Ruby – 如何通过条件最大值的多个键检索数组中的和
原始数组是
[ {"id"=>2, "idx"=>111, "money"=>"4.00", "money1"=>"1.00", "order"=>"001", "order1"=>"1"}, {"id"=>1, "idx"=>112, "money"=>"2.00", "money1"=>"2.00", "order"=>"001", "order1"=>"1"}, {"id"=>3, "idx"=>113, "money"=>"3.00", "money1"=>"1.00", "order"=>"002", "order1"=>"2"} ] 预期的数组应该：
[ {"id"=>2, "idx"=>112, "money"=>"6.00", "money1"=>"3.00","order"=>"001", "order1"=>"1"}, {"id"=>3, "idx"=>113, "money"=>"3.00", "money1"=>"1.00","order"=>"002", "order1"=>"2"} ]
该调用喜欢f_addition(arr, group_fields, sum_fields, max_fields) （例如： f_addition(arr, ["order","order1"], ["money","money1"], ["id", "idx"] ）
P / s：调用方法后不应更改原始文件（它再次用作新group_fields，new sum_fields，new max_fields的参数）
主题Hash of Arrays中的类似问题，group by和sum by with many columns
这样做的一种方法是使用Hash＃update （aka merge! ）的forms，它使用块来确定合并的两个哈希中存在的键的值。
码
def f_addition(arr, group_fields, sum_fields, max_fields) arr.each_with_object({}) do |h,g| g.update( h.values_at(*group_fields) => h ) do |_,gv,hv| gv.merge(hv) do |k,gvv,hvv| case when sum_fields.include?(k) then "%.2f" % (gvv.to_f + hvv.to_f) when max_fields.include?(k) then [gvv, hvv].max else gvv end end end end.values end
例
arr = [{ "id"=>2, "idx"=>111, "money"=>"4.00", "money1"=>"1.00", "order"=>"001", "order1"=>"1", "pet"=>"dog" }, { "id"=>1, "idx"=>112, "money"=>"2.00", "money1"=>"2.00", "order"=>"001", "order1"=>"1", "sport"=>"darts" }, { "id"=>3, "idx"=>113, "money"=>"3.00", "money1"=>"1.00", "order"=>"002", "order1"=>"2" }]
请注意，此数组与问题中给出的数组略有不同。 我已将"pet"=>"dog"到第一个（hash）元素"sport"=>"darts"和第二个哈希。
f_addition(arr, ["order","order1"], ["money","money1"], ["id", "idx"] ) #=> [{ "id"=>2, "idx"=>112, "money"=>"6.00", "money1"=>"3.00", # "order"=>"001", "order1"=>"1", "pet"=>"dog", "sport"=>"darts"}, # { "id"=>3, "idx"=>113, "money"=>"3.00", "money1"=>"1.00", # "order"=>"002", "order1"=>"2" }]
说明
对于上面的例子：
group_fields = ["order", "order1"] sum_fields = ["money", "money1"] max_fields = ["id", "idx"] enum = arr.each_with_object({}) #=> #2, "idx"=>111,..., "pet"=>"dog"}, # {"id"=>1, "idx"=>112,..., "sport"=>"darts"}, # {"id"=>3,"idx"=>113,...,"order1"=>"2"}]:each_with_object({})>
Array＃将每个枚举器的元素传递给块并将其分配给块变量。 传递的第一个元素是：
h, g = enum.next #=> [{ "id"=>2, "idx"=>111, "money"=>"4.00", "money1"=>"1.00", "order"=>"001", "order1"=>"1", "pet"=>"dog" }, {}] h #=> { "id"=>2, "idx"=>111, "money"=>"4.00", "money1"=>"1.00", "order"=>"001", "order1"=>"1", "pet"=>"dog" } g #=> {}
如：
h.values_at(*group_fields) #=> h.values_at(*["order", "order1"]) #=> h.values_at("order", "order1") #=> ["001", "1"]
我们计算：
g.update(["001", "1"] => h) do |k,gv,hv| ... end
这是简写：
g.update({ ["001", "1"] => h }) do |k,gv,hv| ... end
块do |k,gv,hv| ... end do |k,gv,hv| ... end仅在合并的两个哈希值都包含密钥k 。 1由于g = {}包含任何键，因此此时不使用该块：
g.update({ ["001", "1"] => h }) #=> {}.update({ ["001", "1"]=>{ "id"=>2, "idx"=>111, "money"=>"4.00", # "money1"=>"1.00", "order"=>"001", # "order1"=>"1", "pet"=>"dog" } } #=> { ["001", "1"]=>{ "id"=>2, "idx"=>111, "money"=>"4.00", "money1"=>"1.00", # "order"=>"001", "order1"=>"1", "pet"=>"dog" } }
update返回的值是g的新值。
传入块的enum的下一个值是：
h, g = enum.next h #=> { "id"=>1, "idx"=>112, "money"=>"2.00", "money1"=>"2.00", # "order"=>"001", "order1"=>"1", "sport"=>"darts" }, g #=> { ["001", "1"]=>{ "id"=>2, "idx"=>111, "money"=>"4.00", "money1"=>"1.00", # "order"=>"001", "order1"=>"1", "pet"=>"dog" } }]
如：
h.values_at(*group_fields) #=> h.values_at("order", "order1") #=> ["001", "1"]
我们计算：
g.update(["001", "1"] => h) do |k,gv,hv| ... end
由于g和{ ["001", "1"] => h }都包含键[“001”，“1”]，我们必须遵循该块来确定合并散列中该键的值。 我们有：
k = ["001", "1"] gv = { "id"=>2, "idx"=>111, "money"=>"4.00", "money1"=>"1.00", "order"=>"001", "order1"=>"1", "pet"=>"dog" } hv = { "id"=>1, "idx"=>112, "money"=>"2.00", "money1"=>"2.00", "order"=>"001", "order1"=>"1", "sport"=>"darts" }
因此，我们按如下方式评估块（使用merge而不是merge!/update ）：
gv.merge(hv) do |k,gvv,hvv| case when sum_fields.include?(k) then "%.2f" % (gvv.to_f + hvv.to_f) when max_fields.include?(k) then [gvv, hvv].max else gvv end end #=> { "id"=>2, "idx"=>112, "money"=>"6.00", "money1"=>"3.00", # "order"=>"001", "order1"=>"1", "pet"=>"dog", "sport"=>"darts"}
gv不包含关键的“运动”，因此在将"sport"=>"darts"合并为gv时不使用该块。 然而， hvv所有其他键都存在于gvv中，因此我们使用该块来确定合并散列中的值。 对于：
k = "money" gvv = "4.00" hvv = "2.00"
我们发现：
sum_fields.include?(k) #=> ["money", "money1"].include?("money") #=> true
所以case语句返回：
"%.2f" % (gvv.to_f + hvv.to_f) #=> "%.2f" % ("4.00".to_f + "2.00".to_f) #=> "6.00"
将hv其他键的值（合并到gv的哈希值）类似地计算，以给出合并的哈希g的新值。
最后，
{ ["002", "order1"] => { "id"=>3, "idx"=>113, "money"=>"3.00", "money1"=>"1.00", "order"=>"002", "order1"=>"2" }]
被合并到g （不需要使用update的块），并且该方法返回g.values 。
意见
很容易将其概括为传递对，例如：
[["money","money1"], ->(a,b) { "%.2f" % (a.to_f + b.to_f) }] [["id", "idx"], :max]
这可以如下完成：
def f_addition(arr, group_fields, *mods) arr.each_with_object({}) do |h,g| g.update( h.values_at(*group_fields) => h ) do |_,gv,hv| gv.merge(hv) do |k,gvv,hvv| f,op = mods.find { |f,op| f.include?(k) } if f case op when Proc then op.call(gvv,hvv) when Symbol then [gvv, hvv].send(op) end else gvv end end end end.values end f_addition(arr, ["order","order1"], [["money","money1"], ->(a,b) { "%.2f" % (a.to_f + b.to_f) }], [["id", "idx"], :max]) # => [{ "id"=>2, "idx"=>112, "money"=>"6.00", "money1"=>"3.00", # "order"=>"001", "order1"=>"1", "pet"=>"dog", "sport"=>"darts" }, # { "id"=>3, "idx"=>113, "money"=>"3.00", "money1"=>"1.00", # "order"=>"002", "order1"=>"2" }]
我们会发现块中的计算不依赖于块变量`k`。
因此我用局部变量_替换了该变量，以便通知读者。
这应该与你提出的要求非常接近
def f_addition(arr, group_fields, sum_fields, max_fields) arr.group_by do |h| group_fields.map {|k| h[k]} end.values.map do |group| sums = group.each_with_object(Hash.new(0)) do |h, acc| sum_fields.each {|sum_f| acc[sum_f] += h[sum_f].to_f } end group.first.merge(sums) end.flatten end arr = [ {"id"=>2, "idx"=>111, "money"=>"4.00", "money1"=>"1.00", "order"=>"001", "order1"=>"1"}, {"id"=>1, "idx"=>112, "money"=>"2.00", "money1"=>"2.00", "order"=>"001", "order1"=>"1"}, {"id"=>3, "idx"=>113, "money"=>"3.00", "money1"=>"1.00", "order"=>"002", "order1"=>"2"} ] f_addition(arr, ["order","order1"], ["money","money1"], ["id", "idx"] ) # [ # {"id"=>2, "idx"=>111, "money"=>6.0, "money1"=>3.0, "order"=>"001", "order1"=>"1"} # {"id"=>3, "idx"=>113, "money"=>3.0, "money1"=>1.0, "order"=>"002", "order1"=>"2"} # ]
我用过这段代码
def aggregate(arr, group_fields, sum_fields, max_fields) arr.group_by { |x| x.values_at(*group_fields) }.map {|key, hashes| result = hashes[0].clone max_fields.each { |k| hashes.map! {|h| h.merge(h) { |k, v| Integer(v) rescue v }} #hashes.map! { |h| h.each_pair { | k, v | h[k] = Integer(v) rescue v }} result[k] = hashes.max_by { |h| h[k]}[k] } sum_fields.each { |k| result[k] = hashes.inject(0) { |s, x| s + x[k].to_f } } result }end
有一点不好的是将所有字段转换为int（转换为int以进行比较，例如： "id"=>"12"和"id"=>"2" ）。 应该只转换max_fields，但我还没有想到解决方案。 转换代码是：
hashes.map! {|h| h.merge(h) { |k, v| Integer(v) rescue v }} hashes.map! {|h| h.merge(h) { |k, v| Integer(v) rescue v }}或
hashes.map! { |h| h.each_pair { | k, v | h[k] = Integer(v) rescue v
所以，如果有人能够解决这个弱点，那就太好了。