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集合内建方法

Ruby 中使用集合有时候需要 require ‘set' 一下

# Ruby[1] pry(main)> require 'set'=> false[2] pry(main)> s1 = Set[3, 4, 5]=> #
   
    [3] pry(main)> s1 = Set.new([3, 4, 5])=> #
    
     [9] pry(main)> s1 = Set[1, 1, 3, 4, 5]=> #
     
    
   

# Python3>>> set([3, 4, 5]){3, 4, 5}>>> set([1, 1, 3, 4, 5]){1, 3, 4, 5}

set.union  other_set（set | other_set）

# Ruby[10] pry(main)> s1 = Set[3, 4, 5]=> #
   
    [11] pry(main)> s2 = Set[3, 1, 5]=> #
    
     [12] pry(main)> s1.union s2=> #
     
      [13] pry(main)> s1 + s2=> #
      
       [14] pry(main)> s1 | s2=> #
       
      
     
    
   

# Python3>>> s1 = set([3, 4, 5])>>> s2 = set([3, 1, 5])>>> s1.union(s2){1, 3, 4, 5}>>> s1 | s2{1, 3, 4, 5}

set.intersection  other_set（set & other_set）

# Ruby[15] pry(main)> s1 = Set[3, 4, 5]=> #
   
    [16] pry(main)> s2 = Set[3, 1, 5]=> #
    
     [17] pry(main)> s1.intersection s2=> #
     
      [18] pry(main)> s1 & s2=> #
      
     
    
   

# Python3>>> s1 = set([3, 4, 5])>>> s2 = set([3, 1, 5])>>> s1.intersection(s2){3, 5}>>> s1 & s2{3, 5}

set ^ other_set

# Ruby[24] pry(main)> s1 = Set[3, 4, 5]=> #
   
    [25] pry(main)> s2 = Set[3, 1, 5]=> #
    
     [26] pry(main)> s1 ^ s2=> #
     
      [27] pry(main)> (s1 | s2) - (s1 & s2)=> #
      
     
    
   

# Python3>>> s1 = set([3, 4, 5])>>> s2 = set([3, 1, 5])>>> s1 ^ s2{1, 4}>>> (s1 | s2) - (s1 & s2){1, 4}

set - other_set

# Ruby[19] pry(main)> s1 = Set[3, 4, 5]=> #
   
    [20] pry(main)> s2 = Set[3, 1, 5]=> #
    
     [21] pry(main)> s1 - s2=> #
     
      [22] pry(main)> s2 -s1=> #
      
     
    
   

# Python3>>> s1 = set([3, 4, 5])>>> s2 = set([3, 1, 5])>>> s1 - s2{4}>>> s2 -s1{1}

set.add（set.add？）

set.add? item 表示如果元素已经添加过，则执行结果返回值为 nil。当 item 没存在集合中时，执行等价于 set.add。

# Ruby[44] pry(main)> s1 = Set[3, 4, 5]=> #
   
    [45] pry(main)> s1.add 6=> #
    
     [51] pry(main)> s1 << 7=> #
     
      [87] pry(main)> s1.add? 5=> nil[88] pry(main)> s1.add? 6=> #
      
     
    
   

# Python3>>> s1 = set([3, 4, 5])>>> s1.add(6)>>> s1{3, 4, 5, 6}

set.clear

# Ruby[48] pry(main)> s1 = Set[3, 4, 5]=> #
   
    [49] pry(main)> s1.clear=> #
    
   

# Python3>>> s1 = set([3, 4, 5])>>> s1.clear()>>> s1set()

set.count

# Ruby[63] pry(main)> s1 = Set[3, 4, 5]=> #
   
    [64] pry(main)> s1.count=> 3
   

# Python3>>> s1 = set([3, 4, 5])>>> len(s1)3

set.delete（set.delete？）

set.delete? item 表示如果元素已经不存在，则执行结果返回值为 nil。当 item 存在集合中时，执行等价于 set.delete。

# Ruby[90] pry(main)> s1 = Set[3, 4, 5]=> #
   
    [91] pry(main)> s1.delete 5=> #
    
     [92] pry(main)> s1.delete? 5=> nil
    
   

# Python3>>> s1 = set([3, 4, 5])>>> s1.remove(5)>>> s1{3, 4}>>> s1.remove(5)Traceback (most recent call last):  File "
   
    ", line 1, in 
    
     KeyError: 5
    
   

set.empty?

# Ruby[37] pry(main)> s1 = Set[3, 4, 5]=> #
   
    [38] pry(main)> s1.empty?=> false[39] pry(main)> s2 = Set[]=> #
    
     [40] pry(main)> s2.empty?=> true
    
   

# Python3>>> s1 = set([3, 4, 5])>>> len(s1) == 0False>>> s2 = set()>>> len(s2) == 0True

set.include?

# Ruby[28] pry(main)> s1 = Set[3, 4, 5]=> #
   
    [29] pry(main)> s1.include? 3=> true[30] pry(main)> s1.member? 3=> true
   

# Python3>>> s1 = set([3, 4, 5])>>> 3 in s1True

set.subset?（set.superset?）

# Ruby[53] pry(main)> s1 = Set[3, 4, 5]=> #
   
    [54] pry(main)> s2 = Set[3, 4]=> #
    
     [55] pry(main)> s1.subset? s2=> false[56] pry(main)> s2.subset? s1=> true[57] pry(main)> s1.subset? s1=> true[58] pry(main)> s1.proper_subset? s1=> false[59] pry(main)> s2.proper_subset? s1=> true[60] pry(main)> s1.superset? s2=> true
    
   

# Python3>>> s1 = set([3, 4, 5])>>> s2 = set([3, 4])>>> s1.issubset(s2)False>>> s2.issubset(s1)True>>> s1.issubset(s1)True>>> s1.issuperset(s2)True>>> s2.issuperset(s1)False

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