從電子郵件地址列表中選擇最常用的域名

我有一個生成一個隨機的電子郵件地址的功能：從電子郵件地址列表中選擇最常用的域名

def emails 
    names = ["alfred", "daniel", "elisa", "ana", "ramzes"] 
    surnames = ["oak", "leaf", "grass", "fruit"] 
    providers = ["gmail", "yahoo", "outlook", "icloud"] 
    address = "#{names.sample}.#{surnames.sample}#{rand(100..5300)}@#{providers.sample}.com" 
end

鑑於隨機生成的電子郵件地址的列表：

email_list = 100.times.map { emails }

，看起來像這樣：

[email protected] 
[email protected] 
[email protected] 
...

我該如何選擇最常見的提供者（「gmail」，「yahoo」等）？

來源

2017-01-27 Ewelina

@AndreyDeineko：請注意，與「重複」相比有一個轉折點。 –

email_list = 10.times.map { emails } 
    #=> ["[email protected]", "[email protected]", 
    # "[email protected]", "[email protected]", 
    # "[email protected]", "[email protected]", 
    # "[email protected]", "[email protected]", 
    # "[email protected]", "[email protected]"] 

email_list.group_by { |s| s[/@\K.+/] }.max_by { |_,v| v.size }.first 
    #=> "gmail.com"

在正則表達式\K意味着不顧一切至今匹配。或者，@\K可能被替換正反向(?<[email protected])。

步驟如下。

h = email_list.group_by { |s| s[/@\K.+/] } 
    #=> {"gmail.com" =>["[email protected]", "[email protected]", 
    #     "[email protected]"], 
    # "icloud.com" =>["[email protected]", "[email protected]"], 
    # "outlook.com"=>["[email protected]", "[email protected]", 
    #     "[email protected]"], 
    # "yahoo.com" =>["[email protected]", "[email protected]"]} 
a = h.max_by { |_,v| v.size } 
    #=> ["gmail.com", ["[email protected]", "[email protected]", 
    #     "[email protected]"]] 
a.first 
    #=> "gmail.com"

如果像這裏那樣最頻繁出現平局，請修改代碼以獲得所有獲勝者。

h = email_list.group_by { |s| s[/@\K.+/] } 
    # (same as above) 
mx_size = h.map { |_,v| v.size }.max 
    #=> 3 
h.select { |_,v| v.size == mx_size }.keys 
    #=> ["gmail.com", "outlook.com"]

來源

2017-01-27 19:29:31

您的問題與此相似one。然而，有一個轉折：你不想分析電子郵件地址的頻率，但他們的提供者。

def random_email 
    names = ["alfred", "daniel", "elisa", "ana", "ramzes"] 
    surnames = ["oak", "leaf", "grass", "fruit"] 
    providers = ["gmail", "yahoo", "outlook", "icloud"] 
    address = "#{names.sample}.#{surnames.sample}#{rand(100..5300)}@#{providers.sample}.com" 
end 

emails = Array.new(100){ random_email } 

freq = emails.each_with_object(Hash.new(0)) do |email,freq| 
    provider = email.split('@').last 
    freq[provider] += 1 
end 

p freq 
#=> {"outlook.com"=>24, "yahoo.com"=>28, "gmail.com"=>32, "icloud.com"=>16} 

p freq.max_by{|provider, count| count}.first 
#=> "gmail.com"

來源

2017-01-27 12:52:20

非常感謝！對此，我真的非常感激！ – Ewelina

你也可以爲最後一行代碼寫'freq.max_by（＆：last）.first'。 –

從電子郵件地址列表中選擇最常用的域名

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