Convert object with array values into array of object

Question

I do have this kind of params

params = { "people" => 
  {
    "fname" => ['john', 'megan'],
    "lname" => ['doe', 'fox']
  }
}

Wherein I loop through using this code

result = []
params["people"].each do |key, values|
  
  values.each_with_index do |value, i|
    result[i] = {}
    result[i][key.to_sym] = value
  end

end

The problem on my code is that it always gets the last key and value.

[
 { lname: 'doe' },
 { lname: 'fox' }
]

i want to convert it into

[
  {fname: 'john', lname: 'doe'},
  {fname: 'megan', lname: 'fox'}
]

so that i can loop through of them and save to database.

Cary Swoveland · Accepted Answer · 2022-12-30 18:45:03Z

2

Your question has been answered but I'd like to mention an alternative calculation that does not employ indices:

keys, values = params["people"].to_a.transpose
  #=> [["fname", "lname"], [["john", "megan"], ["doe", "fox"]]]

keys = keys.map(&:to_sym)
  #=> [:fname, :lname]

values.transpose.map { |val| keys.zip(val).to_h }
  #=> [{:fname=>"john", :lname=>"doe"},
  #    {:fname=>"megan", :lname=>"fox"}]

edited Dec 30, 2022 at 18:45

answered Dec 30, 2022 at 4:13

Cary Swoveland

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Comments

Silvio Mayolo · Accepted Answer · 2022-12-30 02:45:48Z

1

result[i] = {}

The problem is that you're doing this each loop iteration, which resets the value and deletes any existing keys you already put there. Instead, only set the value to {} if it doesn't already exist.

result[i] ||= {}

answered Dec 30, 2022 at 2:45

Silvio Mayolo

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Comments

LawrenceWebDev · Accepted Answer · 2022-12-30 02:43:50Z

1

In your inner loop, you're resetting the i-th element to an empty hash:

result[i] = {}

So you only end up with the data from the last key-value-pair, i.e. lname.

Instead you can use this to only set it to an empty hash if it doesn't already exist:

result[i] ||= {}

So the first loop through, it gets set to {}, but after that, it just gets set to itself.

Alternatively, you can also use

result[i] = {} if !result[i]

which may or may not be more performant. I don't know.

answered Dec 30, 2022 at 2:43

LawrenceWebDev

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