Sorting IP Addresses

Ruby
1

I’m having to sort some IP addresses that are stored in a hash. I
came up with the code below to accomplish the task. I have two
questions:

  • What if any is a better way to have done this?
  • Given that the hash has to be the mac address, could I store the
    hash values in a different way to make this easier?

require ‘ipaddr’

hash={}
hash[“00:19:d2:7b:55:10”]=[“c”,“12.12.19.11”,3]
hash[“00:18:39:0d:a1:e9”]=[“b”,“12.12.19.83”,1]
hash[“00:11:50:18:0c:0e”]=[“a”,“12.12.19.81”,2]

#the second index is the value item to sort by.
#this particular sort would fail if [1][1] was not an IP address

sortedarray = hash.sort { |a,b|
IPAddr.new(a[1][1]).to_i <=> IPAddr.new(b[1][1]).to_i
}

sortedarray.each { |item|

print “mac:”, item[0] + “\n”
print “IP:”, item[1][1] + “\n”
}

Mike B.

2

print “IP:”, item[1][1] + “\n”
}

sortedarray = hash.sort_by {|mac, settings|
IPAddr.new(settings[1])
}
sortedarray.each { |mac, settings|
puts “mac: #{mac}”
puts “IP: #{settings[1]}”
}

3

On Sep 24, 6:31 pm, barjunk [email protected] wrote:

hash={}
}

sortedarray.each { |item|

print “mac:”, item[0] + “\n”
print “IP:”, item[1][1] + “\n”

}

May not be faster, but doesn’t rely on the ipaddr library

hash={}
hash[“00:19:d2:7b:55:10”]=[“c”,“1.12.19.11”,3]
hash[“00:18:39:0d:a1:e9”]=[“b”,“12.12.19.83”,1]
hash[“00:11:50:18:0c:0e”]=[“a”,“12.12.19.81”,2]
hash[“00:00:00:08:c3:1e”]=[“a”,“2.12.19.81”,2]
hash[“fe:00:00:08:c6:2f”]=[“a”,“204.12.19.81”,2]

sorted_array = hash.sort_by{ |mac, settings|

Break the IP into an array of integers

sort_by will sort by the first entry first, then the second, and

so on
settings[1].split(‘.’).map{ |digits| digits.to_i }
}

Note that you can break apart the array of arrays in the iteration

And use puts and string interpolation for simpler printing

sorted_array.each { |mac, settings|
puts “mac: #{mac}”
puts “IP: #{settings[1]}”
}