Question

关于Codewars的计数笑脸问题，我的代码通过了所有测试，但是始终弹出“退出代码= 1”错误消息，这是什么意思？什么地方出了错？

countSmileys([':)', ';(', ';}', ':-D']);       // should return 2;
countSmileys([';D', ':-(', ':-)', ';~)']);     // should return 3;
countSmileys([';]', ':[', ';*', ':$', ';-D']); // should return 1;

    def count_smileys(arr)
      first = ";:"
      second = "-~"
      third = ")D"
      arr.select{|x|
        third.include?(x[1]) or (second.include?(x[1]) && third.include?(x[2].to_s)) 
      }.count
    end

编辑： 错误消息如下：

main.rb:8:in `include?': no implicit conversion of nil into String (TypeError)
    from main.rb:8:in `block in count_smileys'
    from main.rb:7:in `select'
    from main.rb:7:in `count_smileys'
    from main.rb:16:in `block in <main>'
    from /runner/frameworks/ruby/cw-2.rb:55:in `block in describe'
    from /runner/frameworks/ruby/cw-2.rb:46:in `measure'
    from /runner/frameworks/ruby/cw-2.rb:51:in `describe'
    from main.rb:11:in `<main>'

Answer 1

如消息所述，没有nil到字符串的隐式转换。显式确实存在：

2.3.1 :001 > nil.to_s
=> ""

您可以先为nil解析数组，然后通过select方法将其放入。

def count_smileys(arr)
  first = ";:"
  second = "-~"
  third = ")D"

  # parse your array for nil values here
  arr.map {|x| x.nil? ? "" : x }

  arr.select{|x|
    third.include?(x[1]) or (second.include?(x[1]) && third.include?(x[2].to_s)) 
  }.count
end

Answer 2

我意识到问题出在哪里-有一个测试count_smileys([";", ")", ";*", ":$", "8-D"])，其中x [1]和x [2]对于数组的前2个项无效，因此我需要在内部修复该数组选择方法：

def count_smileys(arr)
  first = ";:"
  second = "-~"
  third = ")D"
  arr.select{|x|
    x[1] = " " if x[1] == nil
    x[2] = "" if x[2] == nil
    (first.include?(x[0]) && third.include?(x[1])) || (first.include?(x[0]) && second.include?(x[1]) && third.include?(x[2])) 
  }.count
end

在需要转换nil的意义上，Joseph Cho是正确的，但是我们应该在迭代中进行转换，并且应该将公共项x [1]设置为带有空格的空字符串，以避免被计数，而x [2]很少见，以至于空字符串都可以工作。

Ruby Count笑脸-退出代码= 1

2 个答案: