98

我有一个像这样的哈希数组:

 [{"testPARAM1"=>"testVAL1"}, {"testPARAM2"=>"testVAL2"}]

我正在尝试将其映射到单个哈希上,如下所示:

{"testPARAM2"=>"testVAL2", "testPARAM1"=>"testVAL1"}

我已经使用

  par={}
  mitem["params"].each { |h| h.each {|k,v| par[k]=v} } 

但我想知道是否有可能以更惯用的方式来做到这一点(最好不使用局部变量)。

我怎样才能做到这一点?

4

5 回答 5

173

你可以创作Enumerable#reduceHash#merge完成你想要的。

input = [{"testPARAM1"=>"testVAL1"}, {"testPARAM2"=>"testVAL2"}]
input.reduce({}, :merge)
  is {"testPARAM2"=>"testVAL2", "testPARAM1"=>"testVAL1"}

减少数组有点像在它的每个元素之间粘贴一个方法调用。

例如[1, 2, 3].reduce(0, :+)就像说0 + 1 + 2 + 3和给予6

在我们的例子中,我们做了类似的事情,但是使用了合并两个哈希的合并函数。

[{:a => 1}, {:b => 2}, {:c => 3}].reduce({}, :merge)
  is {}.merge({:a => 1}.merge({:b => 2}.merge({:c => 3})))
  is {:a => 1, :b => 2, :c => 3}
于 2012-08-08T01:54:19.013 回答
55

怎么样:

h = [{"testPARAM1"=>"testVAL1"}, {"testPARAM2"=>"testVAL2"}]
r = h.inject(:merge)
于 2012-08-08T01:51:51.090 回答
9

使用#inject

hashes = [{"testPARAM1"=>"testVAL1"}, {"testPARAM2"=>"testVAL2"}]
merged = hashes.inject({}) { |aggregate, hash| aggregate.merge hash }
merged # => {"testPARAM1"=>"testVAL1", "testPARAM2"=>"testVAL2"}
于 2012-08-08T01:51:09.853 回答
8

到目前为止,每个答案都建议使用Enumerable#reduce(或inject别名)+ Hash#merge,但要注意,虽然干净、简洁且易于阅读,但该解决方案将非常耗时,并且在大型数组上占用大量内存。

我已经编译了不同的解决方案并对其进行了基准测试。

一些选项

a = [{'a' => {'x' => 1}}, {'b' => {'x' => 2}}]

# to_h
a.to_h { |h| [h.keys.first, h.values.first] }

# each_with_object
a.each_with_object({}) { |x, h| h.store(x.keys.first, x.values.first) }
# each_with_object (nested)
a.each_with_object({}) { |x, h| x.each { |k, v| h.store(k, v) } }
# map.with_object
a.map.with_object({}) { |x, h| h.store(x.keys.first, x.values.first) }
# map.with_object (nested)
a.map.with_object({}) { |x, h| x.each { |k, v| h.store(k, v) } }

# reduce + merge
a.reduce(:merge) # take wayyyyyy to much time on large arrays because Hash#merge creates a new hash on each iteration
# reduce + merge!
a.reduce(:merge!) # will modify a in an unexpected way

基准脚本

使用bmbm很重要,而不是bm因为内存分配和垃圾收集的成本而避免差异。

require 'benchmark'

a = (1..50_000).map { |x| { "a#{x}" => { 'x' => x } } }

Benchmark.bmbm do |x|
  x.report('to_h:') { a.to_h { |h| [h.keys.first, h.values.first] } }
  x.report('each_with_object:') { a.each_with_object({}) { |x, h| h.store(x.keys.first, x.values.first) } }
  x.report('each_with_object (nested):') { a.each_with_object({}) { |x, h| x.each { |k, v| h.store(k, v) } } }
  x.report('map.with_object:') { a.map.with_object({}) { |x, h| h.store(x.keys.first, x.values.first) } }
  x.report('map.with_object (nested):') { a.map.with_object({}) { |x, h| x.each { |k, v| h.store(k, v) } } }
  x.report('reduce + merge:') { a.reduce(:merge) }
  x.report('reduce + merge!:') { a.reduce(:merge!) }
end

注意:我最初使用 1_000_000 个项目数组进行了测试,但是由于reduce + merge花费了成倍的时间,因此需要很长时间才能结束。

基准测试结果

50k 项数组

Rehearsal --------------------------------------------------------------
to_h:                        0.031464   0.004003   0.035467 (  0.035644)
each_with_object:            0.018782   0.003025   0.021807 (  0.021978)
each_with_object (nested):   0.018848   0.000000   0.018848 (  0.018973)
map.with_object:             0.022634   0.000000   0.022634 (  0.022777)
map.with_object (nested):    0.020958   0.000222   0.021180 (  0.021325)
reduce + merge:              9.409533   0.222870   9.632403 (  9.713789)
reduce + merge!:             0.008547   0.000000   0.008547 (  0.008627)
----------------------------------------------------- total: 9.760886sec

                                 user     system      total        real
to_h:                        0.019744   0.000000   0.019744 (  0.019851)
each_with_object:            0.018324   0.000000   0.018324 (  0.018395)
each_with_object (nested):   0.029053   0.000000   0.029053 (  0.029251)
map.with_object:             0.021635   0.000000   0.021635 (  0.021782)
map.with_object (nested):    0.028842   0.000005   0.028847 (  0.029046)
reduce + merge:             17.331742   6.387505  23.719247 ( 23.925125)
reduce + merge!:             0.008255   0.000395   0.008650 (  0.008681)

2M 个项目数组(不包括reduce + merge

Rehearsal --------------------------------------------------------------
to_h:                        2.036005   0.062571   2.098576 (  2.116110)
each_with_object:            1.241308   0.023036   1.264344 (  1.273338)
each_with_object (nested):   1.126841   0.039636   1.166477 (  1.173382)
map.with_object:             2.208696   0.026286   2.234982 (  2.252559)
map.with_object (nested):    1.238949   0.023128   1.262077 (  1.270945)
reduce + merge!:             0.777382   0.013279   0.790661 (  0.797180)
----------------------------------------------------- total: 8.817117sec

                                 user     system      total        real
to_h:                        1.237030   0.000000   1.237030 (  1.247476)
each_with_object:            1.361288   0.016369   1.377657 (  1.388984)
each_with_object (nested):   1.765759   0.000000   1.765759 (  1.776274)
map.with_object:             1.439949   0.029580   1.469529 (  1.481832)
map.with_object (nested):    2.016688   0.019809   2.036497 (  2.051029)
reduce + merge!:             0.788528   0.000000   0.788528 (  0.794186)
于 2021-08-11T15:36:00.213 回答
0

在这里,您可以使用Enumerable类中的注入减少,因为它们都是彼此的别名,因此两者都没有性能优势。

 sample = [{"testPARAM1"=>"testVAL1"}, {"testPARAM2"=>"testVAL2"}]

 result1 = sample.reduce(:merge)
 # {"testPARAM1"=>"testVAL1", "testPARAM2"=>"testVAL2"}

 result2 = sample.inject(:merge)
 # {"testPARAM1"=>"testVAL1", "testPARAM2"=>"testVAL2"}
于 2019-05-24T16:57:10.827 回答