Dictionaries have no order, there is no way to go through them in order in that format without any of the solutions other people have listed above, sorry
Use @D0RYU’s solution. That will work perfectly in your case
2 Likes
Thank you so much! But I still have a lot of dictionaries that have to be converted into an array of dictionaries. Are you kind enough to help me out a bit please? To contact me, you can join my discord server here. Thank you so much again! 
Have a nice day!
Numbers are sorted automatically (from lowest to highest), you can simply do this and use pairs:
local Dict = {
[1] = { Health = 100, Cost = 5000 },
[2] = { Health = 100, Cost = 5000 },
[3] = { Health = 100, Cost = 5000 },
}
for Index, Value in pairs(Dict) do
print(('Index: %d, health: %d, cost: %d'):format(Index, Value.Health, Value.Cost))
end
Thank you so much! But I still have a lot of dictionaries that have to be converted into an array of dictionaries. Are you kind enough to help me out a bit please? Here is an example:
Turn this:
["100"] = 5000;
["110"] = 15000;
["120"] = 35000;
["130"] = 50000;
["145"] = 75000;
["160"] = 125000;
["185"] = 300000;
["200"] = 475000;
["225"] = 750000;
["255"] = 1050000;
["280"] = 1350000;
["320"] = 1850000;
["350"] = "MAX";
Into this:
{Health = 100, Cost = 5000},
{Health = 110, Cost = 15000},
{Health = 120, Cost = 35000},
{Health = 130, Cost = 50000},
{Health = 145, Cost = 75000},
{Health = 160, Cost = 125000},
{Health = 185, Cost = 300000},
{Health = 200, Cost = 475000},
{Health = 225, Cost = 750000},
{Health = 255, Cost = 1050000},
{Health = 280, Cost = 1350000},
{Health = 320, Cost = 1050000},
{Health = 225, Cost = 1850000},
{Health = 350, Cost = "MAX"},
Can you help me please? If you are kind enough, I can send you all my codes
Thank you so much again! Have a nice day!
You can run this in find/replace with regex enabled:
find:
\[\"(.+)\"\] ?= ?(.+);
replace:
{Health = $1, Cost = $2},
Don’t convert it to an array, just use this function…
module.healths = {
["100"] = 5000;
["110"] = 15000;
["120"] = 35000;
["130"] = 50000;
["145"] = 75000;
["160"] = 125000;
["185"] = 300000;
["200"] = 475000;
["225"] = 750000;
["255"] = 1050000;
["280"] = 1350000;
["320"] = 1850000;
...
}
local function getKeys(t)
local keys = {}
for k in next, t do
table.insert(keys, k)
end
table.sort(keys)
return keys
end
for _, k in ipairs(getKeys(module.healths)) do
local v = module.healths[k]
print(k, v)
end
6 Likes
just a note here, this solution will work well, but it depends on the table you are working with
examples:
this works
{
["abc"] = 5000,
["def"] = 15000,
["ghi"] = 35000,
["jkl"] = 50000,
["mno"] = 75000
}
{
["1"] = 5000,
["2"] = 15000,
["3"] = 35000,
["4"] = 50000,
["5"] = 75000
}
this doesn't work
{
["How"] = 5000,
["Are"] = 15000,
["You"] = 35000,
["Today"] = 50000,
["?"] = 75000
}
{
["1"] = 5000,
["2"] = 15000,
[3] = 35000,
["4"] = 50000,
["5"] = 75000
}
{
[true] = 5000,
[false] = 15000
}
you might be wondering, why doesn’t this work with all dictionaries?
well the function sorts alphanumerically, this means that it will work if the number keys are in order or if the text is in alphabetical order
you also can’t have a string and a number as a key in the same table, this is because you can’t compare string and number
anyways since it sorts alphanumerically stuff like this also won’t work right
more stuff that doesn't work
{
["5"] = 5000,
["4"] = 15000,
["3"] = 35000,
["2"] = 50000,
["1"] = 75000
}
{
["mno"] = 5000,
["jkl"] = 15000,
["ghi"] = 35000,
["def"] = 50000,
["abc"] = 75000
}
{
["One"] = 5000,
["1"] = 15000,
["Two"] = 35000,
["2"] = 50000
}
hope some find this reply a bit helpful
2 Likes
I slightly modified a code sample I wrote a few months ago, for your specific situation:
local dictionary = { -- the dictionary in question to be looped through
-- these are example values to be sorted properly
['110'] = 100011001,
['105'] = 20293922,
['230'] = 923929932,
['c'] = 292382,
['a'] = 29382393128,
}
local function loopDictionaryInOrder(dict, func)
local data = {} -- array to hold the indexes
for key, value in pairs(dict) do -- loops though the dictionary provided in the first parameter
table.insert(data, key) -- this adds each key to the 'data' array
end
table.sort(data, function(a, b) -- this sorts the data array by each key's byte (numeric representations)
local toNumA = tonumber(a)
local toNumB = tonumber(b)
if toNumA and toNumB then
-- this checks for 2 integer values and sorts them accordingly
return toNumA < toNumB
end
return tostring(a):lower():byte() < tostring(b):lower():byte()
end)
for _, key in ipairs(data) do
-- loops through the 'data' array and calls the function provided in the second parameter
-- the function gives the key as the first argument and the key's value as the second argument
func(key, dict[key])
end
end
-- this is an example usage of the function
loopDictionaryInOrder(dictionary, function(key, value)
-- print each key and value in order
print(key, value)
end)
I can’t see if this works in Roblox Studio because Roblox is down at the moment, but it works on Lua’s Demo Site.
2 Likes
The regular sort provides the same result in this case.
You could also use the function as an iterator function, which I think is pretty cool:
local function idict(t)
local keys = {}
for k in next, t do
table.insert(keys, k)
end
table.sort(keys)
return function(_, i)
i += 1
local k = keys[i]
if k then
return i, k, t[k]
end
end, keys, 0
end
for _, k, v in idict(module.healths) do
print(k, v)
end
1 Like
you can easily fix this using tostring() in the second argument for table.sort()
table.sort(Table, function(a, b)
return tostring(a) < tostring(b)
end)
now the sorting shouldn’t error like it did before, though like I said it doesn’t sort correctly with every table
1 Like
Thank you very much for all your replies! I should have edited the title, but what I actually want is a function that returns the next index of the given index. Can someone help me with making a function that returns the next index of a given index please? Like for example,
module.NextHealth = function(currentHealth)
-- current health is 145
-- I want to get 160 because it is the next index
-- if currentHealth is already the last index, return the string "MAX"
end
Thank you so much for helping me!
For dictionaries, that works like this.
local nextIndex, nextValue = next(table, currentIndex)
Bear in mind, as I’ve mentioned, that dictionaries don’t really have an order. I think this method does work on arrays and in order, but if it doesn’t then this is all you need.
local function arrayNext(tab, key)
return key+1, tab[key+1] or "MAX"
end
For the next index function in dictionary, can you help me turn it into a function that returns the next index please? Thank you so much!
Unless one of us is confused here, that’s what the first code I posted there will do. Let me know if you need it to do something different.
Its worh noting that Lua cannot guarantee the order of a dictionary that doesn’t have a numeric key sequence. ipairs internally runs through an array from 1 until it hits a nil value.
Since dictionaries dont have a numeric key sequence, it simply cannot work. The only way to do this is to sort the dictionary into a list of key value pairs yourself by passing through it, then ipairs it
{["110"] = "metatablecatgirl"} -> {[1] = {Key = "110", Value = "metatablecatgirl"}}
how do I do it? Can you give me a function that returns the next index of the index i give?
I mean that’s literally the code I gave.
If you want to get the next index and value in a dictionary you can use this
local nextIndex, nextValue = next(table, currentIndex)
So by way of example
local example = {Boat = "Red", Car = "Blue", Truck = "Green"}
local nextIndex, nextValue = next(example, "Car")
print(nextIndex)
print(nextValue)
This will print the index and value after the index “Car”. I really don’t think this is what you want though since you seem to want your dictionary to be ordered. I think you need to refer to my first post in this thread.
module.healths = {
["100"] = 5000;
["110"] = 15000;
["120"] = 35000;
["130"] = 50000;
["145"] = 75000;
["160"] = 125000;
["185"] = 300000;
["200"] = 475000;
["225"] = 750000;
["255"] = 1050000;
["280"] = 1350000;
["320"] = 1850000;
}
module.healthsIndex = {}
for key, value in pairs(module.healths) do
module.healthsIndex[#module.healthsIndex+1] = key
end
table.sort(module.healthsIndex, function(a, b) return module.healths[a] < module.healths[b] end) -- Sort by value rather than key
module.healthsNext = function(key)
local index = table.find(module.healthsIndex, key)
if index then
local newKey = module.healthsIndex[index+1]
if key then return newKey, module.healths[newKey]
else return "NaN", "MAX" end -- If there is nothing larger, return the key NaN and the value MAX
else
warn(("Unable to find key \"%s\" in module.healthsIndex"):format(key))
end
end
print(module.healthsNext("120")) --> 130, 50000
wow thank you so much bro i never new it was that easy!!! Sorry for the confusion but i never knew about the next() function so I made a REALLY COMPLICATED way of using an odered loop to get the next index thanks
just a tip, table.insert is faster performance wise
so I would switch it for this
table.insert(module.healthsIndex, key)
besides table.insert is made for this anyways
Oh shoot I forgot. I kept thinking table.insert was slower just because it shifted all the table entries when you inserted in the middle of the table. But it’s written in C innit?