Question

我正在尝试拆分String并获取每个“单词”的所有起始索引。

例如对于这样的字符串：

"Rabbit jumped over a fence and this Rabbit loves carrots"

如何拆分它以获得每个单词的索引？：

0,7,14,19,21,27,31,36,43,49

Answer 1

你可以这样做

val str="Rabbit jumped over a fence and this Rabbit loves carrots"
val indexArr=str.split(" ").scanLeft(0)((prev,next)=>prev+next.length+1).dropRight(1)

示例输出：

ndexArr: Array[Int] = Array(0, 7, 14, 19, 21, 27, 31, 36, 43, 49)

Answer 2

这个解决方案即使分隔符的宽度不恒定也是有效的（不仅适用于长度为1的分隔符）。

使用前瞻和后瞻FOO的组合，而不是单个分隔符(?<=FOO)|(?=FOO)。
扫描标记和分隔符的长度，累积它们的长度以获得开始索引
扔掉每隔一个条目（分隔符）

在代码中：

val txt = "Rabbit jumped over a fence and this Rabbit loves carrots"
val pieces = txt.split("(?= )|(?<= )")
val startIndices = pieces.scanLeft(0){ (acc, w) => acc + w.size }
val tokensWithStartIndices = (pieces zip startIndices).grouped(2).map(_.head)

tokensWithStartIndices foreach println

结果：

(Rabbit,0)
(jumped,7)
(over,14)
(a,19)
(fence,21)
(and,27)
(this,31)
(Rabbit,36)
(loves,43)
(carrots,49)

以下是一些中间输出，因此您可以更好地了解每个步骤中发生的事情：

scala> val txt = "Rabbit jumped over a fence and this Rabbit loves carrots"
txt: String = Rabbit jumped over a fence and this Rabbit loves carrots

scala> val pieces = txt.split("(?= )|(?<= )")
pieces: Array[String] = Array(Rabbit, " ", jumped, " ", over, " ", a, " ", fence, " ", and, " ", this, " ", Rabbit, " ", loves, " ", carrots)

scala> val startIndices = pieces.scanLeft(0){ (acc, w) => acc + w.size }
startIndices: Array[Int] = Array(0, 6, 7, 13, 14, 18, 19, 20, 21, 26, 27, 30, 31, 35, 36, 42, 43, 48, 49, 56)

Answer 3

即使行以空格开头，或者您有多个空格或制表符分隔某些单词，这也应该是准确的。它遍历String，注意从任何空白字符（空格，制表符，换行符等）到非空格字符的转换。

val txt = "Rabbit jumped over a fence and this Rabbit loves carrots"

txt.zipWithIndex.foldLeft((Seq.empty[Int],true)){case ((s,b),(c,i)) =>
    if (c.isWhitespace) (s,true)
    else if (b) (s :+ i, false)
    else (s,false)
}._1

Answer 4

以下是另一种混合zipWithIndex和collect：

0 :: str.zipWithIndex.collect { case (' ', i) => i + 1 }.toList

预先填写第一个单词的索引并不是非常优雅，它只允许使用长度为1的分隔符;但它的体积极小且易读。

拆分字符串并获取每个段的起始索引

4 个答案: