拆分数组子阵列基于价值

本文介绍了拆分数组子阵列基于价值的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

我一直在寻找一个阵列相当于字符串＃分裂在Ruby中的核心，并惊讶地发现，它并不存在。难道还有比以下基于值的阵列分成多个子阵列更优雅的方式？

I was looking for an Array equivalent String#split in Ruby Core, and was surprised to find that it did not exist. Is there a more elegant way than the following to split an array into sub-arrays based on a value?

class Array def split( split_on=nil ) inject([[]]) do |a,v| a.tap{ if block_given? ? yield(v) : v==split_on a << [] else a.last << v end } end.tap{ |a| a.pop if a.last.empty? } end end p (1..9 ).to_a.split{ |i| i%3==0 }, (1..10).to_a.split{ |i| i%3==0 } #=> [[1, 2], [4, 5], [7, 8]] #=> [[1, 2], [4, 5], [7, 8], [10]]

修改：对于那些有兴趣，这引发了此请求的现实世界的问题，可以在this回答，在这里我用@ FD的回答下面的实现。

Edit: For those interested, the "real-world" problem which sparked this request can be seen in this answer, where I've used @fd's answer below for the implementation.

推荐答案

我试着打高尔夫球了一点，仍然没有一个单一的方法，但：

I tried golfing it a bit, still not a single method though:

(1..9).chunk{|i|i%3==0}.reject{|sep,ans| sep}.map{|sep,ans| ans}

或更快：

(1..9).chunk{|i|i%3==0 || nil}.map{|sep,ans| sep&&ans}pact

此外，可枚举＃块似乎是红宝石1.9+，但它是非常接近你想要什么。

Also, Enumerable#chunk seems to be Ruby 1.9+, but it is very close to what you want.

例如，原始输出将是：

(1..9).chunk{ |i|i%3==0 }.to_a => [[false, [1, 2]], [true, [3]], [false, [4, 5]], [true, [6]], [false, [7, 8]], [true, [9]]]

（即 to_a 是使IRB打印一件好事，因为块给你一个枚举，而不是一个数组）

(The to_a is to make irb print something nice, since chunk gives you an enumerator rather than an Array)

修改：注意上面的优雅的解决方案比最快的实现2-3倍慢：

Edit: Note that the above elegant solutions are 2-3x slower than the fastest implementation:

module Enumerable def split_by result = [a=[]] each{ |o| yield(o) ? (result << a=[]) : (a << o) } result.pop if a.empty? result end end