Fastest way to find a String into an array of string

Question

The script has to verify if one predefined IP is present in a big array of IPs. Currently I code that function like this (saying that "ips" is my array of IP and "ip" is the predefined ip)

ips.each do |existsip|
  if ip == existsip
    puts "ip exists"
    return 1
  end
end
puts "ip doesn't exist"
return nil

Is there a faster way to do the same thing?

Edit : I might have wrongly expressed myself. I can do array.include? but what I'd like to know is : Is array.include? the method that will give me the fastest result?

Solution 1

You can use Set. It is implemented on top of Hash and will be faster for big datasets - O(1).

require 'set'
s = Set.new ['1.1.1.1', '1.2.3.4']
# => #<Set: {"1.1.1.1", "1.2.3.4"}> 
s.include? '1.1.1.1'
# => true 

Solution 2

You could use the Array#include method to return you a true/false.

http://ruby-doc.org/core-1.9.3/Array.html#method-i-include-3F

if ips.include?(ip) #=> true
  puts 'ip exists'
else
  puts 'ip  doesn\'t exist'
end

Solution 3

A faster way would be:

if ips.include?(ip)
  puts "ip exists"
  return 1
else
  puts "ip doesn't exist"
  return nil
end

Solution 4

ips = ['10.10.10.10','10.10.10.11','10.10.10.12']

ip = '10.10.10.10'
ips.include?(ip) => true

ip = '10.10.10.13'
ips.include?(ip) => false

check Documentaion here

Solution 5

have you tried the Array#include? function?

http://ruby-doc.org/core-1.9.3/Array.html#method-i-include-3F

You can see from the source it does almost exactly the same thing, except natively.

View more solutions

Comments

[Phrogz]

Or in your case: s = Set.new(ips)

[Phrogz]

Slightly faster, since the each occurs in C instead of Ruby, but it's still O(n) versus O(1) for a Hash or Set.

[Phrogz]

This is still an O(n) time operation, as it must search through every item in the array (even if it's in C).

[Cocotton]

But is this actually faster than my method? Because the source code of this seems to do practically the same thing as mine.

[Cocotton]

Hello again Alex :) .include method source code seems to do almost the same thing as mine. Or is it actually faster?

[dku.rajkumar]

ofcourse it is faster.. i have using in my project.. moreover when there is a method in ruby, why should we write extra code.

[steenslag]

@Cocotton: Much faster. You could also use a Hash with ip's as keys and 'true' as values.

[Cocotton]

@steenslag,Phrogz : For now I'll leave it like this (.include), but when I'll be done will my script, I'll switch it for a hash

[Marc Talbot]

The obvious caveat here is that the Set is faster, but building the Set can be an expensive operation, so you wouldn't want to build a set to query it a small number of times.

[Peter Ehrlich]

I know that an enum can be sorted, but I don't know how to search such a sorted array. One could make an indexed database column to do the job.

[Phrogz]

Even a binary search is O(log n). Hashing an item and looking it up in a hash table is a constant-time operation that is unrelated to the number of items stored.

[Phrogz]

For an array of 12.4 million short strings: a=('a'..'zzzzz').to_a; time{ a.include?('0') } #=> 0.71s; time{ Set.new(a) } #=> 11.2s; so yes, the overhead of creating the set needs to be worth the performance gains of instantaneous queries.

[Ikon]

@dku.rajkumar wanted to say, it should be faster as .include? is implemented on C level on the Array class.

[Ikon]

Backing up @Phrogz statement, check the source code of both #include? methods and this gist as proof: gist.github.com/vadviktor/66e524c591b2604ce2d7