在多个位置检查值并仅在源唯一的情况下返回匹配项 [英] Checking values across multiple location and returning a match only if the sources are unique

问题描述

让我们说我有一个供应商的列表:Asda，Tesco和Spar.

Lets say I have a list of Vendors: Asda, Tesco, Spar.

我有一个来源(或类似的供应商)列表:Kellogg，Cadbury，Nestle，Johnsons，Pampers，Simple等(定义好的列表大约有20个).

And I have a list of Sources (or suppliers in this analogy): Kellogg, Cadbury, Nestle, Johnsons, Pampers, Simple, etc. (there is a defined list of around 20).

数据流中的其他位置.我要针对多个不同的事物返回一个结果，对于每个供应商来说，是/否.

Elsewhere in the flow of data. I am returning a result, which is Yes/No for each vendor, for multiple different things.

例如:Asda:ukOnly =是"; Spar:ukOnly =否"等等

For example: Asda: ukOnly = "Yes"; Spar: ukOnly = "No" etc.

在此特定部分中，我正在整理结果.

In this specific section, I am collating results.

大多数情况下，供应商的来源是否重叠都没有关系.所以我只能说:

Mostly it doesn't matter if the sources from the vendors overlap. So I can just say:

function concatResults(x) -- concats the result of "x" for each vendor
 local pathAsda = *this is where I call the path location specific to Asda*
 local pathTesco = *this is where I call the path location specific to Tesco*
 local pathSpar = *this is where I call the path location specific to Spar*
   if (pathAsda == "Yes" or pathTesco == "Yes" or pathSpar == "Yes") then
    return "Yes"
   else
    return "No"
   end
end

ukOnlyAgr = concatResults("ukOnly")

太好了！

现在，说我想做些更复杂的事情.

Now, say I want to do something more comple.

我想知道有多少独特的供应商在提供巧克力和谷物.仅当涉及至少两个来源(供应商)并且它们至少必须供应巧克力时，以下示例才被用于进一步生成事实suppliesSweet的过程.这将分别为每个供应商完成(请假设我已经根据输入数据定义了变量:

I want to know how many unique suppliers are providing chocolate and cereal. The example below is being used further up the process to produce a fact suppliesSweet, only if there is at least two sources (suppliers) involved and they must be at least supplying chocolate. This will be done for each vendor separately (please assume I have already defined my variables based on the input data:

if (suppliesChoc > 0 and suppliesCereal > 0 and numSources > 1) or (suppliesChoc > 1) then
  then suppliesSweet = "Yes"
else suppliesSweet = "No"
end

还没有问题.

当我尝试在各个供应商之间汇总这些结果时，问题就来了(就像我以前对ukOnly所做的那样).

The issue comes when I try to aggregate these results across vendors (as I did before with ukOnly).

我已经使用了以下功能:

I already have the following function being used:

table.contains = function(t, value) -- Finds if "value" exists inside the table "t"
    for index = 1, #t do
        if t[index] == value then
            return index    
        end
    end
end

并且正在考虑创建此:

table.overlap = function(t,g) -- Finds if tables "g" and "t" have any overlapping values
    for i=1,#t do
        if table.contains(g,t[i]) then
           return true
        else
           return false
        end
    end
end

但是我不确定从那里去哪里.

But I'm just not sure where to go from there.

您可以假设我已经获得了每个供应商的唯一来源列表，并且我不介意我们是否过于严格. IE.如果两个供应商之间有任何来源重叠，那将使整个结果无效.

You can assume I have already got a list of unique sources for each vendor and I don't mind if we're over restrictive. I.e. if any sources overlap between the two vendors, that would invalidate the entire result.

您还可以假设我有每个事实":分别返回的每个供应商的suppliesChoc，suppliesCereal，numSources和suppliesSweet.

You can also assume I have each "fact": suppliesChoc, suppliesCereal, numSources and suppliesSweet for each vendor returned separately.

推荐答案

我相信您正在寻找两个集合的交集.

I believe your looking for the intersection of two sets.

https://en.wikipedia.org/wiki/Intersection_(set_theory)

一组是您的供应商的供应商，另一组是提供糖果的供应商.

One set being your vendor's suppliers and the other being the suppliers who supply sweets.

local vendors = {
  Asda = {Kellogg = true, Cadbury = true, Nestle = true, Johnsons = true, Pampers = true, Simple = true}, 
  Tesco = {Kellogg = true, Cadbury = true, Nestle = true, Johnsons = true},
  Spar ={Nestle = true, Johnsons = true, Pampers = true, Simple = true}
}

function intersection(s1, s2)
  local output = {}

  for key in pairs(s1) do
    output[#output + 1] = s2[key]
  end

  return output
end

local sweetSuppliers = {Kellogg = true, Cadbury = true, Nestle = true}

for name, suppliers in pairs(vendors) do
  local result = intersection(sweetSuppliers, suppliers)

  print(name .. " has " .. #result .. " sweets suppliers")
end

以下是用于处理集合的库的一些示例:

Here are some examples of a libraries for handling sets:

odkr的properset.lua

Windower的sets.lua

两者都可以使您了解如何使用集合来完成交集之类的事情

Both can give you an idea of how you can use sets to accomplish things like intersection, and much more

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