How do you Compare Key Value Pairs Ruby - ruby

Question

What is the easiest way to compare each key-value pair in a hash in Ruby, to one another?
For example,
I want to sort this code so the highest three values are first. If the third spot has values that are all the same, then I want the one greatest
key to go in that spot.
{"Aa"=>1, "DDD"=>1, "DdD"=>1, "aA"=>1, "aa"=>1, "bb"=>1, "cC"=>1, "cc"=>1, "ddd"=>3, "e"=>7}
I need the above hash to be {"e"=>7, "ddd"=>3, "aa"=>1}

Answer 1

One string - last string makes what u want. Add .reverse after .sort for changing sort direction.
That is a solution:
# two lines for test
m = %w(a a a a a DDD DD ddd Ddd ddd e e e cC cC cC cC b x XXX XXX XXX ZZZ ZZZ ZZZ)
m = %w(n n n KKK KKK KKK KKK KKK LLL LLL LLL kk kk kk kk kk kk kk kk)
m = m.inject(Hash.new(0)) {|h, n| h.update(n => h[n]+1)}
m = m.sort.sort_by {|k, val| -val}.to_h

Answer 2

I would do it in three steps:
(1) Convert your Hash h to an Array of pairs:
kv_array = h.to_a
(2) Sort the array according to your criterion:
kv_array.sort! do
|left, right|
#.... implement your comparision here
end
(3) Turn your sorted array into a Hash
h_sorted = Hash[kv_array]

Answer 3

as suggested by #tadman, using a group_by then sorting the relevant groups will get you what you want, although you will need to tweak to fit your actual need, as a lot of assumptions were made:
m = {"Aa"=>1, "DDD"=>1, "DdD"=>1, "aA"=>1, "aa"=>1, "bb"=>1, "cC"=>1, "cc"=>1, "ddd"=>3, "e"=>7}
m.group_by { |k,v| v }
.each_with_object([]) {|a,x| x << [ a[1].compact.map { |b| b[0] }.min, a[0] ] }
.sort{|a| a[1]}
.to_h
=> {"e"=>7, "ddd"=>3, "Aa"=>1}
explanation:
firstly we group by the value (which returns a hash with the value as the first key, and an array of the hashes that match)
then we collect the grouped hash, and find the "first" key for each grouped value (using min) ... * this is an assumption * ... returning the whole thing as an array
then we sort the new array based on the "value"
then we convert the array to a hash
I added in new-lines to aid readability (hopefully)

Answer 4

I am so excited to get so many great suggestions! Thank you! This entire problem was to take a string and output the top three occurring words in an array. But, no special characters were allowed unless they were an apostrophe that was within the word. If there were multiple values and they were in the top three you had to pick the one that came closest to "a". In the beginning, I used the tally method to add everything up really quickly, and my plan was then to sort the hash by the value, but then when the values were the same I couldn't put the right key-value pair where it had to be if they shared the same value.
So, I came here and asked about sorting a hash, and then realized I needed to scratch my entire approach altogether. In the end, I found that I could sort the hash to an extent, but not to the place I needed/wanted so this is what I came up with!
def top_3_words(str)
str.scan(/[\w]+'?[\w]*/).sort.slice_when{|a, b| a != b}.max(3){|a, b| a.size <=> b.size}.flatten.uniq
end
p top_3_words("a a a b c c d d d d e e e e e") == ["e", "d", "a"]
p top_3_words("e e e e DDD ddd DdD: ddd ddd aa aA Aa, bb cc cC e e e") == ["e", "ddd", "aa"]
p top_3_words(" //wont won't won't ") == ["won't", "wont"]
p top_3_words(" , e .. ") == ["e"]
p top_3_words(" ... ") == []
p top_3_words(" ' ") == []
p top_3_words(" ''' ") == []
p top_3_words("""In a village of La Mancha, the name of which I have no desire to call to
mind, there lived not long since one of those gentlemen that keep a lance
in the lance-rack, an old buckler, a lean hack, and a greyhound for
coursing. An olla of rather more beef than mutton, a salad on most
nights, scraps on Saturdays, lentils on Fridays, and a pigeon or so extra
on Sundays, made away with three-quarters of his income.""") == ["a", "of", "on"]
My thought process with this was to call scan on the str so I could get everything I didn't want in the strings out of them. Then, I called sort on that return value because the last test case was really big and I needed all the like words together. After that, I called slice_when on that return value and said when a doesn't equal b then slice, so there would be a multi d array that I then could call max on, and because I sorted earlier, the values would be alphabetical so if there was a shared value it would give me the right one. I passed 3 to max to get the top three, and then called flatten so I had one array, and uniq to take out the extra characters!!
This is all so different from my original question, but I thought I would share what I was working on in case it could ever help anyone in the future!!