From: "Jan E." <lists@...>
Date: 2012-09-25T23:31:12+09:00
Subject: Re: inject problem

The first parameter of the "inject" block is always the intermediate 
result used for the aggregation. And the second is the current element.

For example:

# calculate sum of 1, 2, ..., 10 (in an inefficient way)
sum = (1..10).inject do |intermediate_sum, integer|
  intermediate_sum + integer
end

Or in your case:

count = [1, 2, 1, 1]
hash = count.inject({}) do |intermediate_hash, number|
  intermediate_hash[number] += 1 rescue intermediate_hash[number] = 1
  intermediate_hash
end
puts hash

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