Regex R:gsub空格之间的单词

我有一个文本如下所示：

a <- "233,236,241 solitude ΔE=1.9"

我尝试了两种方法：

a1 <- strsplit(a,' ',fixed=TRUE)[[1]][2]
a2 <- sapply(strsplit(a, " ", fixed=TRUE), "[", 2)

正确的方法是什么？

试试这个：

gsub("\\s.+$","",gsub("^.+[[:digit:]]\\s","",a))

---

这里有一种使用捕获类（括号内的模式）和字符类（方括号内的模式）的方法

注释第一个捕获类模式：

"(^[^ ]*[ ])([^ ]*)([ ].*$)" , "\\2", a)
         \finds first space
       \ an arbitrary number of times
    \\ inside a character class an '^' as the first character ...
       signals negation of character class. This one with only the space character in it.
  \----- '^' marks the beginning of a character value

"(^[^ ]*[ ])([^ ]*)([ ].*$)" , "\\2", a)
                 \ an arbitrary number of times
              \\negation of character class with only the space character in it.

第二类捕获模式：

"(^[^ ]*[ ])([^ ]*)([ ].*$)" , "\\2", a)
         \finds first space
       \ an arbitrary number of times
    \\ inside a character class an '^' as the first character ...
       signals negation of character class. This one with only the space character in it.
  \----- '^' marks the beginning of a character value

"(^[^ ]*[ ])([^ ]*)([ ].*$)" , "\\2", a)
                 \ an arbitrary number of times
              \\negation of character class with only the space character in it.

第三类：

"(^[^ ]*[ ])([^ ]*)([ ].*$)" , "\\2", a)
                     \ the second space
                        \\anything after second space to end.

---

replacement

中的

“\\\”

条目引用捕获类在

模式中出现的顺序匹配。
strsplit（a，，）[[1]][2]
适用于
aOr，或者可能只是
sub（.*\\s+（.\\\s+.*，“\\1”，a）
我得到“孤独”对于你的每一种方法……两个
gsub
调用有什么意义？第一个将文本剪切到“孤独”之前的空格，第二个将文本从“孤独”之后的空格剪切到结尾。
"(^[^ ]*[ ])([^ ]*)([ ].*$)" , "\\2", a)
                     \ the second space
                        \\anything after second space to end.