Question

我想提取作为123456_Letters_CAPITAL_name_extension形式的foldername的一部分。

名称可以是LETTERS123，LETTERS_123_LETTERS或LETTERS_LETTERS_123。

目前，我正在使用unlist(strsplit(foldername , sep="_"))[4,length(unlist(strsplit(foldername , sep="_")))-1]

提取名称

但是如果_CAPITAL部分不存在我希望能够提取它（它将是3而不是4但我希望有一般的做法它）。

130615_Screen_II_SN_KB_3_lxb/，130615_Screen_II_AL343_lxb/，130615_Screen_II_HK_344_LM_lxb/是完整foldername的代表性示例

我试过但是无法想出任何会这样做的正则表达式。任何想法都会有所帮助。

Answer 1

这个怎么样：

^\d+_[a-zA-Z]+_(?:[A-Z]+_)?([A-Z]+\w+)_[^_]+$

该名称将在第1组中。

用于测试它的perl方法：

my $re = qr~^\d+_[a-zA-Z]+_(?:[A-Z]+_)?([A-Z]+\w+)_[^_]+$~;
while(<DATA>) {
    chomp;
    say $1 if /$re/;
}
__DATA__
130615_Screen_II_SN_KB_3_lxb/
130615_Screen_II_AL343_lxb/
130615_Screen_II_HK_344_LM_lxb/
130615_Screen_HK_344_LM_lxb/

<强>输出：

SN_KB_3
AL343
HK_344_LM
HK_344_LM

<强>解释

The regular expression:

^\d+_[a-zA-Z]+_(?:[A-Z]+_)?([A-Z]+\w+)_[^_]+$

matches as follows:

NODE                     EXPLANATION
----------------------------------------------------------------------
  ^                        the beginning of the string
----------------------------------------------------------------------
  \d+                      digits (0-9) (1 or more times (matching
                           the most amount possible))
----------------------------------------------------------------------
  _                        '_'
----------------------------------------------------------------------
  [a-zA-Z]+                any character of: 'a' to 'z', 'A' to 'Z'
                           (1 or more times (matching the most amount
                           possible))
----------------------------------------------------------------------
  _                        '_'
----------------------------------------------------------------------
  (?:                      group, but do not capture (optional
                           (matching the most amount possible)):
----------------------------------------------------------------------
    [A-Z]+                   any character of: 'A' to 'Z' (1 or more
                             times (matching the most amount
                             possible))
----------------------------------------------------------------------
    _                        '_'
----------------------------------------------------------------------
  )?                       end of grouping
----------------------------------------------------------------------
  (                        group and capture to \1:
----------------------------------------------------------------------
    [A-Z]+                   any character of: 'A' to 'Z' (1 or more
                             times (matching the most amount
                             possible))
----------------------------------------------------------------------
    \w+                      word characters (a-z, A-Z, 0-9, _) (1 or
                             more times (matching the most amount
                             possible))
----------------------------------------------------------------------
  )                        end of \1
----------------------------------------------------------------------
  _                        '_'
----------------------------------------------------------------------
  [^_]+                    any character except: '_' (1 or more times
                           (matching the most amount possible))
----------------------------------------------------------------------
  $                        before an optional \n, and the end of the
                           string
----------------------------------------------------------------------

Answer 2

让我们一步一步解决问题。以下x涵盖了所述的所有可能情况。（OP请确认）

x <- c("123456_Letters_CAPITAL_LETTERS123_extension/", "123456_Letters_CAPITAL_LETTERS_123_LETTERS_extension/", "123456_Letters_CAPITAL_LETTERS_LETTERS_123_extension/", 
    "123456_Letters_LETTERS123_extension/", "123456_Letters_LETTERS_123_LETTERS_extension/", "123456_Letters_LETTERS_LETTERS_123_extension/")
# Lets strip out the parts which we can first..
y <- gsub("[0-9]+_[A-Z]+[a-z]*_(.*)_[a-z]+/", "\\1", x)
y
## [1] "CAPITAL_LETTERS123"          "CAPITAL_LETTERS_123_LETTERS" "CAPITAL_LETTERS_LETTERS_123" "LETTERS123"                 
## [5] "LETTERS_123_LETTERS"         "LETTERS_LETTERS_123" 


#Now we can see that if you have 3 or 1 underscore 
#you need to strip out first part


ifelse(sapply(gregexpr("_", y), FUN = function(X) length(X[X != -1])) %in% c(1, 3), gsub("[A-Z]+_(.*)", "\\1", y), y)
## [1] "LETTERS123"          "LETTERS_123_LETTERS" "LETTERS_LETTERS_123" "LETTERS123"          "LETTERS_123_LETTERS"
## [6] "LETTERS_LETTERS_123"

从具有可变数量的分隔符的字符串中提取名称

2 个答案: