如何将一个字符串分割成给定长度的子串?(复制)

This question already has an answer here:

这个问题已经有了答案:

Chopping a string into a vector of fixed width character elements 10 answers
将一个字符串切成一个固定宽度字符的向量，10个答案。

I have a string such as:

我有一个字符串，例如:

"aabbccccdd"

“aabbccccdd”

I want to break this string into a vector of substrings of length 2 :

我想把这个字符串分解成长度为2的子串的向量:

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

“aa”“bb”“cc”“cc”“dd”

5 个解决方案

#1

Here is one way

这是方法之一

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

or more generally

或更一般的

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

Edit: This is much, much faster

编辑:这要快得多。

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

It first splits the string into characters. Then, it pastes together the even elements and the odd elements.

它首先将字符串分割成字符。然后，它将连元素和奇元素混在一起。

Timings

计时

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

#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)
})

Which gives

这给了

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

#3

There are two easy possibilities:

有两种简单的可能性:

s <- "aabbccccdd"

gregexpr and regmatches:

gregexpr regmatches:

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

strsplit:

strsplit:

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

#4

One can use a matrix to group the characters:

你可以用一个矩阵来对这些字符进行分组:

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

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

Unfortunately, this breaks for an input of odd string length, giving a warning:

不幸的是，这是一个奇怪的字符串长度的输入，给出一个警告:

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]

More unfortunate is that g1 and g2 from @GSee silently return incorrect results for an input of odd string length:

更不幸的是，@GSee的g1和g2会以奇数字符串长度的输入返回错误的结果:

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

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

Here is function in the spirit of s2, taking a parameter for the number of characters in each group, and leaves the last entry short if necessary:

这里是s2的spirit函数，为每个组中的字符数取一个参数，并在必要时保留最后一个条目:

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"

(It is indeed slower than g2, but faster than g1 by about a factor of 7)

(它确实比g2慢，但比g1快1 / 7)

#5

Ugly but works

丑陋但作品

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")

Gives ATG AAT AAA G

给予ATG AAT AAA G。

#1