FAQ: Create a Histogram - Sorting the Hash

This community-built FAQ covers the “Sorting the Hash” exercise from the lesson “Create a Histogram”.

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words.each { |word| frequencies[word] += 1 }

I am really confused by this line of code, can someone please explain it to me. I get that you’re iterating over the words array but what does this part do:
frequencies[word] += 1

Any help would be appreciated :slight_smile:

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words.each { |word| frequencies[word] += 1 }

I believe all that this line of code is doing is increasing the counter that you initially set at ‘0’ (i.e. frequencies = Hash.new(0)).

As per the instructions on the 5th exercise: “This is why our default is 0 . The first time we find the word, it will have a default value of `` 0 that we can increment by 1`”

I’m not sure I understand the solution. Anyone willing to explain it in layman terms?

frequencies = frequencies.sort_by do |word, count|


As we enter this phase of the program, frequencies is a hash of words and their frequency. sort_by will return an array of word/count pairs, so no longer a hash. We choose which to sort by, word or count, and get a sorted list by that key. Since the hash is no longer needed, the output of this method is assigned back onto that variable.

hi! i don’t know what’s happening with my code. Anyone willing to explain me?

puts "How are you? "



words.each { |word| puts word
frequencies [word] +=1
puts frequencies

frequencies=frenquencies.sort_by do |word,count| count end

Could it be that .split() is a method so must be invoked with ()?

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thank you!! was actually a syntaxis error in the last line ahah

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I’m slightly confused as to how the code :

words.each { |word| frequencies[word] += 1 }

knows to increase the word (key’s) value by 1 when it comes across it more than once, instead of simply creating a hash that simply has a key for each and every word, with a value of 1.

Does my question make sense? I would have assumed that we would need some sort of comparison operator or something, to compare the words to the words(keys) that are already in the hash?

Hashes cannot have duplicate keys. If a key is not present it gets inserted with the default value, which is then incremented. When the key exists, its value simply gets incremented.

Perfect! Thank you for this answer!

1 Like

Can someone explain the methodology behind

frequencies = frequencies.sort_by do |word,count|

And are there any other techniques to get the same result?

Not with the same horsepower, afaik.

is able to transform a hash into an array of key-value pairs in their own array.

a = {}
a['word'] = 4
a['phrase'] = 2
a['the'] = 5
a['letter'] = 2
a['and'] = 1
b = a.sort_by do |value, count|
puts a
puts b


{"word"=>4, "phrase"=>2, "the"=>5, "letter"=>2, "and"=>1}
[["and", 1], ["phrase", 2], ["letter", 2], ["word", 4], ["the", 5]]

To reverse the sort we prepend the variable with a negative sign.

b = a.sort_by do |value, count|
puts b


[["the", 5], ["word", 4], ["phrase", 2], ["letter", 2], ["and", 1]]

Bottom line, this is a very powerful method. Any other approach is doubtless going to take more logic.