在前两个冒号上拆分字符串

I would like to split a column of strings on the first two colons, but not on any subsequent colons:

我想在前两个冒号上分割一列字符串,但不在任何后续冒号上:

my.data <- read.table(text='

my.string    some.data
123:34:56:78   -100
87:65:43:21    -200
a4:b6:c8888    -300
11:bbbb:ccccc  -400
uu:vv:ww:xx    -500', header = TRUE)

desired.result <- read.table(text='

my.string1  my.string2  my.string3  some.data
123         34          56:78         -100
87          65          43:21         -200
a4          b6          c8888         -300
11          bbbb        ccccc         -400
uu          vv          ww:xx         -500', header = TRUE)

I have searched extensively and the following question is the closest to my current dilemma:

我进行了广泛的搜索,以下问题最接近我目前的困境:

Split on first comma in string

在字符串中的第一个逗号上拆分

Thank you for any suggestions. I prefer to use base R.

谢谢你的任何建议。我更喜欢使用底座R.

EDIT:

The number of characters before the first colon is not always two and the number of characters between the first two colons is not always two. So, I edited the example to reflect this.

第一个冒号前的字符数不总是两个,前两个冒号之间的字符数不总是两个。所以,我编辑了这个例子以反映这一点。

5 个解决方案

#1

In base R:

在基地R:

> my.data <- read.table(text='
+ 
+ my.string    some.data
+ 123:34:56:78   -100
+ 87:65:43:21    -200
+ a4:b6:c8888    -300
+ 11:bbbb:ccccc  -400
+ uu:vv:ww:xx    -500', header = TRUE,stringsAsFactors=FALSE)
> m <- regexec ("^([^:]+):([^:]+):(.*)$",my.data$my.string)
> my.data$my.string1 <- unlist(lapply(regmatches(my.data$my.string,m),'[',c(2)))
> my.data$my.string2 <- unlist(lapply(regmatches(my.data$my.string,m),'[',c(3)))
> my.data$my.string3 <- unlist(lapply(regmatches(my.data$my.string,m),'[',c(4)))
> my.data
      my.string some.data my.string1 my.string2 my.string3
1  123:34:56:78      -100        123         34      56:78
2   87:65:43:21      -200         87         65      43:21
3   a4:b6:c8888      -300         a4         b6      c8888
4 11:bbbb:ccccc      -400         11       bbbb      ccccc
5   uu:vv:ww:xx      -500         uu         vv      ww:xx

You'll see I've used stringsAsFactors=FALSE to ensure that my.string can be processed as a vector of strings.

你会看到我使用stringsAsFactors = FALSE来确保my.string可以作为字符串向量处理。

#2

Using package stringr:

使用包stringr:

str_match(my.data$my.string, "(.+?):(.+?):(.*)")

     [,1]            [,2]  [,3]   [,4]   
[1,] "123:34:56:78"  "123" "34"   "56:78"
[2,] "87:65:43:21"   "87"  "65"   "43:21"
[3,] "a4:b6:c8888"   "a4"  "b6"   "c8888"
[4,] "11:bbbb:ccccc" "11"  "bbbb" "ccccc"
[5,] "uu:vv:ww:xx"   "uu"  "vv"   "ww:xx"

UPDATE: with latest example (above) and Hadley's comment solution:

更新:使用最新的示例(上图)和Hadley的评论解决方案:

str_split_fixed(my.data$my.string, ":", 3)
     [,1]  [,2]   [,3]   
[1,] "123" "34"   "56:78"
[2,] "87"  "65"   "43:21"
[3,] "a4"  "b6"   "c8888"
[4,] "11"  "bbbb" "ccccc"
[5,] "uu"  "vv"   "ww:xx"

#3

Replace first two ":" with ",", and then split on ",".

将前两个“:”替换为“,”,然后拆分为“,”。

x <- gsub("([[:alnum:]]*):([[:alnum:]]*):(.)","\\1,\\2,\\3","12:34:56:78")

strsplit(x,",")

Applying to data frame

应用于数据框架

a.list <- sapply(my.data$my.string, function(x) strsplit(gsub("([[:alnum:]]*):([[:alnum:]]*):(.)","\\1,\\2,\\3",x),","))
a.vect <- unlist(a.list)
a.df <- as.data.frame(matrix(a.vect,ncol=3,byrow=T), stringsAsFactors = F)
names(a.df) <- c("my.string1",  "my.string2",  "my.string3") 
a.df$some.data <- my.data$some.data
a.df

#4

I'm a bit late to the game. And my solution has much overlap with the earlier answers. Nevertheless, it might be useful someone:

我比赛有点晚了。我的解决方案与之前的答案有很多重叠。然而,有人可能会有用:

# Replace first two colons with commas.
new.string = gsub(pattern="(^[^:]+):([^:]+):(.+$)",
                  replacement="\\1,\\2,\\3",
                  x=my.data$my.string)

# Split on commas, producing a list.
split.data = strsplit(new.string, ",")

# Change list into matrix, then data.frame.
new.data = data.frame(do.call(rbind, split.data))
names(new.data) = paste("my.string", seq(ncol(new.data)), sep="")

my.data$my.string = NULL
my.data = cbind(new.data, my.data)
my.data

#   my.string1 my.string2 my.string3 some.data
# 1        123         34      56:78      -100
# 2         87         65      43:21      -200
# 3         a4         b6      c8888      -300
# 4         11       bbbb      ccccc      -400
# 5         uu         vv      ww:xx      -500

As noted by @topchef, commas (or some other character) must guaranteed to be absent from the data.

正如@topchef所指出的,必须保证数据中不存在逗号(或其他一些字符)。

Also, at least two colons must be present in each string, or else the pattern doesn't match anything and thus no splitting occurs.

此外,每个字符串中必须至少存在两个冒号,否则该模式与任何内容都不匹配,因此不会发生分裂。

#5

Couldn't you just strsplit(sub(":\s*", XX, x), XX) (like the example listed on your link to the other question) on the first colon, take the second half and split on the first colon again?

难道你不能只是strsplit(sub(“:\ s *”,XX,x),XX)(就像你在另一个问题的链接上列出的例子)在第一个冒号上,取下半部分并在第一个冒号上拆分再次冒号?

#1