I want to take a hash with nested hashes and arrays and flatten it out into a single hash with unique values. I keep trying to approach this from different angles, but then I make it way more complex than it needs to be and get myself lost in what's happening.
Example Source Hash:
{
"Name" => "Kim Kones",
"License Number" => "54321",
"Details" => {
"Name" => "Kones, Kim",
"Licenses" => [
{
"License Type" => "PT",
"License Number" => "54321"
},
{
"License Type" => "Temp",
"License Number" => "T123"
},
{
"License Type" => "AP",
"License Number" => "A666",
"Expiration Date" => "12/31/2020"
}
]
}
}
Example Desired Hash:
{
"Name" => "Kim Kones",
"License Number" => "54321",
"Details_Name" => "Kones, Kim",
"Details_Licenses_1_License Type" => "PT",
"Details_Licenses_1_License Number" => "54321",
"Details_Licenses_2_License Type" => "Temp",
"Details_Licenses_2_License Number" => "T123",
"Details_Licenses_3_License Type" => "AP",
"Details_Licenses_3_License Number" => "A666",
"Details_Licenses_3_Expiration Date" => "12/31/2020"
}
For what it's worth, here's my most recent attempt before giving up.
def flattify(hashy)
temp = {}
hashy.each do |key, val|
if val.is_a? String
temp["#{key}"] = val
elsif val.is_a? Hash
temp.merge(rename val, key, "")
elsif val.is_a? Array
temp["#{key}"] = enumerate val, key
else
end
print "=> #{temp}\n"
end
return temp
end
def rename (hashy, str, n)
temp = {}
hashy.each do |key, val|
if val.is_a? String
temp["#{key}#{n}"] = val
elsif val.is_a? Hash
val.each do |k, v|
temp["#{key}_#{k}#{n}"] = v
end
elsif val.is_a? Array
temp["#{key}"] = enumerate val, key
else
end
end
return flattify temp
end
def enumerate (ary, str)
temp = {}
i = 1
ary.each do |x|
temp["#{str}#{i}"] = x
i += 1
end
return flattify temp
end
A hash table is a type of data structure that stores key-value pairs. The key is sent to a hash function that performs arithmetic operations on it. The result (commonly called the hash value or hash) is the index of the key-value pair in the hash table.
We can check if a particular hash contains a particular key by using the method has_key?(key) . It returns true or false depending on whether the key exists in the hash or not.
Hash#flatten () is a Hash class method which returns the one-dimensional flattening hash array. Hash a flatten form : [:a, 100, :b, 200] Hash b flatten form : [:a, 100, :c, [300, 45], :b, 200] Hash c flatten form : [:a, 100]
Using reverse If in our hash the values are unique, that is each value is different, then we can just pass the hash to he reverse function and assign the result to another hash. We will get a new hash in which the former values will be the key, and the former key will be values:
How to exchange keys with values of a hash without losing any data even if there a are duplicate values in the original hash? If in our hash the values are unique, that is each value is different, then we can just pass the hash to he reverse function and assign the result to another hash.
The reason is that the keys of a hash must be unique. As we reverse the hash we have two keys (former values) that are identical. As Perl works through all the key-value pairs the last value of each key wins. Which one is the last can change from execution-to-execution.
Interesting question!
Here's a recursive method to parse your data.
tmp
array.tmp
as key.def recursive_parsing(object, tmp = [])
case object
when Array
object.each.with_index(1).with_object({}) do |(element, i), result|
result.merge! recursive_parsing(element, tmp + [i])
end
when Hash
object.each_with_object({}) do |(key, value), result|
result.merge! recursive_parsing(value, tmp + [key])
end
else
{ tmp.join('_') => object }
end
end
As an example:
require 'pp'
pp recursive_parsing(data)
# {"Name"=>"Kim Kones",
# "License Number"=>"54321",
# "Details_Name"=>"Kones, Kim",
# "Details_Licenses_1_License Type"=>"PT",
# "Details_Licenses_1_License Number"=>"54321",
# "Details_Licenses_2_License Type"=>"Temp",
# "Details_Licenses_2_License Number"=>"T123",
# "Details_Licenses_3_License Type"=>"AP",
# "Details_Licenses_3_License Number"=>"A666",
# "Details_Licenses_3_Expiration Date"=>"12/31/2020"}
Here's a modified version with old-school debugging. It might help you understand what's going on:
def recursive_parsing(object, tmp = [], indent="")
puts "#{indent}Parsing #{object.inspect}, with tmp=#{tmp.inspect}"
result = case object
when Array
puts "#{indent} It's an array! Let's parse every element:"
object.each_with_object({}).with_index(1) do |(element, result), i|
result.merge! recursive_parsing(element, tmp + [i], indent + " ")
end
when Hash
puts "#{indent} It's a hash! Let's parse every key,value pair:"
object.each_with_object({}) do |(key, value), result|
result.merge! recursive_parsing(value, tmp + [key], indent + " ")
end
else
puts "#{indent} It's a leaf! Let's return a hash"
{ tmp.join('_') => object }
end
puts "#{indent} Returning #{result.inspect}\n"
result
end
When called with recursive_parsing([{a: 'foo', b: 'bar'}, {c: 'baz'}])
, it displays:
Parsing [{:a=>"foo", :b=>"bar"}, {:c=>"baz"}], with tmp=[]
It's an array! Let's parse every element:
Parsing {:a=>"foo", :b=>"bar"}, with tmp=[1]
It's a hash! Let's parse every key,value pair:
Parsing "foo", with tmp=[1, :a]
It's a leaf! Let's return a hash
Returning {"1_a"=>"foo"}
Parsing "bar", with tmp=[1, :b]
It's a leaf! Let's return a hash
Returning {"1_b"=>"bar"}
Returning {"1_a"=>"foo", "1_b"=>"bar"}
Parsing {:c=>"baz"}, with tmp=[2]
It's a hash! Let's parse every key,value pair:
Parsing "baz", with tmp=[2, :c]
It's a leaf! Let's return a hash
Returning {"2_c"=>"baz"}
Returning {"2_c"=>"baz"}
Returning {"1_a"=>"foo", "1_b"=>"bar", "2_c"=>"baz"}
Unlike the others, I have no love for each_with_object
:-). But I do like passing a single result hash around so I don't have to merge and remerge hashes over and over again.
def flattify(value, result = {}, path = [])
case value
when Array
value.each.with_index(1) do |v, i|
flattify(v, result, path + [i])
end
when Hash
value.each do |k, v|
flattify(v, result, path + [k])
end
else
result[path.join("_")] = value
end
result
end
(Some details adopted from Eric, see comments)
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