使用丰富的字符串魔法从字符串中提取数字

I want to extract a list of ID of a string pattern in the following: {(2),(4),(5),(100)}

我想在以下内容中提取字符串模式的ID列表:{(2),(4),(5),(100)}

Note: no leading or trailing spaces.

注意:没有前导或尾随空格。

The List can have up to 1000 IDs.

列表最多可包含1000个ID。

I want to use rich string pattern matching to do this. But I tried for 20 minutes with frustration.

我想使用丰富的字符串模式匹配来做到这一点。但我沮丧地试了20分钟。

Could anyone help me to come up with the correct pattern? Much appreciated!

任何人都可以帮我提出正确的模式吗?非常感激!

3 个解决方案

#1

Here's brute force string manipulation.

这是蛮力字符串操作。

scala> "{(2),(4),(5),(100)}".replaceAll("\\(", "").replaceAll("\\)", "").replaceAll("\\{","").replaceAll("\\}","").split(",")

res0: Array[java.lang.String] = Array(2, 4, 5, 100)

Here's a regex as @pst noted in the comments. If you don't want the parentheses change the regular expression to """\d+""".r.

这是@pst在评论中提到的正则表达式。如果您不希望括号将正则表达式更改为“”“\ d +”“”。r。

val num = """\(\d+\)""".r
"{(2),(4),(5),(100)}" findAllIn res0
res33: scala.util.matching.Regex.MatchIterator = non-empty iterator

scala> res33.toList
res34: List[String] = List((2), (4), (5), (100))

#2

"{(2),(4),(5),(100)}".split ("[^0-9]").filter(_.length > 0).map (_.toInt)

Split, where char is not part of a number, and only convert non-empty results.

拆分,其中char不是数字的一部分,只转换非空结果。

Might be modified to include dots or minus signs.

可能会被修改为包括点或减号。

#3

Use Extractor object:

使用Extractor对象:

object MyList {
  def apply(l: List[String]): String =
    if (l != Nil) "{(" + l.mkString("),(") + ")}"
    else "{}"
  def unapply(str: String): Some[List[String]] = 
    Some(
      if (str.indexOf("(") > 0) 
        str.substring(str.indexOf("(") + 1, str.lastIndexOf(")")) split 
          "\\p{Space}*\\)\\p{Space}*,\\p{Space}*\\(\\p{Space}*" toList
      else Nil
    )
}

// test
"{(1),(2)}" match { case MyList(l) => l }
// res23: List[String] = List(1, 2)

#1