Question

我正在尝试创建一个方法来计算它从字典中使用单词的次数，并作为哈希返回。这是我现在的代码：

def substrings(words, dictionary)
  hash = {}
  substrings.downcase!
  dictionary.each do |substring|
    words.each do |word|
      if word.include? substring +=1
      end
    end
  end
 hash.to_s
end

dictionary = ["below", "down", "go", "going", "horn", "how", "howdy", "it", "i", "low", "own", "part", "partner", "sit"]
  words = "below"

substrings(words, dictionary)

我收到了这个错误：

wrong number of arguments (given 0, expected 2)

我正在寻找类似的东西：

=> {"below"=>1, "low"=>1}

我尝试了很多东西，但它从来没有给我那个哈希。我要么得到一个未定义的方法错误，要么：

=> ["below", ["below", "down", "go", "going", "horn", "how", "howdy", "it", "i", "low", "own", "part", "partner", "sit"]]

Answer 1

您的错误是由“substrings.downcase！”行引起的。这是对子串方法的递归调用，它接受两个参数，而您不提供任何参数。如果不是这种情况，您仍然会收到错误，这是由此代码的无限递归引起的堆栈溢出。

Answer 2

这会产生预期的结果，但我会words交换word：

def substrings(word, dictionary)
  word = word.downcase
  dictionary.select { |entry| word.include?(entry.downcase) }
            .group_by(&:itself)
            .map { |k, v| [k, v.size] }.to_h
end

这导致：

>> dictionary = ["below", "down", "go", "going", "horn", "how", "howdy", "it", "i", "low", "own", "part", "partner", "sit"]
>> word = 'below'
>> substrings(word, dictionary)
=> {"below"=>1, "low"=>1}

并计算多个单词的副本，虽然没有明确说明，但大概是你所追求的：

>> dictionary = ["below", "be", "below", "below", "low", "be", "pizza"]
>> word = 'below'
>> substrings(word, dictionary)
=> {"below"=>3, "be"=>2, "low"=>1}

Answer 3

您可以使用#reduce：

def substrings(sentence, dictionary)
  sentence = sentence.downcase
  dictionary.reduce(Hash.new(0)) do |counts,word| 
    counts[word] +=1 if sentence.include?(word.downcase)
    counts
  end
end

dictionary = ["below", "down", "go", "going", "horn", "how", "howdy", "it", "i", "low", "own", "part", "partner", "sit"]
sentence = "below"

substrings(sentence, dictionary) #=> {"below"=>1, "low"=>1}

或#each：

def substrings(sentence, dictionary)
  sentence = sentence.downcase
  counts = Hash.new(0) # Makes the default value `0` instead of `nil`
  dictionary.each do |word| 
    if sentence.include?(word.downcase)
      counts[word] += 1
    end
  end
  counts
end

dictionary = ["below", "down", "go", "going", "horn", "how", "howdy", "it", "i", "low", "own", "part", "partner", "sit"]
sentence = "below"

substrings(sentence, dictionary) #=> {"below"=>1, "low"=>1}

错误的参数和哈希问题

3 个答案: