我希望将字符串拆分为非重叠段，其中段的端点是点域内的数字。我可以使用下面的代码执行此操作。但是，此代码似乎过于复杂，涉及嵌套for-loops。是否有更简单的方法，最好在regex基础中使用R？
以下是一个示例和desired.result。
my.data <- read.table(text = '
my.string cov1 cov2
11....... 1 A
1.1.2.1.1 2 B
1234..... 3 C
1...2...3 4 C
..3..4... 5 D
', header = TRUE, stringsAsFactors = FALSE)
desired.result <- read.table(text = '
my.string cov1 cov2
11....... 1 A
1.1...... 2 B
..1.2.... 2 B
....2.1.. 2 B
......1.1 2 B
12....... 3 C
.23...... 3 C
..34..... 3 C
1...2.... 4 C
....2...3 4 C
..3..4... 5 D
', header = TRUE, stringsAsFactors = FALSE, na.strings = 'NA')
new.data <- data.frame(do.call(rbind, strsplit(my.data$my.string,'')), stringsAsFactors = FALSE)
n.segments <- rowSums(!(new.data[1:ncol(new.data)] == '.')) - 1
my.end.points <- do.call(rbind, gregexpr("[0-9]", my.data$my.string, perl=TRUE))
my.end.point.char <- do.call(rbind, strsplit(my.data$my.string, ""))
my.end.point.char <- t(apply(my.end.point.char, 1, as.numeric))
new.strings <- matrix('.', nrow = sum(n.segments), ncol = max(nchar(my.data$my.string)))
new.cov <- as.data.frame(matrix(NA, nrow = sum(n.segments), ncol = (ncol(my.data) - 1)))
m <- 1
for(i in 1:nrow(new.data)) {
for(j in 1:n.segments[i]) {
for(k in 1:ncol(new.strings)) {
new.strings[m, my.end.points[i, j ] ] <- my.end.point.char[i, my.end.points[i, j ]]
new.strings[m, my.end.points[i, (j+1)] ] <- my.end.point.char[i, my.end.points[i,(j+1)]]
new.cov[m,] <- my.data[i, c(2:ncol(my.data))]
}
m <- m + 1
}
}
my.result <- data.frame(my.string = apply(new.strings, 1, function(x) paste0(x, collapse = '')), stringsAsFactors = FALSE)
my.result <- data.frame(my.result, new.cov)
colnames(my.result) <- names(my.data)
all.equal(desired.result, my.result)
# [1] TRUE
答案 0 :(得分：2)
my.data <- read.table(text = '
my.string cov1 cov2
11....... 1 A
1.1.2.1.1 2 B
1234..... 3 C
1...2...3 4 C
..3..4... 5 D
', header = TRUE, stringsAsFactors = FALSE)
f <- function(x, m) {
if (nchar(gsub('.', '', x, fixed = TRUE)) < 2L) return(x)
y <- gsub('.', '\\.', x)
cs <- attr(m, "capture.start")
cl <- attr(m, "capture.length")
Vectorize(`substr<-`)(y, cs, cl + cs - 1, Vectorize(substr)(x, cs, cl + cs - 1))
}
m <- gregexpr('(?=([0-9][.]*[0-9]))', my.data$my.string, perl = TRUE)
strs <- Map(f, my.data$my.string, m)
tmp <- `rownames<-`(my.data[rep(1:nrow(my.data), lengths(strs)), ], NULL)
tmp$my.string <- unlist(strs)
# my.string cov1 cov2
# 1 11....... 1 A
# 2 1.1...... 2 B
# 3 ..1.2.... 2 B
# 4 ....2.1.. 2 B
# 5 ......1.1 2 B
# 6 12....... 3 C
# 7 .23...... 3 C
# 8 ..34..... 3 C
# 9 1...2.... 4 C
# 10 ....2...3 4 C
# 11 ..3..4... 5 D
identical(tmp, desired.result)
# [1] TRUE
答案 1 :(得分：2)
w <- nchar(my.data$my.string[1L]);
dps <- character(w+1L); dps[1L] <- ''; for (i in seq_len(w)) dps[i+1L] <- paste0(dps[i],'.');
x <- Map(my.data$my.string,gregexpr('[^.]',my.data$my.string),f=function(s,g)
if (length(g)<3L) s else sapply(seq_len(length(g)-1L),function(gi)
paste0(dps[g[gi]],substr(s,g[gi],g[gi+1L]),dps[w-g[gi+1L]+1L])
)
);
res <- transform(my.data[rep(seq_len(nrow(my.data)),sapply(x,length)),],my.string=unlist(x));
res;
## my.string cov1 cov2
## 1 11....... 1 A
## 2 1.1...... 2 B
## 2.1 ..1.2.... 2 B
## 2.2 ....2.1.. 2 B
## 2.3 ......1.1 2 B
## 3 12....... 3 C
## 3.1 .23...... 3 C
## 3.2 ..34..... 3 C
## 4 1...2.... 4 C
## 4.1 ....2...3 4 C
## 5 ..3..4... 5 D
注意：如果你有足够的R版本，你可以用lengths(x)替换sapply(x,length)篇。
library(microbenchmark);
bgoldst <- function(my.data) { w <- nchar(my.data$my.string[1L]); dps <- character(w+1L); dps[1L] <- ''; for (i in seq_len(w)) dps[i+1L] <- paste0(dps[i],'.'); x <- Map(my.data$my.string,gregexpr('[^.]',my.data$my.string),f=function(s,g) if (length(g)<3L) s else sapply(seq_len(length(g)-1L),function(gi) paste0(dps[g[gi]],substr(s,g[gi],g[gi+1L]),dps[w-g[gi+1L]+1L]))); transform(my.data[rep(seq_len(nrow(my.data)),sapply(x,length)),],my.string=unlist(x)); };
rawr <- function(my.data) { f <- function(x, m) { y <- gsub('.', '\\.', x); cs <- attr(m, "capture.start"); cl <- attr(m, "capture.length"); Vectorize(`substr<-`)(y, cs, cl + cs - 1, Vectorize(substr)(x, cs, cl + cs - 1)); }; m <- gregexpr('(?=([0-9][.]*[0-9]))', my.data$my.string, perl = TRUE); strs <- Map(f, my.data$my.string, m); tmp <- `rownames<-`(my.data[rep(1:nrow(my.data), sapply(strs,length)), ], NULL); tmp$my.string <- unlist(strs); tmp; };
carroll <- function(my.data) { strings <- sapply(my.data$my.string, function(x) { stri_match_all_regex(x, "(?=([0-9]{1}\\.*[0-9]{1}))")[[1]][,2]; }); strpos <- lapply(1:length(strings), function(x) { y <- {nchar(sub(perl=T,'^\\.*\\K.*','',my.data$my.string[x]))+c(0, cumsum(nchar(strings[[x]])-1))}; return(y[-length(y)]); }); w <- nchar(my.data$my.string[1L]); output.result <- data.frame(my.string = cbind(unlist(sapply(1:length(strings), function(y) { cbind(sapply(1:length(strings[[y]]), function(x) { leftstr <- paste0(paste0(rep(".", strpos[[y]][[x]]), collapse=""), strings[[y]][x]); rightstr <- paste0(rep(".", w-nchar(leftstr)), collapse=""); paste0(leftstr, rightstr, collapse=""); })); }))), my.data[unlist(sapply(1:length(strings), function(x) { rep(x, sapply(strings, length)[x]); })), c(2,3)], stringsAsFactors=FALSE); row.names(output.result) <- NULL; output.result; };
## OP's sample input
my.data <- read.table(text = '
my.string cov1 cov2
11....... 1 A
1.1.2.1.1 2 B
1234..... 3 C
1...2...3 4 C
..3..4... 5 D
', header = TRUE, stringsAsFactors = FALSE);
ex <- bgoldst(my.data);
all.equal(ex,rawr(my.data),check.attributes=F);
## [1] TRUE
all.equal(ex,carroll(my.data),check.attributes=F);
## [1] TRUE
microbenchmark(bgoldst(my.data),rawr(my.data),carroll(my.data));
## Unit: microseconds
## expr min lq mean median uq max neval
## bgoldst(my.data) 422.094 451.816 483.5305 476.6195 503.775 801.421 100
## rawr(my.data) 1096.502 1160.863 1277.7457 1236.7720 1298.996 3092.785 100
## carroll(my.data) 1130.287 1176.900 1224.6911 1213.2515 1247.249 1525.437 100
## scale test
set.seed(1L);
NR <- 1e4; NS <- 30L; probDot <- 3/4;
x <- c('.',0:9); probs <- c(probDot,rep((1-probDot)/10,10L)); my.data <- data.frame(my.string=do.call(paste0,as.data.frame(replicate(NS,sample(x,NR,T,probs)))),cov1=sample(seq_len(NR)),cov2=sample(make.unique(rep(LETTERS,len=NR))),stringsAsFactors=F);
repeat { w <- which(sapply(gregexpr('[^.]',my.data$my.string),length)==1L); if (length(w)==0L) break; my.data$my.string[w] <- do.call(paste0,as.data.frame(replicate(NS,sample(x,length(w),T,probs)))); }; ## prevent single-digit strings, which rawr and carroll solutions don't support
ex <- bgoldst(my.data);
all.equal(ex,rawr(my.data),check.attributes=F);
## [1] TRUE
all.equal(ex,carroll(my.data),check.attributes=F);
## [1] TRUE
microbenchmark(bgoldst(my.data),rawr(my.data),carroll(my.data),times=1L);
## Unit: milliseconds
## expr min lq mean median uq max neval
## bgoldst(my.data) 904.887 904.887 904.887 904.887 904.887 904.887 1
## rawr(my.data) 2736.462 2736.462 2736.462 2736.462 2736.462 2736.462 1
## carroll(my.data) 108575.001 108575.001 108575.001 108575.001 108575.001 108575.001 1
答案 2 :(得分：1)
这是一个选项。不干净，但问题都没有。
library(stringi)
## isolate the strings, allowing overlap via positive lookaheads
strings <- sapply(my.data$my.string, function(x) {
stri_match_all_regex(x, "(?=([0-9]{1}\\.*[0-9]{1}))")[[1]][,2]
})
确定每组开始时的偏移量。
## identify the . offsets
strpos <- lapply(1:length(strings), function(x) {
y <- {nchar(sub(perl=T,'^\\.*\\K.*','',my.data$my.string[x]))+c(0, cumsum(nchar(strings[[x]])-1))}
return(y[-length(y)])
})
仅使用2个data.frame循环构建sapply。
## collate the results using sapply
w <- nchar(my.data$my.string[1L]);
output.result <- data.frame(
my.string = cbind(unlist(sapply(1:length(strings), function(y) {
cbind(sapply(1:length(strings[[y]]), function(x) {
leftstr <- paste0(paste0(rep(".", strpos[[y]][[x]]), collapse=""), strings[[y]][x])
rightstr <- paste0(rep(".", w-nchar(leftstr)), collapse="")
paste0(leftstr, rightstr, collapse="")
}))
}))),
my.data[unlist(sapply(1:length(strings), function(x) {
rep(x, sapply(strings, length)[x])
})), c(2,3)], stringsAsFactors=FALSE
)
row.names(output.result) <- NULL
output.result
my.string cov1 cov2
1 11....... 1 A
2 1.1...... 2 B
3 ..1.2.... 2 B
4 ....2.1.. 2 B
5 ......1.1 2 B
6 12....... 3 C
7 .23...... 3 C
8 ..34..... 3 C
9 1...2.... 4 C
10 ....2...3 4 C
11 ..3..4... 5 D
identical(desired.result, output.result)
[1] TRUE