Ruby / Rails循环中的魔术第一和最后一个指标？

Question

Ruby / Rails在谈到基本用品的糖时会做很多很酷的事情，我认为有一种非常常见的情况，我想知道是否有人做过帮手或类似的东西。

   a = Array.new(5, 1)

   a.each_with_index do |x, i|
     if i == 0
       print x+1
     elsif i == (a.length - 1)
       print x*10
     else
        print x
     end
   end

原谅丑陋，但这可以达到人们想要的......是否有一种红宝石的方式来对循环的第一个和最后一个做一些事情？

[编辑]我认为理想情况下这将是带参数的数组扩展（数组实例，所有元素函数，第一个元素函数，最后元素函数）......但我对其他想法持开放态度。

Answer 1

如果您愿意，可以抓取第一个和最后一个元素并以不同方式处理它们。

first = array.shift
last = array.pop
process_first_one
array.each { |x| process_middle_bits }
process_last_one

Answer 2

如果第一次和最后一次迭代的代码与其他迭代的代码没有任何共同之处，您还可以执行以下操作：

do_something( a.first )
a[1..-2].each do |x|
  do_something_else( x )
end
do_something_else_else( a.last )

如果不同的情况有一些共同的代码，你的方式很好。

Answer 3

如果你能这样做怎么办？

%w(a b c d).each.with_position do |e, position|
  p [e, position]    # => ["a", :first]
                     # => ["b", :middle]
                     # => ["c", :middle]
                     # => ["d", :last]
end

或这个？

%w(a, b, c, d).each_with_index.with_position do |(e, index), position|
  p [e, index, position]    # => ["a,", 0, :first]
                            # => ["b,", 1, :middle]
                            # => ["c,", 2, :middle]
                            # => ["d", 3, :last]
end

在MRI> = 1.8.7中，只需要这个猴子补丁：

class Enumerable::Enumerator

  def with_position(&block)
    state = :init
    e = nil
    begin
      e_last = e
      e = self.next
      case state
      when :init
        state = :first
      when :first
        block.call(e_last, :first)
        state = :middle
      when :middle
        block.call(e_last, :middle)
      end
    rescue StopIteration
      case state
      when :first
        block.call(e_last, :first)
      when :middle
        block.call(e_last, :last)
      end
      return
    end while true
  end

end

它有一个小的状态引擎，因为它必须向前看一次迭代。

诀窍是each，each_with_index，＆c。 如果没有阻止，则返回枚举器。 枚举器可以完成Enumerable所做的所有事情。 但对我们来说，重要的是我们可以通过猴子修补Enumerator来添加一种迭代方式，“包装”现有的迭代，无论它是什么。

Answer 4

或者一点点领域特定语言：

a = [1, 2, 3, 4]

FirstMiddleLast.iterate(a) do
  first do |e|
    p [e, 'first']
  end
  middle do |e|
    p [e, 'middle']
  end
  last do |e|
    p [e, 'last']
  end
end

# => [1, "first"]
# => [2, "middle"]
# => [3, "middle"]
# => [4, "last"]

和它的代码：

class FirstMiddleLast

  def self.iterate(array, &block)
    fml = FirstMiddleLast.new(array)
    fml.instance_eval(&block)
    fml.iterate
  end

  attr_reader :first, :middle, :last

  def initialize(array)
    @array = array
  end

  def first(&block)
    @first = block
  end

  def middle(&block)
    @middle = block
  end

  def last(&block)
    @last = block
  end

  def iterate
    @first.call(@array.first) unless @array.empty?
    if @array.size > 1
      @array[1..-2].each do |e|
        @middle.call(e)
      end
      @last.call(@array.last)
    end
  end

end

我开始思考，“如果只有你可以将多个块传递给Ruby函数，那么你可以对这个问题有一个灵活而简单的解决方案。” 然后我意识到DSL的玩法几乎就像传递多个块一样。

Answer 5

正如许多人所指出的， each_with_index似乎是关键。 我有这个我喜欢的代码块。

array.each_with_index do |item,index|
  if index == 0
    # first item
  elsif index == array.length-1
    # last item
  else
    # middle items
  end
  # all items
end

要么

array.each_with_index do |item,index|
  if index == 0
    # first item
  end
  # all items
  if index == array.length-1
    # last item
  end
end

或者通过数组扩展

class Array

  def each_with_position
    array.each_with_index do |item,index|
      if index == 0
        yield item, :first
      elsif index == array.length-1
        yield item, :last
      else
        yield item, :middle
      end
    end
  end

  def each_with_index_and_position
    array.each_with_index do |item,index|
      if index == 0
        yield item, index, :first
      elsif index == array.length-1
        yield item, index, :last
      else
        yield item, index, :middle
      end
    end
  end

  def each_with_position_and_index
    array.each_with_index do |item,index|
      if index == 0
        yield item, :first, index
      elsif index == array.length-1
        yield item, :last, index
      else
        yield item, :middle, index
      end
    end
  end

end

Answer 6

如果您愿意添加一些样板文件，可以在数组类中添加以下内容：

class Array
  def each_fl
    each_with_index do |x,i|
      yield [i==0 ? :first : (i==length-1 ? :last : :inner), x]
    end
  end
end

然后你需要的任何地方，你得到以下语法：

[1,2,3,4].each_fl do |t,x|
  case t
    when :first
      puts "first: #{x}"
    when :last
      puts "last: #{x}"
    else
      puts "otherwise: #{x}"
  end
end

对于以下输出：

first: 1
otherwise: 2
otherwise: 3
last: 4

Answer 7

在Ruby中没有“执行此操作（最后|最后）时间”语法。 但如果你正在寻找简洁，你可以这样做：

a.each_with_index do |x, i|
  print (i > 0 ? (i == a.length - 1 ? x*10 : x) : x+1)
end

结果就是你所期望的：

irb(main):001:0> a = Array.new(5,1)
=> [1, 1, 1, 1, 1]
irb(main):002:0> a.each_with_index do |x,i|
irb(main):003:1*   puts (i > 0 ? (i == a.length - 1 ? x*10 : x) : x+1)
irb(main):004:1> end
2
1
1
1
10

Answer 8

有趣的问题，我也想过一个问题。

我认为你必须创建三个不同的块/ procs /无论它们被调用，然后创建一个调用正确的块/ proc /的方法。 （抱歉模糊不清 - 我还不是黑带元程序员）[ 编辑 ：但是，我是从底部的人那里复制的）

class FancyArray
  def initialize(array)
    @boring_array = array
    @first_code = nil
    @main_code = nil
    @last_code = nil
  end

  def set_first_code(&code)
    @first_code = code
  end

  def set_main_code(&code)
    @main_code = code
  end

  def set_last_code(&code)
    @last_code = code
  end

  def run_fancy_loop
    @boring_array.each_with_index do |item, i|
      case i
      when 0 then @first_code.call(item)
      when @boring_array.size - 1 then @last_code.call(item)
      else @main_code.call(item)
      end
    end
  end
end

fancy_array = FancyArray.new(["Matti Nykanen", "Erik Johnsen", "Michael Edwards"])
fancy_array.set_first_code {|item| puts "#{item} came first in ski jumping at the 1988 Winter Olympics"}
fancy_array.set_main_code {|item| puts "#{item} did not come first or last in ski jumping at the 1988 Winter Olympics"}
fancy_array.set_last_code {|item| puts "#{item} came last in ski jumping at the 1988 Winter Olympics"}
fancy_array.run_fancy_loop

产生

Matti Nykanen came first in ski jumping at the 1988 Winter Olympics
Erik Johnsen did not come first or last in ski jumping at the 1988 Winter Olympics
Michael Edwards came last in ski jumping at the 1988 Winter Olympics

编辑 ：Svante对相关问题的回答 （有molf的建议）显示了如何将多个代码块传递给单个方法：

class FancierArray < Array
  def each_with_first_last(first_code, main_code, last_code)
    each_with_index do |item, i|
      case i
        when 0 then first_code.call(item)
        when size - 1 then last_code.call(item)
        else main_code.call(item)
      end
    end
  end
end

fancier_array = FancierArray.new(["Matti Nykanen", "Erik Johnsen", "Michael Edwards"])
fancier_array.each_with_first_last(
  lambda {|person| puts "#{person} came first in ski jumping at the 1988 Winter Olympics"},
  lambda {|person| puts "#{person} did not come first or last in ski jumping at the 1988 Winter Olympics"},
  lambda {|person| puts "#{person} came last in ski jumping at the 1988 Winter Olympics"})

Answer 9

我不时需要这个功能，所以我为此目的制作了一个小课程。

最新版本位于： https ： //gist.github.com/3823837

样品：

("a".."m").to_a.each_pos do |e|
  puts "Char\tfirst?\tlast?\tprev\tnext\twrapped?\tindex\tposition" if e.first?
  print "#{e.item}\t"
  print "#{e.first?}\t"
  print "#{e.last?}\t"
  print "#{e.prev}\t"
  print "#{e.next}\t"
  print "#{e.wrapped?}\t\t"
  print "#{e.index}\t"
  puts  "#{e.position}\t"
end

# Char  first?  last?  prev  next  wrapped?  index  position
# a     true    false        b     false     0      1
# b     false   false  a     c     true      1      2
# c     false   false  b     d     true      2      3
# d     false   false  c     e     true      3      4
# e     false   false  d     f     true      4      5
# f     false   false  e     g     true      5      6
# g     false   false  f     h     true      6      7
# h     false   false  g     i     true      7      8
# i     false   false  h     j     true      8      9
# j     false   false  i     k     true      9      10
# k     false   false  j     l     true      10     11
# l     false   false  k     m     true      11     12
# m     false   true   l           false     12     13



{
  a: "0",
  b: "1",
  c: "2",
  d: "3",
  e: "4",
  f: "5",
  g: "6",
  h: "7",
  i: "8",
  j: "9",
  k: "10",
  l: "11",
  m: "12",
}.each_pos do |(k, v), e|
  puts "KV\tChar\t\tfirst?\tlast?\tprev\t\tnext\t\twrapped?\tindex\tposition" if e.first?
  print "#{k} => #{v}\t"
  print "#{e.item}\t"
  print "#{e.first?}\t"
  print "#{e.last?}\t"
  print "#{e.prev || "\t"}\t"
  print "#{e.next || "\t"}\t"
  print "#{e.wrapped?}\t\t"
  print "#{e.index}\t"
  puts  "#{e.position}\t"
end

# KV      Char        first?  last?   prev        next        wrapped?  index position
# a => 0  [:a, "0"]   true    false               [:b, "1"]   false     0     1
# b => 1  [:b, "1"]   false   false   [:a, "0"]   [:c, "2"]   true      1     2
# c => 2  [:c, "2"]   false   false   [:b, "1"]   [:d, "3"]   true      2     3
# d => 3  [:d, "3"]   false   false   [:c, "2"]   [:e, "4"]   true      3     4
# e => 4  [:e, "4"]   false   false   [:d, "3"]   [:f, "5"]   true      4     5
# f => 5  [:f, "5"]   false   false   [:e, "4"]   [:g, "6"]   true      5     6
# g => 6  [:g, "6"]   false   false   [:f, "5"]   [:h, "7"]   true      6     7
# h => 7  [:h, "7"]   false   false   [:g, "6"]   [:i, "8"]   true      7     8
# i => 8  [:i, "8"]   false   false   [:h, "7"]   [:j, "9"]   true      8     9
# j => 9  [:j, "9"]   false   false   [:i, "8"]   [:k, "10"]  true      9     10
# k => 10 [:k, "10"]  false   false   [:j, "9"]   [:l, "11"]  true      10    11
# l => 11 [:l, "11"]  false   false   [:k, "10"]  [:m, "12"]  true      11    12
# m => 12 [:m, "12"]  false   true    [:l, "11"]              false     12    13

实际课程：

module Enumerable
  # your each_with_position method
  def each_pos &block
    EachWithPosition.each(self, &block)
  end
end

class EachWithPosition
  attr_reader :index

  class << self
    def each *a, &b
      handler = self.new(*a, :each, &b)
    end
  end

  def initialize collection, method, &block
    @index = 0
    @item, @prev, @next = nil
    @collection = collection
    @callback = block
    self.send(method)
  end

  def count
    @collection.count
  end
  alias_method :length, :count
  alias_method :size, :count

  def rest
    count - position
  end

  def first?
    @index == 0
  end

  def last?
    @index == (count - 1)
  end

  def wrapped?
    !first? && !last?
  end
  alias_method :inner?, :wrapped?

  def position
    @index + 1
  end

  def prev
    @prev
  end

  def next
    @next
  end

  def current
    @item
  end
  alias_method :item, :current
  alias_method :value, :current

  def call
    if @callback.arity == 1
      @callback.call(self)
    else
      @callback.call(@item, self)
    end
  end

  def each
    @collection.each_cons(2) do |e, n|
      @prev = @item
      @item = e
      @next = n

      self.call
      @index += 1

      # fix cons slice behaviour
      if last?
        @prev, @item, @next = @item, @next, nil
        self.call
        @index += 1
      end
    end
  end
end

Answer 10

吻

arr.each.with_index do |obj, index|
  p 'first' if index == 0        
  p 'last' if index == arr.count-1                  
end

Answer 11

我无法抗拒:)虽然我认为它不应该比这里的大多数其他答案慢得多，但这并没有针对性能进行调整。 这都是糖！

class Array
  class EachDSL
    attr_accessor :idx, :max

    def initialize arr
      self.max = arr.size
    end

    def pos
      idx + 1
    end

    def inside? range
      range.include? pos
    end

    def nth? i
      pos == i
    end

    def first?
      nth? 1
    end

    def middle?
      not first? and not last?
    end

    def last?
      nth? max
    end

    def inside range
      yield if inside? range
    end

    def nth i
      yield if nth? i
    end

    def first
      yield if first?
    end

    def middle
      yield if middle?
    end

    def last
      yield if last?
    end
  end

  def each2 &block
    dsl = EachDSL.new self
    each_with_index do |x,i|
      dsl.idx = i
      dsl.instance_exec x, &block
    end
  end
end

例1：

[1,2,3,4,5].each2 do |x|
  puts "#{x} is first"  if first?
  puts "#{x} is third"  if nth? 3
  puts "#{x} is middle" if middle?
  puts "#{x} is last"   if last?
  puts
end

# 1 is first
# 
# 2 is middle
# 
# 3 is third
# 3 is middle
# 
# 4 is middle
# 
# 5 is last

例2：

%w{some short simple words}.each2 do |x|
  first do
    puts "#{x} is first"
  end

  inside 2..3 do
    puts "#{x} is second or third"
  end

  middle do
    puts "#{x} is middle"
  end

  last do
    puts "#{x} is last"
  end
end

# some is first
# short is second or third
# short is middle
# simple is second or third
# simple is middle
# words is last

Answer 12

如果你不介意“最后”动作发生在中间的东西之前，那么这个猴子补丁：

class Array

  def for_first
    return self if empty?
    yield(first)
    self[1..-1]
  end

  def for_last
    return self if empty?
    yield(last)
    self[0...-1]
  end

end

允许这个：

%w(a b c d).for_first do |e|
  p ['first', e]
end.for_last do |e|
  p ['last', e]
end.each do |e|
  p ['middle', e]
end

# => ["first", "a"]
# => ["last", "d"]
# => ["middle", "b"]
# => ["middle", "c"]

Answer 13

将数组划分为范围，其中每个范围内的元素应该表现不同。 将如此创建的每个范围映射到块。

class PartitionEnumerator
    include RangeMaker

    def initialize(array)
        @array = array
        @handlers = {}
    end

    def add(range, handler)
        @handlers[range] = handler
    end

    def iterate
        @handlers.each_pair do |range, handler|
          @array[range].each { |value| puts handler.call(value) }
        end
    end
end

可以手动创建范围，但下面的这些帮助使它更容易：

module RangeMaker
  def create_range(s)
    last_index = @array.size - 1
    indexes = (0..last_index)
    return (indexes.first..indexes.first) if s == :first
    return (indexes.second..indexes.second_last) if s == :middle
    return (indexes.last..indexes.last) if s == :last
  end  
end

class Range
  def second
    self.first + 1
  end

  def second_last
    self.last - 1
  end
end

用法：

a = [1, 2, 3, 4, 5, 6]

e = PartitionEnumerator.new(a)
e.add(e.create_range(:first), Proc.new { |x| x + 1 } )
e.add(e.create_range(:middle), Proc.new { |x| x * 10 } )
e.add(e.create_range(:last), Proc.new { |x| x } )

e.iterate

Answer 14

我看到很多hacks非常接近，但都严重依赖于具有固定大小的给定迭代器而不是迭代器。 我还建议在迭代时保存前一个元素，以了解迭代的第一个/最后一个元素。

previous = {}
elements.each do |element|
  unless previous.has_key?(:element)
    # will only execute the first time
  end

  # normal each block here

  previous[:element] = element
end

# the last element will be stored in previous[:element] 

Answer 15

如果您知道数组中的项是唯一的（与此情况不同），您可以这样做：

a = [1,2,3,4,5]

a.each_with_index do |x, i|
  if x == a.first
    print x+1
  elsif x == a.last
    print x*10
  else
    print x
  end
end

Answer 16

有时for循环只是你最好的选择

if(array.count > 0)
   first= array[0]
   #... do something with the first

   cx = array.count -2 #so we skip the last record on a 0 based array
   for x in 1..cx
     middle = array[x]
     #... do something to the middle
   end

   last = array[array.count-1]
   #... do something with the last item. 
end

我知道这个问题已经回答了，但是这个方法没有副作用，也没有检查13,14,15 ...... 10ththth，10,001th ...记录是第一个记录，还是最后一个记录。

以前的答案将在任何数据结构类中失败。