string 如何将字符串拆分为给定长度的子字符串？
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How to split a string into substrings of a given length?
提问by MadSeb
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 :
我想将此字符串分解为长度为 2 的子字符串的向量：
"aa" "bb" "cc" "cc" "dd"
"aa" "bb" "cc" "cc" "dd"
回答by GSee
Here is one way
这是一种方法
substring("aabbccccdd", seq(1, 9, 2), seq(2, 10, 2))
#[1] "aa" "bb" "cc" "cc" "dd"
or more generally
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"
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.
它首先将字符串拆分为字符。然后，它将偶数元素和奇数元素粘贴在一起。
Timings
时间安排
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
回答by Sven Hohenstein
There are two easy possibilities:
有两种简单的可能性：
s <- "aabbccccdd"
gregexprandregmatches:regmatches(s, gregexpr(".{2}", s))[[1]] # [1] "aa" "bb" "cc" "cc" "dd"strsplit:strsplit(s, "(?<=.{2})", perl = TRUE)[[1]] # [1] "aa" "bb" "cc" "cc" "dd"
gregexpr和regmatches：regmatches(s, gregexpr(".{2}", s))[[1]] # [1] "aa" "bb" "cc" "cc" "dd"strsplit：strsplit(s, "(?<=.{2})", perl = TRUE)[[1]] # [1] "aa" "bb" "cc" "cc" "dd"
回答by mindless.panda
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
这使
[1] "aa" "bb" "cc" "cc" "dd"
回答by Matthew Lundberg
One can use a matrix to group the characters:
可以使用矩阵对字符进行分组：
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，对于奇数字符串长度的输入，这会中断，并发出警告：
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 g1and g2from @GSee silently return incorrect results for an input of odd string length:
更不幸的是，g1与g2从@GSee不返回不正确的结果为奇数串长度的输入端：
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:
这是本着 s2 精神的函数，采用每个组中字符数的参数，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 g1by about a factor of 7)
（它确实比 慢g2，但比 快g1约 7 倍）
回答by den2042
Ugly but works
丑但有效
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
