Ruby 集合与容器对象用法详解

插入项目：

a = []
a[0] = "first"
a[0] 是一种语法糖， 实际上是一个方法的调用，脱了它的糖衣应该是：

a.[]=(0, "first")
取回项目：

>> a = [1,2,3,4,5]
=> [1, 2, 3, 4, 5]
>> p a[2]
3
=> 3
a[2] 也是语法糖，脱了糖衣应该是：

a.[](2)
一次多个项目

>> a = ["red", "orange", "yellow", "purple", "gray", "indigo", "violet"]
=> ["red", "orange", "yellow", "purple", "gray", "indigo", "violet"]
>> a[3,2]
=> ["purple", "gray"]
>> a[3,2] = "green", "blue"
=> ["green", "blue"]
>> a
=> ["red", "orange", "yellow", "green", "blue", "indigo", "violet"]
slice 方法跟 [] 一样用。

>> a.values_at(0,3)
=> ["red", "green"]
数组的开头与结尾

unshift 在数组的最前面添加个项目：

>> a = [1,2,3]
=> [1, 2, 3]
>> a.unshift(0)
=> [0, 1, 2, 3]
push 会在数组最后添加项目：

>> a = [1,2,3]
=> [1, 2, 3]
>> a.push(4)
=> [1, 2, 3, 4]
也可以用 << ，像这样：

>> a << 5
=> [1, 2, 3, 4, 5]
push 可以添加多个项目。

用 shift 与 pop 删除数组里的项目：

>> a = [1,2,3,4,5]
=> [1, 2, 3, 4, 5]
>> popped = a.pop
=> 5
>> puts popped
5
=> nil
>> p a
[1, 2, 3, 4]
=> [1, 2, 3, 4]
>> shifted = a.shift
=> 1
>> puts shifted
1
=> nil
>> p a
[2, 3, 4]
=> [2, 3, 4]
合并数组

先试试 concat：

>> [1,2,3].concat([4,5,6])
=> [1, 2, 3, 4, 5, 6]
concat 会改变源数组，如果你想得到一个新的合并以后的数组，可以使用 + ：

>> a = [1,2,3]
=> [1, 2, 3]
>> b = a + [4,5,6]
=> [1, 2, 3, 4, 5, 6]
>> a
=> [1, 2, 3]
用 replace 替换：

>> a = [1,2,3]
=> [1, 2, 3]
>> a.replace([4,5,6])
=> [4, 5, 6]
>> a
=> [4, 5, 6]
再做个实验理解一下 = 与 replace 的区别：

>> a = [1,2,3]
=> [1, 2, 3]
>> b = a
=> [1, 2, 3]
>> a.replace([4,5,6])
=> [4, 5, 6]
>> b
=> [4, 5, 6]
>> a = [7,8,9]
=> [7, 8, 9]
>> b
=> [4, 5, 6]
数组转型

flatten，去掉数组里的层级，在参数里可以指定要去掉的层级数：

>> array = [1,2,[3,4[5]],[6,[7,8]]]
=> [1, 2, [3, 0], [6, [7, 8]]]
>> array.flatten
=> [1, 2, 3, 0, 6, 7, 8]
>> array.flatten(1)
=> [1, 2, 3, 0, 6, [7, 8]]
>> array.flatten(2)
=> [1, 2, 3, 0, 6, 7, 8]
reverse 颠倒：

>> [1,2,3].reverse
=> [3, 2, 1]
join：

>> ["abc", "def", 123].join
=> "abcdef123"
join 的参数：

>> ["abc", "def", 123].join(", ")
=> "abc, def, 123"
还有种合并的方法：

>> a = %w{ one two three }
=> ["one", "two", "three"]
>> a * "-"
=> "one-two-three"
去掉重复：

>> [1,2,3,3].uniq
=> [1, 2, 3]
去掉 nil：

>> zip_codes = ["06511", "08902", "08902", nil, "10027",
"08902", nil, "06511"]
=> ["06511", "08902", "08902", nil, "10027", "08902", nil, "06511"]
>> zip_codes.compact
=> ["06511", "08902", "08902", "10027", "08902", "06511"]
数组查询

a.size，a.length：项目的数量
a.empty?：空白吗？
a.include?(item)：有某个项目吗？
a.count(item)：这个项目有多少？
a.first(n=1)：前几个项目
a.last(n=1)：后几个项目
a.sample(n=1)：随机项目
实验：

>> a
=> ["one", "two", "three"]
>> a.first(n=1)
=> ["one"]
>> a.first(n=2)
=> ["one", "two"]
>> a.last(n=1)
=> ["three"]
>> a.sample(n=1)
=> ["two"]
>> a.sample(n=1)
=> ["one"]
hash

先做个实验，创建一个 hash，根据用户的输入，得到地区的缩写：

p_hash = {
  "北京" => "BJ",
  "上海" => "SH",
  "广州" => "GZ"
}

print "输入地区的名字："
p = gets.chomp
abbr = p_hash[p]
puts "地区的缩写是 #{abbr}"
创建 hash

几个方法可以创建 hash：

字面构造器：{}
Hash.new 方法
Hash.[] 方法
Hash 顶级方法
创建字面 hash

大括号里面是 hash 项目，每个项目都有 key ，还有对应的 value，它们中间用 => 符号连接，项目与项目之间用逗号分隔开。

Hash.new

Hash.[] 类方法

>> Hash["北京", "BJ", "上海", "SH"]
=> {"北京"=>"BJ", "上海"=>"SH"}
>> Hash[ [[1,2], [3,4], [5,6]] ]
=> {1=>2, 3=>4, 5=>6}
Hash 方法

Hash 方法有点特别的行为。调用的时候给它一个空白的数组（[ ]）或 nil，它会返回一个空白的 hash。不然它会在参数上调用 to_hash ，如果数组没 to_hash 方法，就会触发错误。

插入，取回，删除 hash 项目

添加新项目到 hash 里

用 []= 方法：

p_hash["山东"] = "SD"
脱掉糖衣应该是这样的：

p_hash.[]=("山东", "SD")
store 也可以：

p_hash.store("山东", "SD")
做个实验：

>> h = Hash.new
=> {}
>> h["a"] = 1
=> 1
>> h["a"] = 2
=> 2
>> puts h["a"]
2
=> nil
获取值

用 [] 方法：

>> p_hash["北京"]
=> "BJ"
用 fetch 方法也行：

>> p_hash.fetch("北京")
=> "BJ"
[] 与 fetch 的区别是，fetch 不存在的 key 的时候会出现异常，而 [] 会返回 nil 。

>> p_hash.fetch("云南", "Unknown state")
=> "Unknown state"
多个 key ：

>> p_hash.values_at("北京", "上海")
=> ["BJ", "SH"]
默认 hash 值与行为

这是默认的情况：

>> h = Hash.new
=> {}
>> h["no such key!"]
=> nil
设置默认的值：

>> h = Hash.new(0)
=> {}
>> h["no such key!"]
=> 0
这样试一下：

h = Hash.new {|hash,key| hash[key] = 0 }
再这样：

>> h["new key!"]
=> 0
>> h
=> {"new key!"=>0}
合并 hash

update，有破坏性的合并：

>> h1 = {"Smith" => "John", "Jones" => "Jane" }
=> {"Smith"=>"John", "Jones"=>"Jane"}
>> h2 = {"Smith" => "Jim"}
=> {"Smith"=>"Jim"}
>> h1.update(h2)
=> {"Smith"=>"Jim", "Jones"=>"Jane"}
>> puts h1["Smith"]
Jim
=> nil
merge，不具破坏性的合并：

>> h1 = {"Smith" => "John", "Jones" => "Jane" }
=> {"Smith"=>"John", "Jones"=>"Jane"}
>> h2 = {"Smith" => "Jim"}
=> {"Smith"=>"Jim"}
>> h3 = h1.merge(h2)
=> {"Smith"=>"Jim", "Jones"=>"Jane"}
>> p h1["Smith"]
"John"
=> "John"
hash 变型

选择与拒绝元素

select：

>> h = Hash[1,2,3,4,5,6]
=> {1=>2, 3=>4, 5=>6}
>> h.select {|k,v| k > 1}
=> {3=>4, 5=>6}
reject：

>> h.reject {|k,v| k > 1}
=> {1=>2}
keep_if：

>> h
=> {1=>2, 3=>4, 5=>6}
>> h.keep_if {|k,v| k > 1}
=> {3=>4, 5=>6}
>> h
=> {3=>4, 5=>6}
delete_if：

>> h
=> {3=>4, 5=>6}
>> h.delete_if {|k,v| k > 3}
=> {3=>4}
>> h
=> {3=>4}
颠倒

invert：

>> h = { 1 => "one", 2 => "two" }
=> {1=>"one", 2=>"two"}
>> h.invert
=> {"one"=>1, "two"=>2}
注意 key 必须得是唯一的。

>> h = { 1 => "one", 2 => "more than 1", 3 => "more than 1" }
=> {1=>"one", 2=>"more than 1", 3=>"more than 1"}
>> h.invert
=> {"one"=>1, "more than 1"=>3}
清空

clear：

>> {1 => "one", 2 => "two"}.clear
=> {}
替换

>> { 1 => "one", 2 => "two" }.replace({ 10 => "ten", 20 => "twenty"})
=> {10 => "ten", 20 => "twenty"}
查询 hash

h.has_key?(1)
h.include?(1)
h.key?(1)
h.member?(1)
h.has_value?("three")
h.value?("three")
h.empty?
h.size
hash 作为方法的参数

参数列表里的最后一个参数是个 hash，不用 {} 了。

add_to_city_datebase("New York City",
  state: "New York",
  population: 7000000,
  nickname: "Big Apple"
)
方法应该像这样定义：

def add_to_city_datebase(name, info)
  c = City.new
  c.name = name
  c.state = info[:state]
  c.population = info[:population]
  # ...
命名的参数

定义方法的时候这么干：

>> def m(a:, b:)
>>   p a,b
>> end
=> :m
>> m(a: 1, b: 2)
1
2
=> [1, 2]
默认参数：

>> def m(a: 1, b: 2)
>>   p a,b
>> end
=> :m
>> self.m
1
2
=> [1, 2]
>> m(a:10)
10
2
=> [10, 2]
你猜猜 ：

>> def m(a: 1, b: 2, **c)
>> p a,b,c
>> end
=> :m
>> m(x: 1, y: 2)
1
2
{:x=>1, :y=>2}
=> [1, 2, {:x=>1, :y=>2}]
>> def m(x, y, *z, a: 1, b:, **c, &block)
>>   p x,y,z,a,b,c
>> end
=> :m
>> m(1,2,3,4,5,b:10,p:20,q:30)
1
2
[3, 4, 5]
1
10
{:p=>20, :q=>30}
=> [1, 2, [3, 4, 5], 1, 10, {:p=>20, :q=>30}]
Range

Range：范围

包含（inclusion）—— 给的值在一个范围里吗？
枚举（enumeration）—— 把范围可获取的单立项目的集合
创建一个 range

Range.new：

>> r = Range.new(1, 100)
=> 1..100
上面创建的是 inclusive range，再创建一个 exclusive range，用三个点：

>> r = 1...100
=> 1...100
Range-inclusion 逻辑

begin 与 end

>> r = 1..10
=> 1..10
>> r.begin
=> 1
>> r.end
=> 10
exclusive range？

>> r.exclude_end?
=> false
cover?

如果给方法的参数大于 range 的起点，小于终点，range 说它 cover 了这个对象。

>> r = "a".."z"
=> "a".."z"
>> r.cover?("a")
=> true
>> r.cover?("abc")
=> true
>> r.cover?("A")
=> false
"a" >= "a"
"a" <= "z"

"abc" >= "a"
"abc" <= "z"

"A" < "a"
不兼容的情况：

>> r.cover?([])
=> false
include?

include? 把 range 看成是一个值的集合。

>> r.include?("a")
=> true
>> r.include?("abc")
=> false
>> r = 1.0..2.0
=> 1.0..2.0
>> r.include?(1.5)
=> true
Set

使用 Set 先得：

require 'set'
set 是一个集合，它里面的项目都是唯一的。项目可以是任何东西，字符串，整数，数组，其它的 set 等。

创建 set

Set.new 构造器创建 set 。

>> require('set')
=> false
>> n = [1,2,3]
=> [1, 2, 3]
>> n_set = Set.new(n)
=> #
给构造器提供一个代码块：

>> names = ["David", "Yukihiro", "Amy"]
=> ["David", "Yukihiro", "Amy"]
>> name_set = Set.new(names) {|name| name.upcase}
=> #
处理 set 元素

>> n = Set.new([1,2])
=> #
>> n << 3
=> #
>> n << 3
=> #
>> n
=> #
delete

>> n.delete(2)
=> #
add

>> n.add?(2)
=> #
>> n
=> #
>> n.add?(2)
=> nil
intersection，union，difference

intersection，别名 &
union，别名 +  与 |
difference，别名 -
这些方法都会返回新的 set。

>> n = Set.new([1, 2, 3])
=> #
>> n1 = Set.new([3, 4, 5])
=> #
>> n - n1
=> #
>> n + n1
=> #
>> n & n1
=> #
>> n | n1
=> #
^

>> n ^ n1
=> #
merge

>> n
=> #
>> n.object_id
=> 70179526750720
>> n.merge([4])
=> #
>> n.object_id
=> 70179526750720
跟 hash 合并：

>> n
=> #
>> n.merge({"北京" => "BJ"})
=> #
subset 与 superset

subset：子集，superset：超集

>> n = Set.new([1, 2, 3])
=> #
>> n1 = Set_new([1, 2])
>> n1.subset?(n)
=> true
>> n.superset?(n1)
=> true
proper_subset 与 proper_superset。proper 子集至少要比父级少一个项目，如果两个 set 一样，那它们是彼此的 subset，但不是 proper 子集。