local variable vs instance variable Ruby initialize - ruby

Question

I have a class in Ruby where I pass in a Hash of commodity prices. They are in the form
{"date (string)" => price (float), etc, etc}
and in the initialise method I convert the dates to Dates like so:
#data = change_key_format(dates)
But I notice that that method seems to change the original argument as well. Why is that? Here is the code:
def initialize(commodity_name, data)
puts "creating ...#{commodity_name}"
#commodity_name = commodity_name
#data = change_hash_keys_to_dates(data)
#dates = array_of_hash_keys(data)
puts data ######## UNCHANGED
#data = fix_bloomberg_dates(#data, #dates)
puts data ######## CHANGED -------------------- WHY???
#get_price_data
end
def fix_bloomberg_dates(data, dates)
#Fixes the bad date from bloomberg
data.clone.each do |date, price|
#Looks for obvious wrong date
if date < Date.strptime("1900-01-01")
puts dates[1].class
date_gap = (dates[1] - dates[2]).to_i
last_date_day = dates[1].strftime("%a %d %b")
last_date_day = last_date_day.split(" ")
last_date_day = last_date_day[0].downcase
#Correct the data for either weekly or daily prices
#Provided there are no weekend prices
if date_gap == 7 && last_date_day == "fri"
new_date = dates[1] + 7
data[new_date] = data.delete(date)
elsif date_gap == 1 && last_date_day == "thu"
new_date = dates[1] + 4
data[new_date] = data.delete(date)
else
new_date = dates[1] + 1
data[new_date] = data.delete(date)
end
end
end
return data
end
def change_hash_keys_to_dates(hash)
hash.clone.each do |k,v|
date = Date.strptime(k, "%Y-%m-%d")
#Transforms the keys from strings to dates format
hash[date] = hash.delete(k)
end
return hash
end
def array_of_hash_keys(hash)
keys = hash.map do |date, price|
date
end
return keys
end

Answer 1

Because of these lines:
data[new_date] = data.delete(date)
You're modifying the original data object. If you don't want to do this, create a copy of the object:
data2 = data.clone
and then replace all other references to data with data2 in your method (including return data2).