在 Scala 中，如何在不知道长度的情况下获取从第 n 个元素到列表末尾的列表切片? [英] In Scala, how to get a slice of a list from nth element to the end of the list without knowing the length?

问题描述

我正在寻找一种优雅的方法来从元素 n 开始获取列表的一部分，而无需指定列表的长度.假设我们有一个多行字符串，我将其拆分为多行，然后想要获取从第 3 行开始的所有行的列表:

I'm looking for an elegant way to get a slice of a list from element n onwards without having to specify the length of the list. Lets say we have a multiline string which I split into lines and then want to get a list of all lines from line 3 onwards:

string.split("\n").slice(3,X)       // But I don't know what X is...

这里我真正感兴趣的是是否有办法获取由 split 调用返回的列表的引用，以便可以将其长度替换为 X 在 slice 调用时，有点像一个花哨的 _(在这种情况下它会读为 slice(3,_.length)) ?在 python 中，不需要指定切片的最后一个元素.

What I'm really interested in here is whether there's a way to get hold of a reference of the list returned by the split call so that its length can be substituted into X at the time of the slice call, kind of like a fancy _ (in which case it would read as slice(3,_.length)) ? In python one doesn't need to specify the last element of the slice.

当然我可以通过在拆分后使用临时变量来解决这个问题，或者创建一个语法很好的辅助函数，但我只是很好奇.

Of course I could solve this by using a temp variable after the split, or creating a helper function with a nice syntax, but I'm just curious.

推荐答案

只需删除不需要的前 n 个元素:

Just drop first n elements you don't need:

List(1,2,3,4).drop(2)
res0: List[Int] = List(3, 4)

或者在你的情况下:

string.split("\n").drop(2)

还有成对的方法 .take(n) 做相反的事情，你可以把它想象成 .slice(0,n).

There is also paired method .take(n) that do the opposite thing, you can think of it as .slice(0,n).

如果您需要这两个部分，请使用 .splitAt:

In case you need both parts, use .splitAt:

val (left, right) = List(1,2,3,4).splitAt(2)
left: List[Int] = List(1, 2)
right: List[Int] = List(3, 4)

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