詞頻計數

Question

這是我計算詞頻碼

word_arr= ["I", "received", "this", "in", "email", "and", "found", "it", "a", "good", "read", "to", "share......", "Yes,", "Dr", "M.", "Bakri", "Musa", "seems", "to", "know", "what", "is", "happening", "in", "Malaysia.", "Some", "of", "you", "may", "know.", "He", "is", "a", "Malay", "extra horny", "horny nor", "nor their", "their babes", "babes are", "are extra", "extra SEXY..", "SEXY.. .", ". .", ". .It's", ".It's because", "because their", "their CONDOMS", "CONDOMS are", "are Made", "Made In", "In China........;)", "China........;) &&"] 

arr_stop_kwd=["a","and"] 

frequencies = Hash.new(0) 
    word_arr.each { |word| 
     if !arr_stop_kwd.include?(word.downcase) && !word.match('&&') 
     frequencies["#{word.downcase}"] += 1 
     end 
    } 

當我有100K的數據將採取9.03秒，即，S來多少時間我可以計算出任何其它方式

THX提前

Answer 1

看看Facets gem

你可以做這樣的事情使用frequency method

require 'facets' 
frequencies = (word_arr-arr_stop_kwd).frequency 

請注意，可以從word_arr中減去停用詞。參考Array Documentation。