使用gsub或sub函数只获取字符串的一部分？

      Col
WBU-ARGU*06:03:04
WBU-ARDU*08:01:01
WBU-ARFU*11:03:05
WBU-ARFU*03:456

I have a column which has 75 rows of variables such as the col above. I am not quite sure how to use gsub or sub in order to get up until the integers after the first colon.

我有一列有75行变量,如上面的col。我不太确定如何使用gsub或sub来起床直到第一次冒号后的整数。

Expected output:

      Col
WBU-ARGU*06:03
WBU-ARDU*08:01
WBU-ARFU*11:03
WBU-ARFU*03:456

I tried this but it doesn't seem to work:

我试过这个,但它似乎不起作用:

gsub("*..:","", df$col)

3 个解决方案

#1

Following may help you here too.

以下也可以帮到你。

sub("([^:]*):([^:]*).*","\\1:\\2",df$dat)

Output will be as follows.

输出如下。

> sub("([^:]*):([^:]*).*","\\1:\\2",df$dat)
[1] "WBU-ARGU*06:03"   "WBU-ARDU*08:01"   "WBU-ARFU*11:03"   "WBU-ARFU*03:456b"

Where Input for data frame is as follows.

数据帧的输入如下。

dat <- c("WBU-ARGU*06:03:04","WBU-ARDU*08:01:01","WBU-ARFU*11:03:05","WBU-ARFU*03:456b")
df <- data.frame(dat)

Explanation: Following is only for explanation purposes.

说明:以下仅用于说明目的。

sub("      ##using sub for global subtitution function of R here.
([^:]*)     ##By mentioning () we are keeping the matched values from vector's element into 1st place of memory(which we could use later), which is till next colon comes it will match everything.
:           ##Mentioning letter colon(:) here.
([^:]*)     ##By mentioning () making 2nd place in memory for matched values in vector's values which is till next colon comes it will match everything.
.*"         ##Mentioning .* to match everything else now after 2nd colon comes in value.
,"\\1:\\2"  ##Now mentioning the values of memory holds with whom we want to substitute the element values \\1 means 1st memory place \\2 is second memory place's value.
,df$dat)    ##Mentioning df$dat dataframe's dat value.

#2

You may use

你可以用

df$col <- sub("(\\d:\\d+):\\d+$", "\\1", df$col)

See the regex demo

请参阅正则表达式演示

Details

• (\\d:\\d+) - Capturing group 1 (its value will be accessible via \1 in the replacement pattern): a digit, a colon and 1+ digits.

(\\ d:\\ d +) - 捕获组1(其值可通过替换模式中的\ 1访问):数字,冒号和1+位数。

• : - a colon

: - 冒号

• \\d+ - 1+ digits

\\ d + - 1+位数

• $ - end of string.

$ - 结束字符串。

R Demo:

col <- c("WBU-ARGU*06:03:04","WBU-ARDU*08:01:01","WBU-ARFU*11:03:05","WBU-ARFU*03:456")
sub("(\\d:\\d+):\\d+$", "\\1", col)
## => [1] "WBU-ARGU*06:03"  "WBU-ARDU*08:01"  "WBU-ARFU*11:03"  "WBU-ARFU*03:456"

Alternative approach:

df$col <- sub("^(.*?:\\d+).*", "\\1", df$col)

See the regex demo

请参阅正则表达式演示

Here,

• ^ - start of string

^ - 字符串的开头

• (.*?:\\d+) - Group 1: any 0+ chars, as few as possible (due to the lazy *? quantifier), then : and 1+ digits

(。*?:\\ d +) - 第1组:任何0+字符,尽可能少(由于懒惰的*?量词),然后:和1+位数

• .* - the rest of the string.

。* - 字符串的其余部分。

However, it should be used with the PCRE regex engine, pass perl=TRUE:

但是,它应该与PCRE正则表达式引擎一起使用,传递perl = TRUE:

col <- c("WBU-ARGU*06:03:04","WBU-ARDU*08:01:01","WBU-ARFU*11:03:05","WBU-ARFU*03:456")
sub("^(.*?:\\d+).*", "\\1", col, perl=TRUE)
## => [1] "WBU-ARGU*06:03"  "WBU-ARDU*08:01"  "WBU-ARFU*11:03"  "WBU-ARFU*03:456"

See the R online demo.

请参阅R在线演示。

#3

sub("(\\d+:\\d+):\\d+$", "\\1", df$Col)
[1] "WBU-ARGU*06:03"  "WBU-ARDU*08:01"  "WBU-ARFU*11:03"  "WBU-ARFU*03:456"

Alternatively match what you want (instead of subbing out what you don't want) with stringi:

或者用stringi匹配你想要的东西(而不是你不想要的东西):

stringi::stri_extract_first(df$Col, regex = "[A-Z-\\*]+\\d+:\\d+")

Slightly more concise stringr:

更简洁的字符串:

stringr::str_extract(df$Col, "[A-Z-\\*]+\\d+:\\d+")
# or
stringr::str_extract(df$Col, "[\\w-*]+\\d+:\\d+")