如何判断Ruby散列是否是(或包含)另一个散列的子集?
例如:
hash = {a: 1, b: 2, c: 3, d: 4, e: 5, f: 6, g: 7}
hash.include_hash?({}) # true
hash.include_hash?({f: 6, c: 3}) # true
hash.include_hash?({f: 6, c: 1}) # false
答案 0 :(得分:16)
我想到的解决方案使用Hash#merge
方法:
class Hash
def include_hash?(hash)
merge(hash) == self
end
end
hash = {a: 1, b: 2, c: 3, d: 4, e: 5, f: 6, g: 7}
hash.include_hash?({})
# => true
hash.include_hash?(f: 6, c:3)
# => true
hash.include_hash?(f: 6, c:1)
# => false
答案 1 :(得分:13)
从Ruby 2.3开始,您还可以执行以下操作来检查这是否是子集
hash = {a: 1, b: 2, c: 3, d: 4, e: 5, f: 6, g: 7}
{} <= hash # true
{f: 6, c: 3} <= hash # true
{f: 6, c: 1} <= hash # false
答案 2 :(得分:8)
数组差异似乎最简单:
class Hash
def include_hash?(h)
(h.to_a - to_a).empty?
end
end
h = {a: 1, b: 2}
h.include_hash?({b: 2}) #=> true
h.include_hash?({b: 3}) #=> false
答案 3 :(得分:7)
您可以将哈希值转换为sets,然后使用方法subset?
和superset?
(或其各自的别名<=
和>=
)执行检查:
require 'set'
hash.to_set.superset?({}.to_set)
# => true
hash.to_set >= {a: 1}.to_set
# => true
{a: 2}.to_set <= hash.to_set
# => false
更新:建议解决方案的基准:
require 'fruity'
require 'set'
hash = ('aa'..'zz').zip('aa'..'zz').to_h
# {"aa"=>"aa", "ab"=>"ab", ...
find = ('aa'..'zz').zip('aa'..'zz').select { |k, _| k[0] == k[1] }.to_h
# {"aa"=>"aa", "bb"=>"bb", ...
compare(
toro2k: -> { hash.to_set >= find.to_set },
MarekLipka: -> { hash.merge(find) == hash },
CarySwoveland: -> { (find.to_a - hash.to_a).empty? },
ArupRakshit: -> { arr = hash.to_a; find.all? { |pair| arr.include?(pair) } }
)
结果:
Running each test 2 times. Test will take about 1 second.
MarekLipka is faster than toro2k by 3x ± 0.1
toro2k is faster than CarySwoveland by 39.99999999999999% ± 10.0%
CarySwoveland is faster than ArupRakshit by 1.9x ± 0.1
答案 4 :(得分:6)
你可以这样做:
def include_hash?(hash,subset_hash)
arr = hash.to_a
subset_hash.all? { |pair| arr.include?(pair) }
end