Ruby

Question

我正在尝试从 Ruby Monk 网站上解决这个练习，上面写着：

Try implementing a method called occurrences that accepts a string argument and uses inject to build a Hash. The keys of this hash should be unique words from that string. The value of those keys should be the number of times this word appears in that string.

我试过这样做：

def occurrences(str)
  str.split.inject(Hash.new(0)) { |a, i| a[i] += 1 }
end

但我总是得到这个错误：

TypeError: no implicit conversion of String into Integer

同时，这个的解决方案是完全相同的（我认为）：

def occurrences(str)
    str.scan(/\w+/).inject(Hash.new(0)) do |build, word| 
    build[word.downcase] +=1
    build
    end
end

Answer 1

好的，你的问题是你没有从块中返回正确的对象。（在你的情况下 Hash）

#inject 像这样工作

[a,b] 
 ^    -> evaluate block 
 |                      |
  -------return-------- V

在您的解决方案中，这就是正在发生的事情

def occurrences(str)
 str.split.inject(Hash.new(0)) { |a, i| a[i] += 1 }
end
#first pass a = Hash.new(0) and i = word
  #a['word'] = 0 + 1
  #=> 1 
#second pass uses the result from the first as `a` so `a` is now an integer (1). 
#So instead of calling Hash#[] it is actually calling FixNum#[] 
#which requires an integer as this is a BitReference in FixNum.Thus the `TypeError`

简单修复

def occurrences(str)
 str.split.inject(Hash.new(0)) { |a, i| a[i] += 1; a }
end
 #first pass a = Hash.new(0) and i = word
  #a['word'] = 0 + 1; a
  #=> {"word" => 1}

现在块 returns Hash 将再次传递给 a。如您所见，解决方案 returns 块末尾的对象 build 因此解决方案有效。

Ruby - 计算字符串中每个单词的重复次数

Ruby - Count each word repetition in a string

string

algorithm

split

hashmap