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I'm getting data from an API and need to format it differently. I have a car_array that consists of an array of hashes. However, sometimes there will be a sub-array as one of the hash values that contains more than 1 element. In this case, there should be a loop so that each element in the array gets mapped correctly as separate entries.

An example of data, note how prices and options_package are arrays with multiple elements.

[{ dealer_id: 1, dealer_name: "dealership 1", car_make: "jeep", prices: ['30', '32', '35'], options_package: ['A', 'B', 'C'] }, { dealer_id: 2, dealer_name: "dealership 2", car_make: "ford", prices: ['50', '55'], options_package: ['X', 'Y'] }, { dealer_id: 3, dealer_name: "dealership 3", car_make: "dodge", prices: ['70'], options_package: ['A'] }]

I would like to create multiple entries when there are multiple array elements

for example, the data above should be broken out and mapped as:

some_array = [ { dealer_id: 1, dealer_name: "dealership 1", car_make: "jeep", price: '30', options_package: 'A' }, { dealer_id: 1, dealer_name: "dealership 1", car_make: "jeep", price: '32', options_package: 'B' }, { dealer_id: 1, dealer_name: "dealership 1", car_make: "jeep", price: '35', options_package: 'C' }, { dealer_id: 2, dealer_name: "dealership 2", car_make: "ford", price: '50', options_package: 'X' }, { dealer_id: 2, dealer_name: "dealership 2", car_make: "ford", price: '55', options_package: 'Y' }, { dealer_id: 3, dealer_name: "dealership 3", car_make: "dodge", price: '70', options_package: 'A' } ]

Here's what I've got so far:

car_arr.each do |car| if car['Prices'].length > 1 # if there are multiple prices/options loop through each one and create a new car car.each do |key, value| if key == 'Prices' value.each do |price| formatted_car_array << { dealer_id: car['dealer_id'], dealer_name: car['dealer_name'], car_make: car['make'], options_package: ???????, price: price, } end end end else # there's only element for price and options_package formatted_car_array << { dealer_id: car['dealer_id'], dealer_name: car['dealer_name'], car_make: car['make'], options_package: car['options_package'], price: car['prices'] } end end
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4 Answers 4

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Consider just one hash to start with, and how this problem can be solved for this simpler problem.

h = { dealer_id: 1, dealer_name: "dealership 1", car_make: "jeep", prices: ['30', '32', '35'], options_package: ['A', 'B', 'C'] }

Let's get combinations of prices and options packages using #zip.

h[:prices].zip(h[:options_package]) # => [["30", "A"], ["32", "B"], ["35", "C"]]

The length of 3 for this array corresponds with how many hashes we expect to get from it, so let's map over those, building a new hash each time.

h[:prices].zip(h[:options_package]).map do |price, pkg| { dealer_id: h[:dealer_id], dealer_name: h[:dealer_name], car_make: h[:car_make], price: price, options_package: pkg, } end # => [{:price=>"30", :options_package=>"A", :dealership=>nil, :dealer_id=>1, :car_make=>"jeep"}, # {:price=>"32", :options_package=>"B", :dealership=>nil, :dealer_id=>1, :car_make=>"jeep"}, # {:price=>"35", :options_package=>"C", :dealership=>nil, :dealer_id=>1, :car_make=>"jeep"}]

Now you just need to #flat_map this over your array.

car_arr.flat_map do |h| h[:prices].zip(h[:options_package]).map do |price, pkg| { dealer_id: h[:dealer_id], dealer_name: h[:dealer_name], car_make: h[:car_make], price: price, options_package: pkg, } end end # => [{:price=>"30", :options_package=>"A", :dealership=>nil, :dealer_id=>1, :car_make=>"jeep"}, # {:price=>"32", :options_package=>"B", :dealership=>nil, :dealer_id=>1, :car_make=>"jeep"}, # {:price=>"35", :options_package=>"C", :dealership=>nil, :dealer_id=>1, :car_make=>"jeep"}, # {:price=>"50", :options_package=>"X", :dealership=>nil, :dealer_id=>2, :car_make=>"ford"}, # {:price=>"55", :options_package=>"Y", :dealership=>nil, :dealer_id=>2, :car_make=>"ford"}, # {:price=>"70", :options_package=>"A", :dealership=>nil, :dealer_id=>3, :car_make=>"dodge"}]
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Comments

2

If arr is the array of hashes given in the question one may write the following.

arr.flat_map do |h| h[:prices].zip(h[:options_package]).map do |p,o| h.reject { |k,_| k == :prices }.merge(price: p, options_package: o) end end #=> [{:dealer_id=>1, :dealer_name=>"dealership 1", :car_make=>"jeep", # :options_package=>"A", :price=>"30"}, # {:dealer_id=>1, :dealer_name=>"dealership 1", :car_make=>"jeep", # :options_package=>"B", :price=>"32"}, # {:dealer_id=>1, :dealer_name=>"dealership 1", :car_make=>"jeep", # :options_package=>"C", :price=>"35"}, # {:dealer_id=>2, :dealer_name=>"dealership 2", :car_make=>"ford", # :options_package=>"X", :price=>"50"}, # {:dealer_id=>2, :dealer_name=>"dealership 2", :car_make=>"ford", # :options_package=>"Y", :price=>"55"}, # {:dealer_id=>3, :dealer_name=>"dealership 3", :car_make=>"dodge", # :options_package=>"A", :price=>"70"}]

Note that this code need not be changed if key-value pairs are added to the hash or if the keys :dealer_id and :dealer_name are deleted or renamed.

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2

My answer is an extension upon the answer of Chris.

car_arr.flat_map do |h| h[:prices].zip(h[:options_package]).map do |price, pkg| { dealer_id: h[:dealer_id], dealer_name: h[:dealer_name], car_make: h[:car_make], price: price, options_package: pkg, } end end

This answer could be shortened if you're using the newest Ruby version.

Ruby 3.0 introduced Hash#except, which lets you easily create a copy of a hash without the specified key(s). This allows us to reduce the answer to:

cars.flat_map do |car| car[:prices].zip(car[:options_package]).map do |price, pkg| car.except(:prices).merge(price: price, options_package: pkg) end end

car.except(:prices) will create a new hash without the :prices key/value-pair, we don't need to remove :options_package, since merge will overwrite the the old :options_package value with a new value.

Ruby 3.1 introduced { x:, y: } as syntax sugar for { x: x, y: y }. This allows us to "reduce" the answer further to:

cars.flat_map do |car| car[:prices].zip(car[:options_package]).map do |price, options_package| car.except(:prices).merge(price:, options_package:) end end
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1 Comment

It does make the answer a bit longer, since 2 * len(options_package) = 30 and len(options_package) + 2 * len(pkg) = 21, but it is more expressive and probably improves readability. I've updated the answer.
1

Input

input = [{ dealer_id: 1, dealer_name: "dealership 1", car_make: "jeep", prices: ['30', '32', '35'], options_package: ['A', 'B', 'C'] }, { dealer_id: 2, dealer_name: "dealership 2", car_make: "ford", prices: ['50', '55'], options_package: ['X', 'Y'] }, { dealer_id: 3, dealer_name: "dealership 3", car_make: "dodge", prices: ['70'], options_package: ['A'] }]

Code

result = input.map do |h| prices = h[:prices] options_package = h[:options_package] h[:options_package].count.times.map do h[:prices] = prices.shift h[:options_package] = options_package.shift h.dup end end p result.flatten

Output

[{:dealer_id=>1, :dealer_name=>"dealership 1", :car_make=>"jeep", :prices=>"30", :options_package=>"A"}, {:dealer_id=>1, :dealer_name=>"dealership 1", :car_make=>"jeep", :prices=>"32", :options_package=>"B"}, {:dealer_id=>1, :dealer_name=>"dealership 1", :car_make=>"jeep", :prices=>"35", :options_package=>"C"}, {:dealer_id=>2, :dealer_name=>"dealership 2", :car_make=>"ford", :prices=>"50", :options_package=>"X"}, {:dealer_id=>2, :dealer_name=>"dealership 2", :car_make=>"ford", :prices=>"55", :options_package=>"Y"}, {:dealer_id=>3, :dealer_name=>"dealership 3", :car_make=>"dodge", :prices=>"70", :options_package=>"A"}]
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2 Comments

Bear in mind this will mutate the original hash, which is not necessarily wrong but does need to be considered. Also hash may be a slightly misleading name for an array of hashes.
hash name might be misleading. Yes. Thanks for letting me know.

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