我正在尝试在 Ruby 上做一个股票选择器方法，但我的代码中有一些问题

Question

我正在尝试做一种选股方法，它可以接收一系列股票价格，每个假设日都有一个。它应该返回一对代表最佳买入日和最佳卖出日的日期。天数从 0 开始。

def stock_picker stocks
  pair = []

  if stocks.size < 2
    return "Please enter an array with a valid number of stocks"
  else
    buy_day = 0
    sell_day = 0
    profit = 0

    stocks.each_with_index do |buy, index|
      i = index
      while (i < stocks[index..-1].size)
        if ((buy - stocks[i]) > profit)
          profit = buy - stocks[i]
          buy_day = stocks.index(buy)
          sell_day = i
        end
        i+= 1
      end

    end
    pair = [buy_day,sell_day]
    return pair.inspect
  end
end

stock_picker([17,3,6,9,15,8,6,1,10])

它应该返回 [1,4] 而不是 [0,7]

Answer 1

您可以在 stock_prices 数组中循环选择具有最大正差的日子。您的 while 条件需要更改。

#steps
#sets value of biggest_profit to 0(biggest_loss if looking for loss)
#sets most_profitable_days to [nil,nil]
#loops through array
#takes buy day
#loops through remainder of array
#if current day-first day>biggest_profit (first_day-current_day for loss)
#make >= for shortest holding period
#reassign biggest_profit
#most_profitable_days.first=buy_day, most_profitable_days.last=sell_day
#sell_day & buy_day are values of indices

#tests
#must accept only array
#must return array
#must return correct array

def stock_picker(arr)

    #checks to make sure array inputs only are given
    raise 'Only arrays allowed' unless arr.instance_of?(Array)

    #sets value of biggest_profit to 0(biggest_loss if looking for loss)
    biggest_profit=0

    #sets most_profitable_days to [nil,nil]
    most_profitable_days=[nil,nil]

    #loops through array
    arr.each_with_index do |starting_price, buy_day|

        #takes buy day
        arr.each_with_index do |final_price,sell_day|

            #loops through remainder of array
            next if sell_day<=buy_day

            #if current day-first day>biggest_profit (first_day-current_day for loss)
            #make '>=' for shortest holding period
            if final_price-starting_price>=biggest_profit

              #reassign biggest_profit
              biggest_profit=final_price-starting_price

              #most_profitable_days.first=buy_day, 
              most_profitable_days[0]=buy_day#+1 #to make it more user friendly

              #most_profitable_days.last=sell_day
              most_profitable_days[-1]=sell_day#+1 #to make it more user friendly
            end
        end
    end
    
  #return most_profitable_days
  most_profitable_days
end
p stock_picker([3,2,5,4,12,3]) #[1,4]

Answer 2

另一种选择是在迭代数组时对数组进行切片以找到最佳利润：

res = ary.each_with_index.with_object([]) do |(buy_val, i), res|
  highest_val = ary[i..].max
  highest_idx = ary[i..].each_with_index.max[1] + i
  res << [highest_val - buy_val, i, highest_idx]
end.max_by(&:first)

#=> [12, 1, 4]

其中12 是利润，1 是买入指数，4 是卖出指数。

---

要了解它是如何工作的，请运行这个扩展版本，它比任何书面解释都更有价值：

res = []
ary.each_with_index do |buy_val, i|
  p buy_val
  p ary[i..]
  p highest_val = ary[i..].max
  p highest_idx = ary[i..].each_with_index.max[1] + i
  res << [highest_val - buy_val, i, highest_idx]
  p '----'
end

res #=> [[0, 0, 0], [12, 1, 4], [9, 2, 4], [6, 3, 4], [0, 4, 4], [2, 5, 8], [4, 6, 8], [9, 7, 8], [0, 8, 8]]

来自 Ruby 标准库，我使用了 Enumerable#each_with_index、Enumerable#each_with_object、Enumerable#max 和 Enumerable#max_by。

---

为了获得最大值的索引，我好心从 Chuck (https://stackoverflow.com/a/2149874) 那里偷了，谢谢和 +1。我没有寻找更好的选择。

根据链接帖子中 Cary Swoveland 的评论：

[..] a.index(a.max) 将返回第一个和 a.each_with_index.max[1] 将返回最后一个 [..] 的索引

所以，也许您想使用第一个选项来缩短买卖之间的时间。

Answer 3

使用Array#combination:

stocks.
    each_with_index.
    to_a.
    combination(2).
    select { |(_, idx1), (_, idx2)| idx2 > idx1 }.
    reduce([-1, [-1, -1]]) do |(val, acc), ((v1, idx1), (v2, idx2))|
  val < v2 - v1 ? [v2 - v1, [idx1, idx2]] : [val, acc]
end
#⇒ [ 12, [1, 4] ]