I'm trying to build a simple method to look at about 100 entries in a database for a last name and pull out all the ones that match above a specific percentage of letters. My current approach is:
This works, but I feel like there must be some cool ruby/regex/active record method of doing this more efficiently. I have googled quite a bit but can't find anything.
To comment on the merit of the measure you suggested would require speculation, which is out-of-bounds at SO. I therefore will merely demonstrate how you might implement your proposed approach.
Code
First define a helper method:
class Array
def difference(other)
h = other.each_with_object(Hash.new(0)) { |e,h| h[e] += 1 }
reject { |e| h[e] > 0 && h[e] -= 1 }
end
end
In short, if
a = [3,1,2,3,4,3,2,2,4]
b = [2,3,4,4,3,4]
then
a - b #=> [1]
whereas
a.difference(b) #=> [1, 3, 2, 2]
This method is elaborated in my answer to this SO question . I've found so many uses for it that I've proposed it be added to the Ruby Core .
The following method produces a hash whose keys are the elements of names
(strings) and whose values are the fractions of the letters in the target
string that are contained in each string in names
.
def target_fractions(names, target)
target_arr = target.downcase.scan(/[a-z]/)
target_size = target_arr.size
names.each_with_object({}) do |s,h|
s_arr = s.downcase.scan(/[a-z]/)
target_remaining = target_arr.difference(s_arr)
h[s] = (target_size-target_remaining.size)/target_size.to_f
end
end
Example
target = "Jimmy S. Bond"
and the names you are comparing are given by
names = ["Jill Dandy", "Boomer Asad", "Josefine Simbad"]
then
target_fractions(names, target)
#=> {"Jill Dandy"=>0.5, "Boomer Asad"=>0.5, "Josefine Simbad"=>0.8}
Explanation
For the above values of names
and target
,
target_arr = target.downcase.scan(/[a-z]/)
#=> ["j", "i", "m", "m", "y", "s", "b", "o", "n", "d"]
target_size = target_arr.size
#=> 10
Now consider
s = "Jill Dandy"
h = {}
then
s_arr = s.downcase.scan(/[a-z]/)
#=> ["j", "i", "l", "l", "d", "a", "n", "d", "y"]
target_remaining = target_arr.difference(s_arr)
#=> ["m", "m", "s", "b", "o"]
h[s] = (target_size-target_remaining.size)/target_size.to_f
#=> (10-5)/10.0 => 0.5
h #=> {"Jill Dandy"=>0.5}
The calculations are similar for Boomer and Josefine.
The technical post webpages of this site follow the CC BY-SA 4.0 protocol. If you need to reprint, please indicate the site URL or the original address.Any question please contact:yoyou2525@163.com.