词频计数

Question

这是我计算词频码

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 
    } 

当我有100K的数据将采取9.03秒，即，S来多少时间我可以计算出任何其它方式

THX提前

Answer 1

看看Facets gem

你可以做这样的事情使用frequency method

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

请注意，可以从word_arr中减去停用词。参考Array Documentation。