我正试图找出一种有效的方法来分割像
这样的字符串"111110000011110000111000"
进入矢量
[1] "11111" "00000" "1111" "0000" "111" "000"
其中“0”和“1”可以是任何交替的字符。
答案 0 :(得分：88)
尝试
strsplit(str1, '(?<=1)(?=0)|(?<=0)(?=1)', perl=TRUE)[[1]]
#[1] "11111" "00000" "1111" "0000" "111" "000"
使用stri_extract_all_regex
library(stringi)
stri_extract_all_regex(str1, '(?:(\\w))\\1*')[[1]]
#[1] "11111" "00000" "1111" "0000" "111" "000"
stri_extract_all_regex(x1, '(?:(\\w))\\1*')[[1]]
#[1] "11111" "00000" "222" "000" "3333" "000" "1111" "0000" "111"
#[10] "000"
stri_extract_all_regex(x2, '(?:(\\w))\\1*')[[1]]
#[1] "aaaaa" "bb" "ccccccc" "bbb" "a" "d" "11111"
#[8] "00000" "222" "aaa" "bb" "cc" "d" "11"
#[15] "D" "aa" "BB"
library(stringi)
set.seed(24)
x3 <- stri_rand_strings(1, 1e4)
akrun <- function() stri_extract_all_regex(x3, '(?:(\\w))\\1*')[[1]]
#modified @thelatemail's function to make it bit more general
thelate <- function() regmatches(x3,gregexpr("(?:(\\w))\\1*", x3,
perl=TRUE))[[1]]
rawr <- function() strsplit(x3, '(?<=(\\w))(?!\\1)', perl=TRUE)[[1]]
ananda <- function() unlist(read.fwf(textConnection(x3),
rle(strsplit(x3, "")[[1]])$lengths,
colClasses = "character"))
Colonel <- function() with(rle(strsplit(x3,'')[[1]]),
mapply(function(u,v) paste0(rep(v,u), collapse=''), lengths, values))
Cryo <- function(){
res_vector=rep(NA_character_,nchar(x3))
res_vector[1]=substr(x3,1,1)
counter=1
old_tmp=''
for (i in 2:nchar(x3)) {
tmp=substr(x3,i,i)
if (tmp==old_tmp) {
res_vector[counter]=paste0(res_vector[counter],tmp)
} else {
res_vector[counter+1]=tmp
counter=counter+1
}
old_tmp=tmp
}
res_vector[!is.na(res_vector)]
}
richard <- function(){
cs <- cumsum(
rle(stri_split_boundaries(x3, type = "character")[[1L]])$lengths
)
stri_sub(x3, c(1, head(cs + 1, -1)), cs)
}
nicola<-function(x) {
indices<-c(0,which(diff(as.integer(charToRaw(x)))!=0),nchar(x))
substring(x,indices[-length(indices)]+1,indices[-1])
}
richard2 <- function() {
cs <- cumsum(rle(strsplit(x3, NULL)[[1L]])[[1L]])
stri_sub(x3, c(1, head(cs + 1, -1)), cs)
}
system.time(akrun())
# user system elapsed
# 0.003 0.000 0.003
system.time(thelate())
# user system elapsed
# 0.272 0.001 0.274
system.time(rawr())
# user system elapsed
# 0.397 0.001 0.398
system.time(ananda())
# user system elapsed
# 3.744 0.204 3.949
system.time(Colonel())
# user system elapsed
# 0.154 0.001 0.154
system.time(Cryo())
# user system elapsed
# 0.220 0.005 0.226
system.time(richard())
# user system elapsed
# 0.007 0.000 0.006
system.time(nicola(x3))
# user system elapsed
# 0.190 0.001 0.191
稍微大一点的字符串，
set.seed(24)
x3 <- stri_rand_strings(1, 1e6)
system.time(akrun())
#user system elapsed
#0.166 0.000 0.155
system.time(richard())
# user system elapsed
# 0.606 0.000 0.569
system.time(richard2())
# user system elapsed
# 0.518 0.000 0.487
system.time(Colonel())
# user system elapsed
# 9.631 0.000 9.358
library(microbenchmark)
microbenchmark(richard(), richard2(), akrun(), times=20L, unit='relative')
#Unit: relative
# expr min lq mean median uq max neval cld
# richard() 2.438570 2.633896 2.365686 2.315503 2.368917 2.124581 20 b
#richard2() 2.389131 2.533301 2.223521 2.143112 2.153633 2.157861 20 b
# akrun() 1.000000 1.000000 1.000000 1.000000 1.000000 1.000000 20 a
注意：尝试运行其他方法，但需要很长时间
str1 <- "111110000011110000111000"
x1 <- "1111100000222000333300011110000111000"
x2 <- "aaaaabbcccccccbbbad1111100000222aaabbccd11DaaBB"
答案 1 :(得分：25)
主题的变化：
x <- "111110000011110000111000"
regmatches(x,gregexpr("1+|0+",x))[[1]]
#[1] "11111" "00000" "1111" "0000" "111" "000"
答案 2 :(得分：21)
您可以使用substr或read.fwf以及rle（尽管它不太可能像任何基于正则表达式的解决方案一样高效）：
x <- "111110000011110000111000"
unlist(read.fwf(textConnection(x),
rle(strsplit(x, "")[[1]])$lengths,
colClasses = "character"))
# V1 V2 V3 V4 V5 V6
# "11111" "00000" "1111" "0000" "111" "000"
这种方法的一个优点是，它可以工作，比如说：
x <- paste(c(rep("a", 5), rep("b", 2), rep("c", 7),
rep("b", 3), rep("a", 1), rep("d", 1)), collapse = "")
x
# [1] "aaaaabbcccccccbbbad"
unlist(read.fwf(textConnection(x),
rle(strsplit(x, "")[[1]])$lengths,
colClasses = "character"))
# V1 V2 V3 V4 V5 V6
# "aaaaa" "bb" "ccccccc" "bbb" "a" "d"
答案 3 :(得分：20)
另一种方法是在交替数字之间添加空格。这适用于任何两个，而不仅仅是1和0。然后在空白处使用strsplit：
x <- "111110000011110000111000"
(y <- gsub('(\\d)(?!\\1)', '\\1 \\2', x, perl = TRUE))
# [1] "11111 00000 1111 0000 111 000 "
strsplit(y, ' ')[[1]]
# [1] "11111" "00000" "1111" "0000" "111" "000"
或者更为简洁，正如@akrun所指出的那样：
strsplit(x, '(?<=(\\d))(?!\\1)', perl=TRUE)[[1]]
# [1] "11111" "00000" "1111" "0000" "111" "000"
同时将\\d更改为\\w也可以
x <- "aaaaabbcccccccbbbad"
strsplit(x, '(?<=(\\w))(?!\\1)', perl=TRUE)[[1]]
# [1] "aaaaa" "bb" "ccccccc" "bbb" "a" "d"
x <- "111110000011110000111000"
strsplit(x, '(?<=(\\w))(?!\\1)', perl=TRUE)[[1]]
# [1] "11111" "00000" "1111" "0000" "111" "000"
您还可以使用\K（而不是明确使用捕获组，\\1和\\2），我不会看到它们使用过多，也不知道如何使用解释一下：}
AFAIK \\K重置报告的匹配的起点，并且不再包含任何以前消耗的字符，基本上丢弃了与该点匹配的所有内容。
x <- "1111100000222000333300011110000111000"
(z <- gsub('(\\d)\\K(?!\\1)', ' ', x, perl = TRUE))
# [1] "11111 00000 222 000 3333 000 1111 0000 111 000 "
答案 4 :(得分：14)
原创方法：这是一种包含rle()的 stringi 方法。
x <- "111110000011110000111000"
library(stringi)
cs <- cumsum(
rle(stri_split_boundaries(x, type = "character")[[1L]])$lengths
)
stri_sub(x, c(1L, head(cs + 1L, -1L)), cs)
# [1] "11111" "00000" "1111" "0000" "111" "000"
或者，您可以使用length
stri_sub()参数
rl <- rle(stri_split_boundaries(x, type = "character")[[1L]])
with(rl, {
stri_sub(x, c(1L, head(cumsum(lengths) + 1L, -1L)), length = lengths)
})
# [1] "11111" "00000" "1111" "0000" "111" "000"
针对效率进行了更新：在意识到base::strsplit()比stringi::stri_split_boundaries()更快之后，这是我之前使用基本功能的答案的更高效版本。
set.seed(24)
x3 <- stri_rand_strings(1L, 1e6L)
system.time({
cs <- cumsum(rle(strsplit(x3, NULL)[[1L]])[[1L]])
substring(x3, c(1L, head(cs + 1L, -1L)), cs)
})
# user system elapsed
# 0.686 0.012 0.697
答案 5 :(得分：11)
另一种方法，使用mapply：
x="111110000011110000111000"
with(rle(strsplit(x,'')[[1]]),
mapply(function(u,v) paste0(rep(v,u), collapse=''), lengths, values))
#[1] "11111" "00000" "1111" "0000" "111" "000"
答案 6 :(得分：8)
它并不是OP所寻求的（简洁的R代码），但我想在Rcpp中尝试一下，结果相对简单，大约比其快5倍最快的基于R的答案。
library(Rcpp)
cppFunction(
'std::vector<std::string> split_str_cpp(std::string x) {
std::vector<std::string> parts;
int start = 0;
for(int i = 1; i <= x.length(); i++) {
if(x[i] != x[i-1]) {
parts.push_back(x.substr(start, i-start));
start = i;
}
}
return parts;
}')
并测试这些
str1 <- "111110000011110000111000"
x1 <- "1111100000222000333300011110000111000"
x2 <- "aaaaabbcccccccbbbad1111100000222aaabbccd11DaaBB"
提供以下输出
> split_str_cpp(str1)
[1] "11111" "00000" "1111" "0000" "111" "000"
> split_str_cpp(x1)
[1] "11111" "00000" "222" "000" "3333" "000" "1111" "0000" "111" "000"
> split_str_cpp(x2)
[1] "aaaaa" "bb" "ccccccc" "bbb" "a" "d" "11111" "00000" "222" "aaa" "bb" "cc" "d" "11"
[15] "D" "aa" "BB"
基准测试显示它比R解决方案快5-10倍。
akrun <- function(str1) strsplit(str1, '(?<=1)(?=0)|(?<=0)(?=1)', perl=TRUE)[[1]]
richard1 <- function(x3){
cs <- cumsum(
rle(stri_split_boundaries(x3, type = "character")[[1L]])$lengths
)
stri_sub(x3, c(1, head(cs + 1, -1)), cs)
}
richard2 <- function(x3) {
cs <- cumsum(rle(strsplit(x3, NULL)[[1L]])[[1L]])
stri_sub(x3, c(1, head(cs + 1, -1)), cs)
}
library(microbenchmark)
library(stringi)
set.seed(24)
x3 <- stri_rand_strings(1, 1e6)
microbenchmark(split_str_cpp(x3), akrun(x3), richard1(x3), richard2(x3), unit = 'relative', times=20L)
比较：
Unit: relative
expr min lq mean median uq max neval
split_str_cpp(x3) 1.000000 1.000000 1.000000 1.000000 1.000000 1.000000 20
akrun(x3) 9.675613 8.952997 8.241750 8.689001 8.403634 4.423134 20
richard1(x3) 5.355620 5.226103 5.483171 5.947053 5.982943 3.379446 20
richard2(x3) 4.842398 4.756086 5.046077 5.389570 5.389193 3.669680 20
答案 7 :(得分：2)
简单for循环解决方案
x="aaaaabbcccccccbbbad1111100000222aaabbccd11DaaBB"
res_vector=substr(x,1,1)
for (i in 2:nchar(x)) {
tmp=substr(x,i,i)
if (tmp==substr(x,i-1,i-1)) {
res_vector[length(res_vector)]=paste0(res_vector[length(res_vector)],tmp)
} else {
res_vector[length(res_vector)+1]=tmp
}
}
res_vector
#[1] "aaaaa" "bb" "ccccccc" "bbb" "a" "d" "11111" "00000" "222" "aaa" "bb" "cc" "d" "11" "D" "aa" "BB"
或者使用预先分配的结果向量可能稍快一点
x="aaaaabbcccccccbbbad1111100000222aaabbccd11DaaBB"
res_vector=rep(NA_character_,nchar(x))
res_vector[1]=substr(x,1,1)
counter=1
old_tmp=''
for (i in 2:nchar(x)) {
tmp=substr(x,i,i)
if (tmp==old_tmp) {
res_vector[counter]=paste0(res_vector[counter],tmp)
} else {
res_vector[counter+1]=tmp
counter=counter+1
}
old_tmp=tmp
}
res_vector[!is.na(res_vector)]
答案 8 :(得分：1)
这个怎么样：
s <- "111110000011110000111000"
spl <- strsplit(s,"10|01")[[1]]
l <- length(spl)
sapply(1:l, function(i) paste0(spl[i],i%%2,ifelse(i==1 | i==l, "",i%%2)))
# [1] "11111" "00000" "1111" "0000" "111" "000"