Ruby中的Array.prototype.splice [英] Array.prototype.splice in Ruby

问题描述

一个朋友问我，使用Ruby的最佳方式来实现Ruby中JavaScript的 splice 方法的效果.这意味着在数组本身或副本上没有迭代.

A friend asked me the Ruby best and performant way to achieve the effect of JavaScript's splice method in Ruby. This means no iteration on the Array itself or copies.

从索引开始处开始，删除长度项目并(可选)插入元素.最后将已删除项目返回到数组中."<<这具有误导性，请参见下面的JS示例.

"begin at index start, remove length items and (optionally) insert elements. Finally return the removed items in an array." << This is misleading, see the JS example below.

http://www.mennovanslooten.nl/blog/post/41

没有可选替代项的快速hack:

Quick hack that doesn't have the optional substitution:

from_index     = 2
for_elements   = 2
sostitute_with = :test
initial_array  = [:a, :c, :h, :g, :t, :m]
# expected result: [:a, :c, :test, :t, :m]
initial_array[0..from_index-1] + [sostitute_with] + initial_array[from_index + for_elements..-1]

你是什么人?一行更好.

What's yours? One line is better.

更新:

// JavaScript
var a = ['a', 'c', 'h', 'g', 't', 'm'];
var b = a.splice(2, 2, 'test'); 
> b is now ["h", "g"]
> a is now ["a", "c", "test", "t", "m"]

我需要生成的'a'数组，而不是'b'.

I need the resulting 'a' Array, not 'b'.

推荐答案

使用 Array#[] = .

a = [1, 2, 3, 4, 5, 6]
a[2..4] = [:foo, :bar, :baz, :wibble]
a # => [1, 2, :foo, :bar, :baz, :wibble, 6]

# It also supports start/length instead of a range:
a[0, 3] = [:a, :b]
a # => [:a, :b, :bar, :baz, :wibble, 6]

对于返回已删除的元素， [] = 不会执行此操作...您可以编写自己的帮助方法来实现:

As for returning the removed elements, []= doesn't do that... You could write your own helper method to do it:

class Array
  def splice(start, len, *replace)
    ret = self[start, len]
    self[start, len] = replace
    ret
  end
end

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