Question

我需要数一串中的桃子和茄子的数量，然后说出更多的数量。我尝试过：

def counting(eggplant_peaches)
  eggplants_counting = 0
  peaches_counting = 0
  (0..eggplant_peaches.length).each do |i|
    if eggplant_peaches[i] == 
      eggplants_counting = eggplants_counting + 1
    elsif eggplant_peaches[i] == 
      peaches_counting = peaches_counting + 1
    end
  end
  if eggplants_counting > peaches_counting
    puts “More ”
  elsif peaches_counting > eggplants_counting
    puts “More ”
  end
end

我得到一个错误：

undefined local variable or method '' for main:Object

我该如何计算代码并使代码少一点？

Answer 1

您的茄子和桃子必须用引号引起来，以使字符串正确，

if eggplant_peaches[i] == ""

现在，当您询问如何缩短代码长度时，您可以执行以下操作：

def counting(array)
  winner = array.group_by(&:itself).sort_by {|k,v| v.size }.last.first
  # The steps here are:
  # %w[  ].group_by(&:itself)
  #  => {""=>[""], ""=>["", ""]}
  # .sort_by { |k,v| v.size}
  #  => [["", [""]], ["", ["", ""]]]
  # .last
  #  => ["", ["", ""]]
  # .first
  #  => ""
  puts "More #{winner}"
end

counting(%w[  ])
 => More

作为奖励，以上代码也适用于香蕉：

counting(%w[     ])
 => More

Answer 2

更快的方法（频率计数）：

BitBlt

尽管您也可以在Marcin Kołodziej的决定中使用：max_by使其足够快：

array.each.with_object(Hash.new(0)) do |i, res|
  res[i] += 1
end.max_by(&:last).first

如何计算字符

2 个答案: