如何在Ruby中对同一日期的列进行求和

Question

我有一个包含这些类型数据的数组，并且我需要总结具有相同日期的列。

[["01-04-2013", 100.0, 110.0, 120, 0, 0, 0], ["02-04-2013", 100.0, 110.0, 130, 0, 0, 0], ["03-04-2013", 100.0, 110.0, 120, 0, 0, 0], ["10-04-2013", 100.0, 110.0, 100, 0, 0, 0], ["02-04-2013", 100.0, 140.0, 0, 70, 0, 0], ["10-04-2013", 100.0, 140.0, 0, 100, 0, 0], ["11-04-2013", 100.0, 140.0, 0, 110, 0, 0], ["12-04-2013", 100.0, 140.0, 0, 120, 0, 0], ["09-04-2013", 0.0, 0.0, 0, 0, 130, 0], ["17-04-2013", 0.0, 0.0, 0, 0, 30, 0], ["15-04-2013", 100.0, 130.0, 0, 0, 0, 17], ["17-04-2013", 100.0, 130.0, 0, 0, 0, 90], ["18-04-2013", 100.0, 130.0, 0, 0, 0, 100]] 

我该怎么做红宝石？我的意思是将同一日期的行整理成一行，如果没有重复的日期，请保留旧的。评论后

Answer 1

aggregated_rows = rows.group_by(&:first).map do |date, rows_by_date| 
    values = rows_by_date.transpose.drop(1).map { |xs| xs.reduce(:+) } 
    [date, values] 
end 

#[["01-04-2013", [100.0, 110.0, 120, 0, 0, 0]], 
# ["02-04-2013", [200.0, 250.0, 130, 70, 0, 0]], 
... 
# ["18-04-2013", [100.0, 130.0, 0, 0, 0, 100]]] 

Answer 2

require 'pp' 
require 'matrix' 

d = [["01-04-2013", 100.0, 110.0, 120, 0, 0, 0], ["02-04-2013", 100.0, 110.0, 130, 0, 0, 0], ["03-04-2013", 100.0, 110.0, 120, 0, 0, 0], ["10-04-2013", 100.0, 110.0, 100, 0, 0, 0], ["02-04-2013", 100.0, 140.0, 0, 70, 0, 0], ["10-04-2013", 100.0, 140.0, 0, 100, 0, 0], ["11-04-2013", 100.0, 140.0, 0, 110, 0, 0], ["12-04-2013", 100.0, 140.0, 0, 120, 0, 0], ["09-04-2013", 0.0, 0.0, 0, 0, 130, 0], ["17-04-2013", 0.0, 0.0, 0, 0, 30, 0], ["15-04-2013", 100.0, 130.0, 0, 0, 0, 17], ["17-04-2013", 100.0, 130.0, 0, 0, 0, 90], ["18-04-2013", 100.0, 130.0, 0, 0, 0, 100]] 

pp(
    d.group_by(&:first).values.reject do |v| 
    v.size <= 1 
    end.map do |e| 
    e.inject do |m, e| 
     (Vector.[](*m) + Vector.[](*e)).to_a 
    end 
    end 
) 

更新：

d.group_by(&:first).values.map do |e| 
    e.inject do |m, e| 
     [e[0], (Vector.[](*m[1..-1]) + Vector.[](*e[1..-1])).to_a].flatten 
    end 
    end.sort 

规格更改警报：

def v m 
    Vector.[](*m.drop(1)) 
end 

d.group_by(&:first).values.map do |group| 
    r = group.inject do |m, e| 
    [e[0], *(v(m) + v(e)).to_a] 
    end 
    r[1] /= group.size 
    r[2] /= group.size 
    r 
end.sort 

---

^{注意。
我不是说这是作业，但是在很多情况下，应该很明显的是，当我们为学生做的时候，我们并没有真正地对他们做任何好处，对吧？另外，这个解决方案是在一个公共站点上提供的，该站点可以立即被谷歌索引，并且在世界的前100个站点中，这对于教授或者分级者来说并不是秘密。如果学校使用像​​这样的国家数据库怎么办？我想他们可以检查公共代码片段，如果他们想。最后，还有一些相当精心编写的代码，发布在SO，由嗜好爱好者。如果我自己这样说，我不确定它通常可以通过低级别的入门课程原创作品。 :-)}

Answer 3

require 'pp' 

a = [["01-04-2013", 100.0, 110.0, 120, 0, 0, 0], ["02-04-2013", 100.0, 110.0, 130, 0, 0, 0], ["03-04-2013", 100.0, 110.0, 120, 0, 0, 0], ["10-04-2013", 100.0, 110.0, 100, 0, 0, 0], ["02-04-2013", 100.0, 140.0, 0, 70, 0, 0], ["10-04-2013", 100.0, 140.0, 0, 100, 0, 0], ["11-04-2013", 100.0, 140.0, 0, 110, 0, 0], ["12-04-2013", 100.0, 140.0, 0, 120, 0, 0], ["09-04-2013", 0.0, 0.0, 0, 0, 130, 0], ["17-04-2013", 0.0, 0.0, 0, 0, 30, 0], ["15-04-2013", 100.0, 130.0, 0, 0, 0, 17], ["17-04-2013", 100.0, 130.0, 0, 0, 0, 90], ["18-04-2013", 100.0, 130.0, 0, 0, 0, 100]] 
h = {} 
a.group_by(&:first).each{|k,v| v.flatten!.delete(k); h[k] = v.inject(:+)} 
pp h 

输出：

{"01-04-2013"=>330.0, 
"02-04-2013"=>650.0, 
"03-04-2013"=>330.0, 
"10-04-2013"=>650.0, 
"11-04-2013"=>350.0, 
"12-04-2013"=>360.0, 
"09-04-2013"=>130.0, 
"17-04-2013"=>350.0, 
"15-04-2013"=>247.0, 
"18-04-2013"=>330.0} 

---

pp a.group_by(&:first).map{|k,v| v.flatten!.uniq!} 

输出：

[["01-04-2013", 100.0, 110.0, 120, 0], 
["02-04-2013", 100.0, 110.0, 130, 0, 140.0, 70], 
["03-04-2013", 100.0, 110.0, 120, 0], 
["10-04-2013", 100.0, 110.0, 100, 0, 140.0], 
["11-04-2013", 100.0, 140.0, 0, 110], 
["12-04-2013", 100.0, 140.0, 0, 120], 
["09-04-2013", 0.0, 0, 130], 
["17-04-2013", 0.0, 0, 30, 100.0, 130.0, 90], 
["15-04-2013", 100.0, 130.0, 0, 17], 
["18-04-2013", 100.0, 130.0, 0, 100]] 

---

pp a.group_by(&:first).map{|k,v| v.transpose.map!{|a| a.inject(:+)}} 

输出：

[["01-04-2013", 100.0, 110.0, 120, 0, 0, 0], 
["02-04-201302-04-2013", 200.0, 250.0, 130, 70, 0, 0], 
["03-04-2013", 100.0, 110.0, 120, 0, 0, 0], 
["10-04-201310-04-2013", 200.0, 250.0, 100, 100, 0, 0], 
["11-04-2013", 100.0, 140.0, 0, 110, 0, 0], 
["12-04-2013", 100.0, 140.0, 0, 120, 0, 0], 
["09-04-2013", 0.0, 0.0, 0, 0, 130, 0], 
["17-04-201317-04-2013", 100.0, 130.0, 0, 0, 30, 90], 
["15-04-2013", 100.0, 130.0, 0, 0, 0, 17], 
["18-04-2013", 100.0, 130.0, 0, 0, 0, 100]] 

Answer 4

a = [["01-04-2013", 100.0, 110.0, 120, 0, 0, 0], ["02-04-2013", 100.0, 110.0, 130, 0, 0, 0], ["03-04-2013", 100.0, 110.0, 120, 0, 0, 0], ["10-04-2013", 100.0, 110.0, 100, 0, 0, 0], ["02-04-2013", 100.0, 140.0, 0, 70, 0, 0], ["10-04-2013", 100.0, 140.0, 0, 100, 0, 0], ["11-04-2013", 100.0, 140.0, 0, 110, 0, 0], ["12-04-2013", 100.0, 140.0, 0, 120, 0, 0], ["09-04-2013", 0.0, 0.0, 0, 0, 130, 0], ["17-04-2013", 0.0, 0.0, 0, 0, 30, 0], ["15-04-2013", 100.0, 130.0, 0, 0, 0, 17], ["17-04-2013", 100.0, 130.0, 0, 0, 0, 90], ["18-04-2013", 100.0, 130.0, 0, 0, 0, 100]] 

require "pp" 

def group_and_sum_rows_by_date_string(a) 
    # instantiate a hash that returns an empty array for a key 
    # that doesn't exist 
    h = Hash.new([]) 
    a.each do |row| 
     # populate the hash with date string as key, and array of 
     # arrays of the values for that date string 
     h[k=row.shift] = ([row] + h[k]).compact 
    end 
    # add up all the corresponding values in each element's array 
    # arrays, and return the result as an array 
    h.map{|k, v| [k, v.transpose.map{|x| x.inject(:+)}]} 
end 

pp group_and_sum_rows_by_date_string(a) 

[["15-04-2013", [100.0, 130.0, 0, 0, 0, 17]], 
["03-04-2013", [100.0, 110.0, 120, 0, 0, 0]], 
["02-04-2013", [200.0, 250.0, 130, 70, 0, 0]], 
["17-04-2013", [100.0, 130.0, 0, 0, 30, 90]], 
["18-04-2013", [100.0, 130.0, 0, 0, 0, 100]], 
["09-04-2013", [0.0, 0.0, 0, 0, 130, 0]], 
["01-04-2013", [100.0, 110.0, 120, 0, 0, 0]], 
["12-04-2013", [100.0, 140.0, 0, 120, 0, 0]], 
["10-04-2013", [200.0, 250.0, 100, 100, 0, 0]], 
["11-04-2013", [100.0, 140.0, 0, 110, 0, 0]]]