Ruby将第一个数组中的元素与第二个数组进行匹配

Question

I have two arrays. The first one will be an array of string with a name and the amount. The second is an array of letters.

a1 = ["ASLK 50", "BSKD 150", "ZZZZ 100", "BSDF 50"]
a2 = ["B", "Z"]

I want to create a third array to sort the contents from a1 based off a2 and return the number based on the information from first array. Since a2 has "B" and "Z", I need to scan first array for all entry starting with letter B and Z and add up the numbers.

My ultimate goal is to return the sum on third array, something like this:

a3 = ["B = 200", "Z = 100"]

Since "A" was not on a2, it is not counted.

I was able to extract the information from a1:

arr = a1.map{|el| el[0] + " : " + el.gsub(/\D/, '\1')}
#=> ["A : 50", "B : 150", "Z : 100", "B : 50"]

I am having trouble comparing a1with a2. I have tried different methods, such as:

a1.find_all{|i| i[0] == a2[0]} #=> only returns the first element of a2. How can I iterate through a2?

alternatively,

i = 0
arr_result = []
while i < (arr.length + 1)
    #(append to arr_result the number from  a1 if a1[i][0] matches a2[i])

I think either would solve it, but I can't put neither idea down to working code. How can I implement either method? Is there a more efficient way to do it?

Answer 1

Here is how I would do this:

a1 = ["ASLK 50", "BSKD 150", "ZZZZ 100", "BSDF 50"]
a2 = ["B", "Z"]
a3 = a1.each_with_object(Hash.new(0)) do |a,obj|
  obj[a[0]] += a.split.last.to_i if a2.include?(a[0]) 
end.map {|a| a.join(" = ")}
#=>["B = 200", "Z = 100"]

First step adds them all into a Hash by summing the values by each first letter that is contained in the second Array.

Second step provides the desired output

If you want a Hash instead just take off the last call to map and you'll have.

{"B" => 200, "Z" => 100}

Answer 2

Running with your requirements, that you want to turn this:

a1 = ["ASLK 50", "BSKD 150", "ZZZZ 100", "BSDF 50"]
a2 = ["B", "Z"]

into this: a3 = ["B = 200", "Z = 100"]

a3 = a2.map do |char|
  sum = a1.reduce(0) do |sum, item|
    name, price = item.split(" ")
    sum += price.to_i if name[0].eql?(char)
    sum
  end
  "#{char} = #{sum}"
end

Answer 3

Rather than defining an Array like ["B = 200", "Z = 100"] , it would be more sensible to define this as a mapping - ie the following Hash object:

{"B" => 200, "Z" => 100}

As for the implementation, there are many ways to do it. Here is just one approach:

a1 = ["ASLK 50", "BSKD 150", "ZZZZ 100", "BSDF 50"]
a2 = ["B", "Z"]
result = a2.map do |letter|
  [
    letter,
    a1.select {|str| str[0] == letter}
      .inject(0) {|sum, str| sum += str[/\d+/].to_i}
  ]
end.to_h

puts result # => {"B"=>200, "Z"=>100}

Explanation:

• At the top level, I've used Array#to_h to convert the array of pairs: [["B", 200], ["Z", 100]] into a Hash : {"B" => 200, "Z" => 100} .
• a1.select {|str| str[0] == letter} a1.select {|str| str[0] == letter} selects only the elements from a1 whose first letter is that of the hash key.
• inject(0) {|sum, str| sum += str[/\\d+/].to_i} inject(0) {|sum, str| sum += str[/\\d+/].to_i} adds up all the numbers, with safe-guards to default to zero (rather than having nil thrown around unexpectedly).

Answer 4

This is quite similar to @engineersmnky's answer.

r = /
    \A[A-Z] # match an upper case letter at the beginning of the string
    |       # or
    \d+     # match one or more digits
    /x      # free-spacing regex definition mode

a1.each_with_object(Hash.new(0)) do |s,h|
  start_letter, value = s.scan(r)
  h[start_letter] += value.to_i if a2.include?(start_letter)
end.map { |k,v| "#{k} = #{v}" }
  #=> ["B = 200", "Z = 100"] 

Hash.new(0) is often referred to as a "counting hash". See the doc for the class method Hash::new for an explanation.

The regex matches the first letter of the string or one or more digits. For example,

"ASLK 50".scan(r)
  #=> ["A", "50"]