如何在Ruby中結合相似數據的數組？

我正在研究一個將比較兩個.csv文件的程序。將相關數據從一個csv文件中提取到一個數組數組後，我需要組合相關條目。例如，我希望把這個數組：如何在Ruby中結合相似數據的數組？

[["11/13/15", ["4001", "1392"], "INBOUND"], 
["11/13/15", ["4090", "540"], "INBOUND"], 
["11/13/15", ["1139", "162"], "INBOUND"], 
["11/13/15", ["1158", "64"], "INBOUND"], 
["11/13/15", ["4055", "352"], "OUTBOUND"], 
["11/13/15", ["4055", "448"], "OUTBOUND"], 
["11/13/15", ["4055", "352"], "OUTBOUND"], 
["11/13/15", ["1139", "162"], "OUTBOUND"], 
["11/13/15", ["1158", "64"], "OUTBOUND"], 
["11/13/15", ["4091", "520"], "OUTBOUND"]]

成這樣：

[["11/13/15", ["4001", "1392"], "INBOUND"], 
["11/13/15", ["4090", "540"], "INBOUND"], 
["11/13/15", ["1139", "162"], "INBOUND"], 
["11/13/15", ["1158", "64"], "INBOUND"], 
["11/13/15", ["4055", "1152"], "OUTBOUND"], 
["11/13/15", ["1139", "162"], "OUTBOUND"], 
["11/13/15", ["1158", "64"], "OUTBOUND"], 
["11/13/15", ["4091", "520"], "OUTBOUND"]]

對於數組的某些元素，如果項目在[0]，[1][0]和[2]比賽的另外一個，然後創建一個新項目（數組），其項目爲[1][1]，爲[1][1]上所有項目的總和，並刪除舊數組。如果它會更容易，我可以改變提取相關數據的方式，以便[1]的項目不是數組，而每行有4項而不是3個。

來源

2011-11-15 Sean Lerner

是連續結合這些元素？ – tokland

數據將被排序，以便它看起來像頂部數組如果打印，所以是的（如果我理解你的問題）。 –

假設托克蘭的回答真的是你想要的，你的問題就是結果數組中'[4] [1] [1]'的值有一個拼寫錯誤，這是唯一的關鍵值。它應該是1152，而不是1115.我必須說你的問題很sl。。 – sawa

我假設要分組的元素是連續的我們可以使用Enumerable#chunk。功能的方法：

grouped_xs = xs.chunk { |date, (id1, id2), direction| [date, id1, direction] } 
grouped_xs.map do |(date, id1, direction), ary| 
    id2_sum = ary.map { |date, (id1, id2), direction| id2.to_i }.inject(:+) 
    [date, id1, id2_sum.to_s, direction] 
end

輸出（你想要的輸出數組中4元，對吧？）：

[["11/13/15", "4001", "1392", "INBOUND"], 
["11/13/15", "4090", "540", "INBOUND"], 
["11/13/15", "1139", "162", "INBOUND"], 
["11/13/15", "1158", "64", "INBOUND"], 
["11/13/15", "4055", "1152", "OUTBOUND"], 
["11/13/15", "1139", "162", "OUTBOUND"], 
["11/13/15", "1158", "64", "OUTBOUND"], 
["11/13/15", "4091", "520", "OUTBOUND"]]

來源

2011-11-15 21:55:25 tokland

這是完美的。謝謝！ –

@肖恩：不客氣。就像一般性建議一樣，我認爲最好不要急於選擇一個答案太快，有人可能會想出一個更好的解決方案:-) – tokland

這是一些非常小巧的Ruby @tokland :) –

這應做到：

def lookup(list, id, direction) 
    index = nil 
    list.each_with_index do |e, i| 
    if (id == e[1][0]) and (e[2] == direction) 
     index = i 
     break 
    end 
    end 
    index 
end 

b = [] 

a.each do |e| 
    id = e[1][0] 
    direction = e[2] 
    i = lookup(b, id, direction) 
    if i.nil? 
    b << e 
    else 
    count = e[1][1].to_i 
    sum = count + b[i][1][1].to_i 
    b[i][1][1] = sum.to_s 
    end 
end 

b.each{|e| p e}

輸出：

["11/13/15", ["4001", "1392"], "INBOUND"] 
["11/13/15", ["4090", "540"], "INBOUND"] 
["11/13/15", ["1139", "162"], "INBOUND"] 
["11/13/15", ["1158", "64"], "INBOUND"] 
["11/13/15", ["4055", "1152"], "OUTBOUND"] 
["11/13/15", ["1139", "162"], "OUTBOUND"] 
["11/13/15", ["1158", "64"], "OUTBOUND"] 
["11/13/15", ["4091", "520"], "OUTBOUND"]

來源

2011-11-15 21:57:39

h = Hash.new(0) 
[["11/13/15", ["4001", "1392"], "INBOUND"], 
["11/13/15", ["4090", "540"], "INBOUND"], 
["11/13/15", ["1139", "162"], "INBOUND"], 
["11/13/15", ["1158", "64"], "INBOUND"], 
["11/13/15", ["4055", "352"], "OUTBOUND"], 
["11/13/15", ["4055", "448"], "OUTBOUND"], 
["11/13/15", ["4055", "352"], "OUTBOUND"], 
["11/13/15", ["1139", "162"], "OUTBOUND"], 
["11/13/15", ["1158", "64"], "OUTBOUND"], 
["11/13/15", ["4091", "520"], "OUTBOUND"]] 
.each{|a, (b, c), d| h[[a, b, d]] += c.to_i} 
p h.map{|(a, b, d), c| [a, [b, c], d]}

會給：

[["11/13/15", ["4001", 1392], "INBOUND"], 
["11/13/15", ["4090", 540], "INBOUND"], 
["11/13/15", ["1139", 162], "INBOUND"], 
["11/13/15", ["1158", 64], "INBOUND"], 
["11/13/15", ["4055", 1152], "OUTBOUND"], 
["11/13/15", ["1139", 162], "OUTBOUND"], 
["11/13/15", ["1158", 64], "OUTBOUND"], 
["11/13/15", ["4091", 520], "OUTBOUND"]]

來源

2011-11-15 22:30:23 sawa

而只是舉例 - 我的一行（與兩個1.8和1.9紅寶石作品）：

table = [["11/13/15", ["4001", "1392"], "INBOUND"], 
["11/13/15", ["4090", "540"], "INBOUND"], 
["11/13/15", ["1139", "162"], "INBOUND"], 
["11/13/15", ["1158", "64"], "INBOUND"], 
["11/13/15", ["4055", "352"], "OUTBOUND"], 
["11/13/15", ["4055", "448"], "OUTBOUND"], 
["11/13/15", ["4055", "352"], "OUTBOUND"], 
["11/13/15", ["1139", "162"], "OUTBOUND"], 
["11/13/15", ["1158", "64"], "OUTBOUND"], 
["11/13/15", ["4091", "520"], "OUTBOUND"]] 

result = table.group_by {|a, (b, c), d| [a, [b], d]}.map {|k, v| k[1] << v.map {|a| a[1][1].to_i}.inject(:+).to_s; k}

來源

2011-11-15 23:11:27

我投這個，但有一個錯字 - 最後] – pguardiario

更正，謝謝！ –

我特別喜歡「group_by {| a，（b，c），d |」 - 非常好。 – pguardiario

如何在Ruby中結合相似數據的數組？

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