I want a table like the following:
local Users = {}
local function GetUsers (user)
--cycle through all Users
local Id = GetUserID (user)
local Age = GetAge (user)
local Type = GetType (user)
--Id returns ID of User (value about 8 char length: 27296654)
table.insert (Users, {[Id] = {Age = Age, Type = Type}}
end
This is working as it should but
#Users == 0
if I call Users[Id].Age it returns correct value.
How to make the # work?
As I want to cycle trough all Users to check if a User is multiple times in or missing.
They need to be sorted via IDS.
I also thought on transforming the IDs to word with string.char()
As words will be counted as I want it to be.
I want to make it with
for i = 1, #Users do
An example table looks like this:
Users = {
[12345678] = {Age = 18, Type = 1}
[62952766] = {Age = 22, Type = 1}
[23456788] = {Age = 33, Type = 1}
}
So #Users have to be 3 for me but it shows 0.
But I can call the Age and type out of the table, that means they are in.
Does the table have Problems when the index is such a high number?
Look at # operator documentation:
The length of a table t is defined to be any integer index n such that t[n] is not nil and t[n+1] is nil; moreover, if t1 is nil, n can be zero. For a regular array, with non-nil values from 1 to a given n, its length is exactly that n, the index of its last value. If the array has "holes" (that is, nil values between other non-nil values), then #t can be any of the indices that directly precedes a nil value (that is, it may consider any such nil value as the end of the array).
You are using dictionary, so the # operator doesn't work as you expect it. The only way is to iterate the whole table with pairs(..).
function getTableLength(T)
local count = 0
for _ in pairs(T) do
count = count + 1
end
return count
end
Related
I have some lua code like this:
_table = {
stuff = {
item1 = {Name="Stack",Rarity="Over"};
item2 = {Name="Flow",Rarity="Com"}
};
};
print("placeholder") -- example thing
_stuff = _table.stuff
for i = 1, #_stuff do
print(_stuff[i].Name)
end
The output is this:
placeholder
I've tried to look at stuff but I don't think it was related to my problem.
Lua tables conceptually are "maps": They map keys to values. Your table _table maps the key stuff to the {item1 = ..., item2 = ...} table. Your stuff table maps the string "item1" to the {Name="Stack",Rarity="Over"} table and the string "item2" to the {Name="Flow",Rarity="Com"} table.
The length operator # completely ignores string keys. It returns an integer i such that t[i] ~= nil and t[i+1] == nil. In particular, it returns 0 if there are no integer keys in the table. This is the case for your stuff table: You only have string keys. Thus your loop never runs, as the limit is 0 and i = 1 > 0 already is the case before the first loop iteration.
Put simply, you're trying to iterate over a "dictionary" (a table with string keys) as if it were a list. Lua tables have both a "list part" (integer keys from 1 to n, where usually n = #t) and a "hash part" (of all other keys). The "list part" is empty in your case, so there's nothing for you to iterate over.
To iterate over all key-value pairs of any table (list or dictionary) in an undefined order, use pairs (rather than ipairs or iterating over indices i, which only works for lists):
for itemname, item in pairs(_stuff) do
print(item.Name)
end
If you want a defined order of iteration, either build a second table of keys...
local order = {"item1", "item2"}
for _, key in ipairs(order) do
print(_stuff[key].Name)
end
... or use the integer keys ("list part") of the table:
_table = {
stuff = {
{key = "item1", Name="Stack", Rarity="Over"};
{key = "item2", Name="Flow", Rarity="Com"};
};
};
_stuff = _table.stuff
for _, key in ipairs(order) do
for i = 1, #_stuff do
print(_stuff[i].Name)
end
end
then works as expected.
I have already seen many threads with examples of how to do this, the problem is, I still can't do it.
All the examples have tables with extra data. For example somethings like this
lines = {
luaH_set = 10,
luaH_get = 24,
luaH_present = 48,
}
or this,
obj = {
{ N = 'Green1' },
{ N = 'Green' },
{ N = 'Sky blue99' }
}
I can code in a few languages but I'm very new to Lua, and tables are really confusing to me. I can't seem to work out how to adapt the code in the examples to be able to sort a simple table.
This is my table:
local players = {"barry", "susan", "john", "wendy", "kevin"}
I want to sort these names alphabetically. I understand that Lua tables don't preserve order, and that's what's confusing me. All I essentially care about doing is just printing these names in alphabetical order, but I feel I need to learn this properly and know how to index them in the right order to a new table.
The examples I see are like this:
local function cmp(a, b)
a = tostring(a.N)
b = tostring(b.N)
local patt = '^(.-)%s*(%d+)$'
local _,_, col1, num1 = a:find(patt)
local _,_, col2, num2 = b:find(patt)
if (col1 and col2) and col1 == col2 then
return tonumber(num1) < tonumber(num2)
end
return a < b
end
table.sort(obj, cmp)
for i,v in ipairs(obj) do
print(i, v.N)
end
or this:
function pairsByKeys (t, f)
local a = {}
for n in pairs(t) do table.insert(a, n) end
table.sort(a, f)
local i = 0 -- iterator variable
local iter = function () -- iterator function
i = i + 1
if a[i] == nil then return nil
else return a[i], t[a[i]]
end
end
return iter
end
for name, line in pairsByKeys(lines) do
print(name, line)
end
and I'm just absolutely thrown by this as to how to do the same thing for a simple 1D table.
Can anyone please help me to understand this? I know if I can understand the most basic example, I'll be able to teach myself these harder examples.
local players = {"barry", "susan", "john", "wendy", "kevin"}
-- sort ascending, which is the default
table.sort(players)
print(table.concat(players, ", "))
-- sort descending
table.sort(players, function(a,b) return a > b end)
print(table.concat(players, ", "))
Here's why:
Your table players is a sequence.
local players = {"barry", "susan", "john", "wendy", "kevin"}
Is equivalent to
local players = {
[1] = "barry",
[2] = "susan",
[3] = "john",
[4] = "wendy",
[5] = "kevin",
}
If you do not provide keys in the table constructor, Lua will use integer keys automatically.
A table like that can be sorted by its values. Lua will simply rearrange the index value pairs in respect to the return value of the compare function. By default this is
function (a,b) return a < b end
If you want any other order you need to provide a function that returs true if element a comes befor b
Read this https://www.lua.org/manual/5.4/manual.html#pdf-table.sort
table.sort
Sorts the list elements in a given order, in-place, from list[1] to
list[#list]
This example is not a "list" or sequence:
lines = {
luaH_set = 10,
luaH_get = 24,
luaH_present = 48,
}
Which is equivalent to
lines = {
["luaH_set"] = 10,
["luaH_get"] = 24,
["luaH_present"] = 48,
}
it only has strings as keys. It has no order. You need a helper sequence to map some order to that table's element.
The second example
obj = {
{ N = 'Green1' },
{ N = 'Green' },
{ N = 'Sky blue99' }
}
which is equivalent to
obj = {
[1] = { N = 'Green1' },
[2] = { N = 'Green' },
[3] = { N = 'Sky blue99' },
}
Is a list. So you could sort it. But sorting it by table values wouldn't make too much sense. So you need to provide a function that gives you a reasonable way to order it.
Read this so you understand what a "sequence" or "list" is in this regard. Those names are used for other things as well. Don't let it confuse you.
https://www.lua.org/manual/5.4/manual.html#3.4.7
It is basically a table that has consecutive integer keys starting at 1.
Understanding this difference is one of the most important concepts while learning Lua. The length operator, ipairs and many functions of the table library only work with sequences.
This is my table:
local players = {"barry", "susan", "john", "wendy", "kevin"}
I want to sort these names alphabetically.
All you need is table.sort(players)
I understand that LUA tables don't preserve order.
Order of fields in a Lua table (a dictionary with arbitrary keys) is not preserved.
But your Lua table is an array, it is self-ordered by its integer keys 1, 2, 3,....
To clear up the confusing in regards to "not preserving order": What's not preserving order are the keys of the values in the table, in particular for string keys, i.e. when you use the table as dictionary and not as array. If you write myTable = {orange="hello", apple="world"} then the fact that you defined key orange to the left of key apple isn't stored. If you enumerate keys/values using for k, v in pairs(myTable) do print(k, v) end then you'd actually get apple world before orange hello because "apple" < "orange".
You don't have this problem with numeric keys though (which is what the keys by default will be if you don't specify them - myTable = {"hello", "world", foo="bar"} is the same as myTable = {[1]="hello", [2]="world", foo="bar"}, i.e. it will assign myTable[1] = "hello", myTable[2] = "world" and myTable.foo = "bar" (same as myTable["foo"]). (Here, even if you would get the numeric keys in a random order - which you don't, it wouldn't matter since you could still loop through them by incrementing.)
You can use table.sort which, if no order function is given, will sort the values using < so in case of numbers the result is ascending numbers and in case of strings it will sort by ASCII code:
local players = {"barry", "susan", "john", "wendy", "kevin"}
table.sort(players)
-- players is now {"barry", "john", "kevin", "susan", "wendy"}
This will however fall apart if you have mixed lowercase and uppercase entries because uppercase will go before lowercase due to having lower ASCII codes, and of course it also won't work properly with non-ASCII characters like umlauts (they will go last) - it's not a lexicographic sort.
You can however supply your own ordering function which receives arguments (a, b) and needs to return true if a should come before b. Here an example that fixes the lower-/uppercase issues for example, by converting to uppercase before comparing:
table.sort(players, function (a, b)
return string.upper(a) < string.upper(b)
end)
I am a hobbyest making mods in TableTop Simulator using LUA and have a question that I can not seam to work out.
I have a number of "objects" which is a table in TTS that contains various data for those objects. For example.. obj.position = {x,y,z}... and can be accessed at the axis level as well.
obj.position = {5,10,15} -- x,y,z
obj.position.x == 5
This is an example. The makers of TTS have made it so you can access all the parts like that. So I can acess the object.. and then its various parts. There is a heap, like name, mesh, difuse and a ton more. roations{x,y,z} etc etc
Anyway. I have a table of objects... and would like to order those objects based on the positional data of the x axis.. so highest to lowest. So if I have a table and obj1 in that table is x=3 and obj2 is x=1 and obj3 = x=2 it would be sorted as obj2,obj3,obj1
Pseudo code:
tableOfObjects = {obj1,obj2,obj3}
--[[
tableOfObjectsp[1] == obj1
tableOfObjectsp[2] == obj2
tableOfObjectsp[3] == obj3
tableOfObjectsp[1].position.x == 3
tableOfObjectsp[2].position.x == 1
tableOfObjectsp[4].position.x == 2
--]]
---After Sort it would look this list
tableOfObjects = {obj1,obj3,obj2}
--[[
tableOfObjectsp[1] == obj1
tableOfObjectsp[2] == obj3
tableOfObjectsp[3] == obj2
tableOfObjectsp[1].position.x == 3
tableOfObjectsp[2].position.x == 2
tableOfObjectsp[3].position.x == 1
--]]
I hope I am making sense. I am self taught in the last few months!
So basically I have a table of objects and want to sort the objects in that table based on a single value attached to each individual object in the table. In this case the obj.position.x
Thanks!
You need table.sort. The first argument is the table to sort, the second is a function to compare items.
Example:
t = {
{str = 42, dex = 10, wis = 100},
{str = 18, dex = 30, wis = 5}
}
table.sort(t, function (k1, k2)
return k1.str < k2.str
end)
This article has more information
table.sort(tableOfObjects, function(a, b) return a.position.x > b.position.x end)
This line will sort your table tableOfObjects in descending order by the x-coordinate.
To reverse order, replace > by <.
From the Lua reference manual:
table.sort (list [, comp])
Sorts list elements in a given order, in-place, from list[1] to
list[#list]. If comp is given, then it must be a function that
receives two list elements and returns true when the first element
must come before the second in the final order (so that, after the
sort, i < j implies not comp(list[j],list[i])). If comp is not given,
then the standard Lua operator < is used instead.
Note that the comp function must define a strict partial order over
the elements in the list; that is, it must be asymmetric and
transitive. Otherwise, no valid sort may be possible.
The sort algorithm is not stable: elements considered equal by the
given order may have their relative positions changed by the sort.
So in other words table.sort will sort a table in ascending order by its values.
If you want to order descending or by something other than the table value (like the x-coordinate of your table value's position in your case) you have to provide a function that tells Lua which element will come first.
you can create a function that handles this exact thing:
local function fix_table(t)
local x_data = {};
local inds = {};
local rt = {};
for i = 1, #t do
x_data[#x_data + 1] = t[i].position.x;
inds[t[i].position.x] = t[i];
end
local min_index = math.min(table.unpack(x_data));
local max_index = math.max(table.unpack(x_data));
for i = min_index, max_index do
if inds[i] ~= nil then
rt[#rt + 1] = inds[i];
end
end
return rt;
end
local mytable = {obj1, obj2, obj3};
mytable = fix_table(mytable);
fix_table first takes in every x value inside of the given table, and also places a new index inside the table inds according to each x value (so that they will be ordered from least to greatest), then it gets the smallest value in the x_data array table, which is used to traverse the inds table in order. fix_table checks to make sure that inds[i] is not equal to nil before it increases the size of the return table rt so that every value in rt is ordered from greatest to least, starting at index 1, and ending at index #rt, finally rt is returned.
I hope this helped.
Let's suppose I want to store a list of element. Including some nil values. The position of the values is significant, and I need to represent the absence of a value in the list at a given position.
Here is a problem:
a = {1,2,3,nil,4}
for k,v in ipairs(a) do
print(k,v)
end
print(a[4])
print(a[5])
The for loop will only print elements 1,2 and 3. It stops at nil. The first print statement prints nil, but I'm not sure if it is actually stored in the table or not. (Who knows?) The second print statement prints 4 - as expected.
So here is the question: how to represent a list of elements in a table, and iterate through them efficiently? Given the conditions above, e.g. the position is significant, and some of the positions are "empty". In other words: have no value, but the absence of that value at that position has a meaning.
This is module "null.lua"
local function null(...)
local t, n = {...}, select('#', ...)
for k = 1, n do
local v = t[k]
if v == null then t[k] = nil
elseif v == nil then t[k] = null
end
end
return (table.unpack or unpack)(t, 1, n)
end
_G.null = null
Use null() as encoder and decoder
require("null")
a = {null(1,2,3,nil,4)}
-- the same could be done element-by-element
-- a = {null(1),null(2),null(3),null(nil),null(4)}
for k,v in ipairs(a) do
v = null(v)
print(k,v)
end
print(null(a[4]))
print(null(a[5]))
Lua tables can be used to create any Abstract Data Structure, in your case you indicated that you want a "list". A Lua table is a data structure that combines numeric index based access with key:value access.
Based on your example, you are using the numeric index feature of tables that let you iterate (with ipairs()) through those values. You will not be able to put nil into the table since the numeric index stops at the first nil entry. The remaining values in the table are stored as key:value pairs.
There are several work-arounds, but it depends on why you want a nil in the list. The simplest approach is to use the string "nil" rather than the native data type nil.
a = {1, 2, 3, "nil", 4}
for k,v in ipairs(a) do
print(k,v)
end
The result of this code is:
1 1
2 2
3 3
4 nil
5 4
Because of the way Lua implements strings, there is not a performance penalty for comparing to the string "nil" versus comparing to the native type nil.
The issue of "holes" (caused by nil) in an array are discussed in Programming in Lua, Chapter 5 Tables. Roberto Ierusalimschy recommendation is to track the size of the array to avoid problems with holes.
The following code shows an Object Oriented approach to tracking the size of the list. There are many possible variations on this theme.
function makeList(...)
local list = table.pack(...)
list.length =
function(self) return self.n
end
list.append =
function(self, value)
self.n = self.n + 1
self[self.n] = value
end
list.print =
function(self)
for i = 1, self.n do print(i, self[i]) end
end
return list
end
a = makeList(1, 2, 3, nil, 4)
a:append(5)
a:print()
print(a:length())
The result is:
1 1
2 2
3 3
4 nil
5 4
6 5
6
Note that the function table.pack creates a field 'n' which contains the correct number of items even when 'nil' is present. See PIL chapter 6.2, Variadic Functions for a complete explanation.
Don't just hack something together, write your own datastructure for this. If you "overload" ipairs (by writing an appropriate iterator) you can use it as a table:
function create(...)
local t = table.pack(...)
local self = {
num = t.n,
elements = { ... }
}
return self
end
function elements(t)
local f = function(s, i)
i = i + 1
if i <= s.num then
return i, s.elements[i]
end
end
return f, t, 0
end
local seq = create(1, 2, nil, 3)
print(seq.num)
for i, e in elements(seq) do
print(i, e)
end
-- results:
-- 4
-- 1 1
-- 2 2
-- 3 nil
-- 4 3
You could know define a metatable for this structure and have it use its own ipairs, so you don't even have to change the name.
Well, you can't store nil in the table without issues.
The most simple solution here would be to introduce your own unique value.
local mynil = {} -- every new table is unique!
a = {1,2,3,mynil,4}
for k,v in ipairs(a) do
if (v == mynil) then
v = nil
end
print(k,v)
end
No more issues with "nil" string that might be stored in the table as well, the minor issue is one more comparison. ipairs or any other iterator will show that the key with mynil value exists. That means you can separate mynil key existence with missing key =nil.
P.S. If you want to shift your list, you may consider table.remove(list, key) function.
The answer to this is rather simple, and these "workaround a" suggested is definitely overkill. Just keep track of the number of items in your table whenever it's changed (note: do not use #, you have too keep track manually to deal with nil values) and use a numeric for loop to iterate over it.
Currently I'm stuck on this:
t = {['79402d'] = {'-5.4','5','1.6'}, ['5813g1'] = {'3','0.15','18'}}
Now i need to loop through this table to check if name == t[1], but how can i do so?
I tried doing something like: for i=1,#t,1 do print(t[i]) but it doesn't seem to work.
I hope you can help me guys ;)
Not sure why it didn't worked first time but i solved my problem with:
for a,b in pairs(t) do
print(a, b[1], b[2], b[3])
end
Please note that the length operator # will give you the correct number of elements in a table only in a special case. in your case #t will return 0, hence your for loop does nothing.
Please refer to https://www.lua.org/manual/5.3/manual.html Section 3.4.7 – The Length Operator for details on how to use the lenght operator.
For a regular array, with non-nil values from 1 to a given n, its length is exactly that n, the index of its last value. so #t will be zero if you have no t[1] or n-1 if t[n] is the first nil value in your table starting from t[1].
As you have no t[1] but only a t['79402d'] and a t['5813g1'] your for loop indexing t[i] would not work anyway.
Please read 3.4.9 – Table Constructors on how table construction works.
t = {"hello", "world"}
is the same as
t = {}
t[1] = "hello"
t[2] = "world"
(here t[1] is "hello" and #t is 2
whereas t = {['key1'] = "hello", ['key2'] = "world"}
equals
t = {}
t['key1'] = "hello"
t['key2'] = "world"
so t[1] here is nil and #t is 0
# operator returns lenght of array part of the table. Your table is not an array (i.e. a table with non-nil values from index 1 to a given n). Because of that your loop is not iterating any elements.
Use pairs to iterate over all keys in the table regardless of what they are.