我有一个字符串,例如:
"aabbccccdd"
我想把这个字符串分解成一个长度为2的子字符串的向量:
"aa" "bb" "cc" "cc" "dd"
发布于 2012-07-24 04:09:11
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)
})
这给了我们
[1] "aa" "bb" "cc" "cc" "dd"
发布于 2013-02-19 01:45:00
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"
不幸的是,对于奇数字符串长度的输入,这会中断,并给出一个警告:
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]
更不幸的是,对于奇数字符串长度的输入,@GSee中的g1
和g2
会默默返回错误的结果:
g1('abc')
## [1] "ab"
g2('abc')
## [1] "ab" "cb"
以下是遵循s2精神的函数,它接受每个组中的字符数的参数,并在必要时保留最后一项:
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"
(它确实比g2
慢,但比g1
快大约7倍)
发布于 2014-04-24 15:28:27
虽然丑陋,但很管用
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")
提供ATG AAT AAA G
https://stackoverflow.com/questions/11619616
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