LUA - Most common item in the table - lua

Question

I have a table like below, I don't need to know which product sells how much, I just want to know which product is the most popular. Do you have any idea which method is the best way to do this?
Like in the example below, knowing that because 3 of the product "blue" have been sold, it is the most popular
local products = {}
products["430373bb5b7a40a04f9"] = "red"
products["0dce4263af4b5cfe0de"] = "red"
products["cf2559afb736c1eb1bc"] = "green"
products["abc4d248541c3386c88"] = "blue"
products["bb9386c65270948ebee"] = "blue"
products["b193fba741cd646a9c0"] = "blue"

Answer 1

this example will count the number of sales in a single pass.
local products = {}
products["430373bb5b7a40a04f9"] = "red"
products["0dce4263af4b5cfe0de"] = "red"
products["cf2559afb736c1eb1bc"] = "green"
products["abc4d248541c3386c88"] = "blue"
products["bb9386c65270948ebee"] = "blue"
products["b193fba741cd646a9c0"] = "blue"
local pop = {}
for k,v in pairs(products) do
pop[v] = (pop[v] or 0) + 1
end
-- if you need to sort by sales then:
local pop_s = {}
for k,v in pairs(pop) do
pop_s[#pop_s+1] = { item=k, sales=v}
end
table.sort(pop_s, function (a,b) return a.sales>b.sales end)
for k,v in pairs(pop_s) do
print(v.item,v.sales)
end
result:
blue 3
red 2
green 1

Answer 2

To expand on Mike's answer: You could even sort the counts in linear (vs. O(n log n)) time since you know that the counts range from 1 to n; thus you can use a "bucket" approach where you use an index of products by count sold:
local products = {
["430373bb5b7a40a04f9"] = "red",
["0dce4263af4b5cfe0de"] = "red",
["cf2559afb736c1eb1bc"] = "green",
["abc4d248541c3386c88"] = "blue",
["bb9386c65270948ebee"] = "blue",
["b193fba741cd646a9c0"] = "blue",
}
local counts = {}
for _, product in pairs(products) do
counts[product] = (counts[product] or 0) + 1
end
local prods_by_cnt = {}
local max_cnt = 0
for product, count in pairs(counts) do
prods_by_cnt[count] = prods_by_cnt[count] or {}
table.insert(prods_by_cnt[count], product)
max_cnt = math.max(max_cnt, count)
end
local prods_sorted = {}
for cnt = 1, max_cnt do
for _, prod in ipairs(prods_by_cnt[cnt]) do
table.insert(prods_sorted, prod)
end
end
Side note: There is no need to build tables {item = ..., sales = ...} as long as you have the counts table: You can just keep a table of item names and use the item names to index the counts table. This may be slightly slower in practice because the counts table is larger and thus presumably more expensive to index - on the other hand, it is more memory-efficient; in theory, both are equally fast and memory efficient, requiring a constant time for lookup of the sales / counts and constant space to store each count.

Answer 3

Try this:
local products = {}
products["430373bb5b7a40a04f9"] = "red"
products["0dce4263af4b5cfe0de"] = "red"
products["cf2559afb736c1eb1bc"] = "green"
products["abc4d248541c3386c88"] = "blue"
products["bb9386c65270948ebee"] = "blue"
products["b193fba741cd646a9c0"] = "blue"
function GetCommonItem(arr)
local Count = 0
local Product = null
for i, v in pairs(arr) do -- Loops through the table
local Temp = 0
for e, f in pairs(arr) do -- Loops again to count the number of common products
if f = v then
Temp += 1
if Temp >= Count then -- Picks the product if it's the most common one
Count = Temp
Product = f
end
end
end
end
return Product
end
print("Most Common Product: ".GetCommonItem(products))
Didn't test so not sure if it works! Let me know the results after u try it!