在ruby数组中计数重复项，并将计数存储在hash中

Question

我有一本单词词典，我想检查给定的字符串是否有这些单词。我希望它们存储在一个散列中，密钥是重复的单词，值是它发生的次数。

目前，它只会存储完整的字符串匹配(下面不算作包含单词low(，并且实际上不会增加重复的计数器。

给我指正确的方向？：(

dictionary = ["below","down","go","going","horn","how","howdy","it","i","low","own","part","partner","sit"]
def substringer(string, dict)
string_array = string.split(/W+/)
final_hash = {}
count = 0
dict.each do |entry|
if string_array.include?(entry)
final_hash = {entry => +1}
p final_hash
end
end
end
substringer("below, below, how's it goin?", dictionary)

结果

{"below"=>1}
{"how"=>1}
{"it"=>1}

Answer 1

这是我的"一个衬垫"；解决方案：

dictionary = ["below", "down", "go", "going", "horn", "how", "howdy", "it", "i", "low", "own", "part", "partner", "sit"]
str = "below, below, how's it goin?"
str.split(/W+/).tally.slice(*dictionary) #=> {"below"=>2, "how"=>1, "it"=>1}

Answer 2

string_array.include?(entry)返回true，无论该单词在给定数组中出现多少次。相反，使用string_array.count(entry)，它会告诉你它出现的确切次数。

dict.each do |entry|
count = string_array.count(entry)
final_hash[entry] = count if count > 0
end

然而，这并不是最有效的方法，因为每个字典单词都需要遍历string_array一次(我认为string_aarray可能比dictionary大得多(。试着考虑一种在字符串数组上迭代的方法。

另外，考虑一下如何处理以大写字母开头的单词！

Answer 3

让我们反转代码并循环string_array中的元素，然后检查是否包含在dict中。

同样使用#each_with_object构建散列，Hash.new(0)创建一个散列，其中键的默认值为0。

dictionary = ["below", "down", "go", "going", "horn", "how", "howdy", "it", "i", "low", "own", "part", "partner", "sit"]
def substringer(string, dict)
string_array = string.split(/W+/)
final_hash = string_array.each_with_object(Hash.new(0)) do |word, hsh|
hsh[word] += 1 if dict.include?(word)
end
p final_hash
end

测试：

substringer("below, below, how's it goin?", dictionary)
# {"below"=>2, "how"=>1, "it"=>1}

Answer 4

如果要评估多个字符串，我们可以编写以下内容。

require 'set'
dictionary = %w|
below down go going horn how howdy
it i low own part partner sit
|
DICTIONARY = dictionary.to_set
#=> #<Set: {"below", "down", "go", "going", "horn", "how", "howdy",
#           "it", "i", "low", "own", "part", "partner", "sit"}>

def doit(str)
str.scan(/w+/).map(&:downcase).tally.select { |word| DICTIONARY.include?(word) }
end

doit "Below, below, how's it goin?"
#=> {"below"=>2, "how"=>1, "it"=>1}

doit "The sub's skipper said 'howdy' before going below then the sub went down below"
#=>{"howdy"=>1, "going"=>1, "below"=>2, "down"=>1}

请参阅String#扫描、Enumerable#计数和Hash#选择。

Set查找非常快，特别是语句的执行

DICTIONARY.include?(word)