rails array of hashes calculate one column - ruby-on-rails

Question

I have an array and it has many columns and I want to change one value of my one column.
My array is:
[
{
id: 1,
Districts: "Lakhisarai",
Area: 15.87,
Production: 67.77,
Productivity: 4271,
Year: 2015,
Area_Colour: "Red",
Production_Colour: "Orange",
Productivity_Colour: "Dark_Green",
created_at: "2018-07-24T11:24:13.000Z",
updated_at: "2018-07-24T11:24:13.000Z"
},
{
id: 29,
Districts: "Begusarai",
Area: 18.53,
Production: 29.35,
Productivity: 1584,
Year: 2015,
Area_Colour: "Red",
Production_Colour: "Red",
Productivity_Colour: "Orange",
created_at: "2018-07-24T11:24:13.000Z",
updated_at: "2018-07-24T11:24:13.000Z"
},
...
]
This is my sample array and I want my Productivity to be divided by 100 for that I am using one empty array and pushing these hashes to my array like:
j = []
b.map do |k|
if k.Productivity
u = k.Productivity/100
j.push({id: k.id, Productivity: u })
else
j.push({id: k.id, Productivity: k.Productivity })
end
Is there any simple way where I can generate this kind of array and reflect my changes to to one column. Is there any way where I don't need to push name of column one by one in push method.
I want to generate exact same array with one modification in productivity

Answer 1

let's say your array is e, then:
e.each { |item| item[:Productivity] = item[:Productivity]/100}
Example:
e = [{p: 12, d: 13}, {p:14, d:70}]
e.each { |item| item[:p] = item[:p]/10}
output: [{:p=>1, :d=>13}, {:p=>1, :d=>70}]

Answer 2

You could take help of map method here to create a new array from your original array, but with the mentioned changes.
ary.map do |elem|
h = elem.slice(:id)
h[:productivity] = elem[:Productivity] / 100 if elem[:Productivity]
h
end
=> [{:id=>1, :productivity=>42}, {:id=>29, :productivity=>15}]
Note, Hash#slice returns a new hash with only the key-value pairs for the keys passed in argument e.g. here, it returns { id: 1 } for first element.
Also, we are assigning the calculated productivity to the output only when it is set on original hash. Hence, the if condition there.