Ruby : About_Scoring_Project (精确地说，是.map方法的迭代器)

Question

我是新来的鲁比。当我通过Ruby，About_Scoring_Project工作时，我遇到了一些问题。.map方法根据短语的顺序而不是变量的名称返回值。我很困惑..。

def score(dice)
  total_score = 0
  side = 6
  count = Array.new
  side.times{count << 0} # [0, 0, 0, 0, 0, 0]

  # Tell the occured time of each side
  while i = dice.pop
    count[i-1] += 1
  end

  # Calculating total score

  i = 0

  count.map! do |item|
    i += 1
    if item >= 3 && i != 1
      total_score = i*100
      item - 3
    elsif item >= 3
      total_score = 1000
      item - 3
    else
      item
    end
  end

  # Count the rest
  total_score += ( count[0]*100 + count[4]*50 )

  total_score # return the total score
end

虽然这个可以用，但是当我最初写的时候：

  elsif item >= 3
  item - 3
  total_score = 1000

执行时，数组计数为1000、0、0、0、0、0、0。

  score([1,1,1]) # score([1,1,1]) = 101000 which should be 1000

也就是说，它给了item的值1000，而不是给total_value。但是，当我改变两个短语的顺序时，它是正确的，就像上面指出的那样。能帮我个忙吗。我是Ruby和编程方面的新手。原谅我破碎的英语..。

项目背景：

# Greed is a dice game where you roll up to five dice to accumulate
# points.  The following "score" function will be used to calculate the
# score of a single roll of the dice.
#
# A greed roll is scored as follows:
#
# * A set of three ones is 1000 points
#
# * A set of three numbers (other than ones) is worth 100 times the
#   number. (e.g. three fives is 500 points).
#
# * A one (that is not part of a set of three) is worth 100 points.
#
# * A five (that is not part of a set of three) is worth 50 points.
#
# * Everything else is worth 0 points.
#
#
# Examples:
#
# score([1,1,1,5,1]) => 1150 points
# score([2,3,4,6,2]) => 0 points
# score([3,4,5,3,3]) => 350 points
# score([1,5,1,2,4]) => 250 points
#
# More scoring examples are given in the tests below:
#
# Your goal is to write the score method.

Answer 1

.map方法根据短语的顺序而不是变量的名称返回值。我很困惑..。

或多或少，是的。一个块只返回最后一个表达式的值，非常类似于一个方法。

如果你有：

count.map! do |item|
  item - 3
  total_score = 1000
end

然后，计算为total_score = 1000的1000是块的返回值。

如果你有：

count.map! do |item|
  total_score = 1000
  item - 3
end

然后item - 3是块的返回值。

map!依次创建一个新数组，其中包含来自块的值。

Answer 2

另一种方法是：

def three_of_a_kind(dice, count)
  count * (dice == 1 ? 1000 : 100 * dice)
end

def no_set(dice, count)
  case dice
  when 1
    count * 100
  when 5
    count * 50
  else
    0
  end
end

def score(rolls)
  rolls.group_by(&:itself).map do |dice, throws|
    n = throws.size
    three_of_a_kind(dice, n / 3) + no_set(dice, n % 3)
  end.inject(:+)
end

p score([1, 1, 1, 5, 1]) == 1150
p score([2, 3, 4, 6, 2]) == 0
p score([3, 4, 5, 3, 3]) == 350
p score([1, 5, 1, 2, 4]) == 250

顺便问一下：

  side = 6
  count = Array.new
  side.times{count << 0}

只是：

Array.new(6){ 0 }