将数组转换为 Ruby 中的索引哈希

Question

I have an array, and I want to make a hash so I can quickly ask "is X in the array?".

In perl, there is an easy (and fast) way to do this:

my @array = qw( 1 2 3 );
my %hash;
@hash{@array} = undef;

This generates a hash that looks like:

{
    1 => undef,
    2 => undef,
    3 => undef,
}

The best I've come up with in Ruby is:

array = [1, 2, 3]
hash = Hash[array.map {|x| [x, nil]}]

which gives:

{1=>nil, 2=>nil, 3=>nil}

Is there a better Ruby way?

EDIT 1

No, Array.include? is not a good idea. Its slow . It does a query in O(n) instead of O(1). My example array had three elements for brevity; assume the actual one has a million elements. Let's do a little benchmarking:

#!/usr/bin/ruby -w
require 'benchmark'

array = (1..1_000_000).to_a
hash = Hash[array.map {|x| [x, nil]}]

Benchmark.bm(15) do |x|
    x.report("Array.include?") { 1000.times { array.include?(500_000) } }
    x.report("Hash.include?") { 1000.times { hash.include?(500_000) } }
end

Produces:

                     user     system      total        real
Array.include?  46.190000   0.160000  46.350000 ( 46.593477)
Hash.include?    0.000000   0.000000   0.000000 (  0.000523)

Answer 1

If all you need the hash for is membership, consider using a Set :

Set

Set implements a collection of unordered values with no duplicates. This is a hybrid of Array's intuitive inter-operation facilities and Hash's fast lookup.

Set is easy to use with Enumerable objects (implementing each ). Most of the initializer methods and binary operators accept generic Enumerable objects besides sets and arrays. An Enumerable object can be converted to Set using the to_set method.

Set uses Hash as storage, so you must note the following points:

• Equality of elements is determined according to Object#eql? and Object#hash .
• Set assumes that the identity of each element does not change while it is stored. Modifying an element of a set will render the set to an unreliable state.
• When a string is to be stored, a frozen copy of the string is stored instead unless the original string is already frozen.

Comparison

The comparison operators < , > , <= and >= are implemented as shorthand for the {proper_,}{subset?,superset?} methods. However, the <=> operator is intentionally left out because not every pair of sets is comparable. ({x,y} vs. {x,z} for example)

Example

require 'set' s1 = Set.new [1, 2] # -> #<Set: {1, 2}> s2 = [1, 2].to_set # -> #<Set: {1, 2}> s1 == s2 # -> true s1.add("foo") # -> #<Set: {1, 2, "foo"}> s1.merge([2, 6]) # -> #<Set: {1, 2, "foo", 6}> s1.subset? s2 # -> false s2.subset? s1 # -> true

[...]

Public Class Methods

new(enum = nil)

Creates a new set containing the elements of the given enumerable object.

If a block is given, the elements of enum are preprocessed by the given block.

Answer 2

try this one:

a=[1,2,3]
Hash[a.zip]

Answer 3

You can do this very handy trick:

Hash[*[1, 2, 3, 4].map {|k| [k, nil]}.flatten]
=> {1=>nil, 2=>nil, 3=>nil, 4=>nil}

Answer 4

If you want to quickly ask "is X in the array?" you should use Array#include? .

Edit (in response to addition in OP):

If you want speedy look up times, use a Set. Having a Hash that points to all nil s is silly. Conversion is an easy process too with Array#to_set .

require 'benchmark'
require 'set'

array = (1..1_000_000).to_a
set = array.to_set

Benchmark.bm(15) do |x|
    x.report("Array.include?") { 1000.times { array.include?(500_000) } }
    x.report("Set.include?") { 1000.times { set.include?(500_000) } }
end

Results on my machine:

                     user     system      total        real
Array.include?  36.200000   0.140000  36.340000 ( 36.740605)
Set.include?     0.000000   0.000000   0.000000 (  0.000515)

You should consider just using a set to begin with, instead of an array so that a conversion is never necessary.

Answer 5

I'm fairly certain that there isn't a one-shot clever way to construct this hash. My inclination would be to just be explicit and state what I'm doing:

hash = {}
array.each{|x| hash[x] = nil}

It doesn't look particularly elegant, but it's clear, and does the job.

FWIW, your original suggestion (under Ruby 1.8.6 at least) doesn't seem to work. I get an "ArgumentError: odd number of arguments for Hash" error. Hash.[] expects a literal, even-lengthed list of values:

Hash[a, 1, b, 2] # => {a => 1, b => 2}

so I tried changing your code to:

hash = Hash[*array.map {|x| [x, nil]}.flatten]

but the performance is dire:

#!/usr/bin/ruby -w
require 'benchmark'

array = (1..100_000).to_a

Benchmark.bm(15) do |x|
  x.report("assignment loop") {hash = {}; array.each{|e| hash[e] = nil}}
  x.report("hash constructor") {hash = Hash[*array.map {|e| [e, nil]}.flatten]}
end

gives

                     user     system      total        real
assignment loop  0.440000   0.200000   0.640000 (  0.657287)
hash constructor  4.440000   0.250000   4.690000 (  4.758663)

Unless I'm missing something here, a simple assignment loop seems the clearest and most efficient way to construct this hash.

Answer 6

Rampion beat me to it. Set might be the answer.

You can do:

require 'set'
set = array.to_set
set.include?(x)

Answer 7

Your way of creating the hash looks good. I had a muck around in irb and this is another way

>> [1,2,3,4].inject(Hash.new) { |h,i| {i => nil}.merge(h) }
=> {1=>nil, 2=>nil, 3=>nil, 4=>nil}

Answer 8

I think chrismear 's point on using assignment over creation is great. To make the whole thing a little more Ruby-esque, though, I might suggest assigning something other than nil to each element:

hash = {}
array.each { |x| hash[x] = 1 } # or true or something else "truthy"
...
if hash[376]                   # instead of if hash.has_key?(376)
  ...
end

The problem with assigning to nil is that you have to use has_key? instead of [] , since [] give you nil (your marker value) if the Hash doesn't have the specified key. You could get around this by using a different default value, but why go through the extra work?

# much less elegant than above:
hash = Hash.new(42)
array.each { |x| hash[x] = nil }
...
unless hash[376]
  ...
end

Answer 9

If you're not bothered what the hash values are

irb(main):031:0> a=(1..1_000_000).to_a ; a.length
=> 1000000
irb(main):032:0> h=Hash[a.zip a] ; h.keys.length
=> 1000000

Takes a second or so on my desktop.

Answer 10

Maybe I am misunderstanding the goal here; If you wanted to know if X was in the array, why not do array.include?("X") ?

Answer 11

Doing some benchmarking on the suggestions so far gives that chrismear and Gaius's assignment-based hash creation is slightly faster than my map method (and assigning nil is slightly faster than assigning true). mtyaka and rampion's Set suggestion is about 35% slower to create.

As far as lookups, hash.include?(x) is a very tiny amount faster than hash[x] ; both are twice as a fast as set.include?(x) .

                user     system      total        real
chrismear   6.050000   0.850000   6.900000 (  6.959355)
derobert    6.010000   1.060000   7.070000 (  7.113237)
Gaius       6.210000   0.810000   7.020000 (  7.049815)
mtyaka      8.750000   1.190000   9.940000 (  9.967548)
rampion     8.700000   1.210000   9.910000 (  9.962281)

                user     system      total        real
times      10.880000   0.000000  10.880000 ( 10.921315)
set        93.030000  17.490000 110.520000 (110.817044)
hash-i     45.820000   8.040000  53.860000 ( 53.981141)
hash-e     47.070000   8.280000  55.350000 ( 55.487760)

Benchmarking code is:

#!/usr/bin/ruby -w
require 'benchmark'
require 'set'

array = (1..5_000_000).to_a

Benchmark.bmbm(10) do |bm|
    bm.report('chrismear') { hash = {}; array.each{|x| hash[x] = nil} }
    bm.report('derobert')  { hash = Hash[array.map {|x| [x, nil]}] }
    bm.report('Gaius')     { hash = {}; array.each{|x| hash[x] = true} }
    bm.report('mtyaka')    { set = array.to_set }
    bm.report('rampion')   { set = Set.new(array) }
end

hash = Hash[array.map {|x| [x, true]}]
set = array.to_set
array = nil
GC.start

GC.disable
Benchmark.bmbm(10) do |bm|
    bm.report('times')  { 100_000_000.times { } }
    bm.report('set')    { 100_000_000.times { set.include?(500_000) } }
    bm.report('hash-i') { 100_000_000.times { hash.include?(500_000) } }
    bm.report('hash-e') { 100_000_000.times { hash[500_000] } }
end
GC.enable

Answer 12

This preserves 0's if your hash was [0,0,0,1,0]

  hash = {}
  arr.each_with_index{|el, idx| hash.merge!({(idx + 1 )=> el }) }

Returns :

  # {1=>0, 2=>0, 3=>0, 4=>1, 5=>0}

Answer 13

If you're looking for an equivalent of this Perl code:

grep {$_ eq $element} @array

You can just use the simple Ruby code:

array.include?(element)

Answer 14

Here's a neat way to cache lookups with a Hash:

a = (1..1000000).to_a
h = Hash.new{|hash,key| hash[key] = true if a.include? key}

Pretty much what it does is create a default constructor for new hash values, then stores "true" in the cache if it's in the array (nil otherwise). This allows lazy loading into the cache, just in case you don't use every element.