ruby - 快速 Ruby 哈希访问

Question

检查 Ruby 哈希对象的所有键是否都指向空数组的最快方法是什么？

我目前的做法：

h = {"a" => [1,2,3], "b" => []}
var = "something" if h.values.flatten.size > 0

另一种方法：

h = {"a" => [1,2,3], "b" => []}
var = "something" if h.values.flatten.empty?

还有其他明显更快的方法吗？

Answer 1

Flatten 将导致您的代码重新分配一个新的、更大的缓冲区。您可以像这样跳过展平：

h.values.all? &:empty?

基准：

Benchmark.measure {100000.times{ h.values.all? &:empty? }}
# =>   0.100000   0.000000   0.100000 (  0.096073)

Benchmark.measure {100000.times{ h.values.flatten.empty? }}
# =>   0.140000   0.000000   0.140000 (  0.143457)

更大的基准，包括h.all? {|_,v| v.empty?}

h = {}
(1...10000).each {|i| h[i] = []}  # Pathological case

Benchmark.measure {1000.times{ h.values.flatten.empty? }}
# =>   1.880000   0.000000   1.880000 (  1.882853)
Benchmark.measure {1000.times{ h.values.all? &:empty? }}
# =>   1.750000   0.000000   1.750000 (  1.748415)
Benchmark.measure {1000.times{ h.all? {|_,v| v.empty?} }}
# =>   4.140000   0.000000   4.140000 (  4.137548)

Answer 2

让我们贪婪：

all_empty = true

h.each_value do |value|
  unless value.empty?
    all_empty = false
    break
  end
end

基准：

h = {}
(1...10000).each {|i| h[i] = []}

Benchmark.measure {1000.times{ h.values.flatten.empty? }}
=>   2.020000   0.000000   2.020000 (  2.026274)
Benchmark.measure {1000.times{ h.values.all? &:empty? }}
=>   1.750000   0.000000   1.750000 (  1.750908)
Benchmark.measure {1000.times{ h.all? {|_,v| v.empty?} }}
=>   3.570000   0.000000   3.570000 (  3.570945)
Benchmark.measure {1000.times{ <code above> }} # Worst case
=>   1.530000   0.000000   1.530000 (  1.529857)