紅寶石 - 計數每個單詞的重複在一個字符串

Question

我試圖解決這一運動從紅寶石僧網站，該網站說：

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.

我一直試圖做這樣的：

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

但我總是得到這樣的錯誤：

TypeError: no implicit conversion of String into Integer 

同時，該解決方案FO這一點是相當相同的（我認爲）：

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

Answer 1

好吧，所以你的問題是，你是不是從塊返回正確的對象。 （在你的情況下，Hash）

#inject是這樣的

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

在您的解決方案，這是發生了什麼事

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` 

簡單的解決

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} 

現在塊11返回Hash傳遞給再次。正如你所看到的解決方案返回塊的結尾處的對象build，因此該解決方案起作用。