccdd" 

1. gregexpr和regmatches：

   regmatches(s, gregexpr(".{2}", s))[[1]] 
   # [1] "aa" "bb" "cc" "cc" "dd" 
   

2. strsplit：

   strsplit(s, "(?<=.{2})", perl = TRUE)[[1]] 
   # [1] "aa" "bb" "cc" "cc" "dd"