ruby – 打印文件中最常用的n个单词(字符串)

发布时间：2020-12-17 03:06:09 所属栏目：百科来源：网络整理

导读：目标： Write a function that takes two parameters: (1) a String representing a text document and (2) an integer providing the number of items to return. Implement the function such that it returns a list of Strings ordered by word frequenc

目标：

Write a function that takes two parameters: (1) a String representing a text document and (2) an integer providing the number of items to return. Implement the function such that it returns a list of Strings ordered by word frequency,the most frequently occurring word first. Use your best judgement to decide how words are separated. Your solution should run in O(n) time where n is the number of characters in the document.

我的想法是,在最坏的情况下,函数的输入可以是文档中的单词总数,从而减少了按照频率对单词进行排序的问题.这使我认为如果我使用比较排序方法,时间复杂度的下限将是O(n log n).所以,我的想法是,最好的方法是实现计数排序.这是我的代码.

我想告诉我,我的分析是否正确,我已经用我对时间复杂度的概念注释了代码,但它肯定是不正确的.这段代码的实际时间和空间复杂度是多少？如果有任何替代方法可以在实践中使用,我还想听听这是否是一个好方法.

### n is number of characters in string,k is number of words ###
def word_frequencies(string,n)
  words = string.split(/s/)  # O(n)
  max = 0
  min = Float::INFINITY
  frequencies = words.inject(Hash.new(0)) do |hash,word|  # O(k)
    occurrences = hash[word] += 1                     # O(1)
    max = occurrences if occurrences > max            # O(1)
    min = occurrences if occurrences < min            # O(1)
    hash;                                             # O(1)  
  end

  ### perform a counting sort ###
  sorted = Array.new(max + words.length)

  delta = 0

  frequencies.each do |word,frequency|   #O(k)
    p word + "--" + frequency.to_s
    index = frequency
    if sorted[index]
      sorted[index] = sorted[index].push(word)  # ??? I think O(1).
    else
      sorted[index] = [word]                    # O(1)
    end
  end

  return sorted.compact.flatten[-n..-1].reverse   
  ### Compact is O(k).  Flatten is O(k).  Reverse is O(k). So O(3k)
end

### Total --- O(n + 5k) = O(n).  Correct? 
### And the space complexity is O(n) for the hash + O(2k) for the sorted array.  
### So total O(n).


text = "hi hello hi my name is what what hi hello hi this is a test test test test hi hi hi what hello these are some words these these"

p word_frequencies(text,4)

解决方法

两种方式：

def word_counter(string,max)
  string.split(/s+/)
        .group_by{|x|x}
        .map{|x,y|[x,y.size]} 
        .sort_by{|_,size| size} # Have to sort =/
        .last(max)
end

def word_counter(string,max)

  # Create a Hash and a List to store values in.
  word_counter,max_storage = Hash.new(0),[]

  #Split the string an and add each word to the hash:
  string.split(/s+/).each{|word| word_counter[word] += 1}

  # Take each word and add it to the list (so that the list_index = word_count)
  # I also add the count,but that is not really needed
  word_counter.each{|key,val| max_storage[val] = [*max_storage[val]] << [key,val]}

  # Higher count will always be at the end,remove nils and get the last "max" elements.
  max_storage.compact.flatten(1).last(max)

end

（编辑：李大同）

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