Check if table contains number - lua

I need to check if a table contains a specific number that is known.
e.g. My variable is {1, 2, 3, 4, 5} and I want to know how I can check if that variable contains the number 3.
Thanks in advance.

Just loop over the values of your table and compare them against the number:
function contains(table, number)
for key, value in pairs(table) do if value == number then return true end end
return false
end
this will search through the values of both hash & list part. Usage based on your example: var = {1, 2, 3, 4, 5}; assert(contains(var, 3)).

If your list is sorted, you can use a binary search:
function contains(t, num)
local upper = #t
local lower = 1
while upper >= lower do
local mid = math.floor((lower + upper) / 2)
local v = t[mid]
if v < num then
lower = mid + 1
elseif v > num then
upper = mid - 1
else -- v == num
return true
end
end
return false
end

Related

How could i loop through every OTHER element in a table in Roblox.lua

I am trying to loop through every other element in an table but I cannot find a way to do so.
Any help is appreciated.
Thank you.
It depends what kind of table you're working with. If you have an array-like table, you can use a simple for-loop :
local t = {1, 2, 3, 4, 5, 6, 7, 8}
-- start at 1, loop until i > the length of t, increment i by 2 every loop
for i = 1, #t, 2 do
local val = t[i]
print(val) -- will print out : 1, 3, 5, 7
end
But if you have a dictionary-like table, you will need something to track which key to skip. You could use a simple boolean to keep track, but be aware that there is no guaranteed order of a dictionary-like table.
local t = {
a = 1,
b = 2,
c = 3,
d = 4,
}
local shouldPrint = true
for k, v in pairs(t) do
-- only print the value when it's true
if shouldPrint then
print(k, v) -- will probably print a, 1 and c, 3
end
-- swap shouldPrint's value every loop
shouldPrint = !shouldPrint
end
Maybe try this
local count = 0
for i = 1 , #table/2 do
table[count + i].value = value
count = count + 1
end
Add to the count as you go down

Calling table.sort returns nil

I am trying to sort a table 'array' that contains tables with two keys called 'pt' and 'angle'. I want to sort the 'array' elements regarding their 'angle' value. To my understanding of table.sort this code snippet should do the trick:
local array = {}
-- Some code that calls
-- table.insert(array, {pt = somePt, angle = someAngle})
-- multiple times
local sorted_table = table.sort(array, function(a,b) return a.angle < b.angle end)
However, sorted_table is always nil. Am I doing something wrong here?
table.sort sorts the array part of a table in place. It does not return a new array. If you need to keep the original, you first need to copy to a temporary array.
So, try something like this:
table.sort(array,function(a,b) return a.angle < b.angle end)
table.sort sorts the table in place; that is, it changes the table that you give it and doesn't return a new one.
If you want a sorted copy, you'd first have to make a copy of the table yourself, then sort it.
This could look like this:
local function sorted_copy(tab, func)
local tab = {table.unpack(tab)}
table.sort(tab, func)
return tab
end
That will create a copy of the table (at least the numeric indices up to some random border) and sort it.
there is no secret, there is an algorithm for this sorting, which has been used many times:
function quicksort(t, sortname, start, endi)
start, endi = start or 1, endi or #t
sortname = sortname or 1
if(endi - start < 1) then return t end
local pivot = start
for i = start + 1, endi do
if t[i][sortname] <= t[pivot][sortname] then
local temp = t[pivot + 1]
t[pivot + 1] = t[pivot]
if(i == pivot + 1) then
t[pivot] = temp
else
t[pivot] = t[i]
t[i] = temp
end
pivot = pivot + 1
end
end
t = quicksort(t, sortname, start, pivot - 1)
return quicksort(t, sortname, pivot + 1, endi)
end
local array = {}
table.insert(array, {pt = 1, angle = 2})
table.insert(array, {pt = 4, angle = 9})
table.insert(array, {pt = 1, angle = 5})
table.insert(array, {pt = 2, angle = 7})
table.insert(array, {pt = 2, angle = 1})
table.insert(array, {pt = 5, angle = 2})
local s_t = quicksort(array, "angle")
for k,v in pairs(s_t) do
print(k, "v=", v.pt, v.angle)
end
Output:
1 v= 2 1
2 v= 5 2
3 v= 1 2
4 v= 1 5
5 v= 2 7
6 v= 4 9

Lua #Table returning 0, despite Table containing 3 elements (tables)? [duplicate]

Sounds like a "let me google it for you" question, but somehow I can't find an answer. The Lua # operator only counts entries with integer keys, and so does table.getn:
tbl = {}
tbl["test"] = 47
tbl[1] = 48
print(#tbl, table.getn(tbl)) -- prints "1 1"
count = 0
for _ in pairs(tbl) do count = count + 1 end
print(count) -- prints "2"
How do I get the number of all entries without counting them?
You already have the solution in the question -- the only way is to iterate the whole table with pairs(..).
function tablelength(T)
local count = 0
for _ in pairs(T) do count = count + 1 end
return count
end
Also, notice that the "#" operator's definition is a bit more complicated than that. Let me illustrate that by taking this table:
t = {1,2,3}
t[5] = 1
t[9] = 1
According to the manual, any of 3, 5 and 9 are valid results for #t. The only sane way to use it is with arrays of one contiguous part without nil values.
You can set up a meta-table to track the number of entries, this may be faster than iteration if this information is a needed frequently.
The easiest way that I know of to get the number of entries in a table is with '#'. #tableName gets the number of entries as long as they are numbered:
tbl={
[1]
[2]
[3]
[4]
[5]
}
print(#tbl)--prints the highest number in the table: 5
Sadly, if they are not numbered, it won't work.
There's one way, but it might be disappointing: use an additional variable (or one of the table's field) for storing the count, and increase it every time you make an insertion.
count = 0
tbl = {}
tbl["test"] = 47
count = count + 1
tbl[1] = 48
count = count + 1
print(count) -- prints "2"
There's no other way, the # operator will only work on array-like tables with consecutive keys.
function GetTableLng(tbl)
local getN = 0
for n in pairs(tbl) do
getN = getN + 1
end
return getN
end
You're right. There are no other way to get length of table
You could use penlight library. This has a function size which gives the actual size of the table.
It has implemented many of the function that we may need while programming and missing in Lua.
Here is the sample for using it.
> tablex = require "pl.tablex"
> a = {}
> a[2] = 2
> a[3] = 3
> a['blah'] = 24
> #a
0
> tablex.size(a)
3
local function CountedTable(x)
assert(type(x) == 'table', 'bad parameter #1: must be table')
local new_t = {}
local mt = {}
-- `all` will represent the number of both
local all = 0
for k, v in pairs(x) do
all = all + 1
end
mt.__newindex = function(t, k, v)
if v == nil then
if rawget(x, k) ~= nil then
all = all - 1
end
else
if rawget(x, k) == nil then
all = all + 1
end
end
rawset(x, k, v)
end
mt.__index = function(t, k)
if k == 'totalCount' then return all
else return rawget(x, k) end
end
return setmetatable(new_t, mt)
end
local bar = CountedTable { x = 23, y = 43, z = 334, [true] = true }
assert(bar.totalCount == 4)
assert(bar.x == 23)
bar.x = nil
assert(bar.totalCount == 3)
bar.x = nil
assert(bar.totalCount == 3)
bar.x = 24
bar.x = 25
assert(bar.x == 25)
assert(bar.totalCount == 4)
I stumbled upon this thread and want to post another option. I'm using Luad generated from a block controller, but it essentially works by checking values in the table, then incrementing which value is being checked by 1. Eventually, the table will run out, and the value at that index will be Nil.
So subtract 1 from the index that returned a nil, and that's the size of the table.
I have a global Variable for TableSize that is set to the result of this count.
function Check_Table_Size()
local Count = 1
local CurrentVal = (CueNames[tonumber(Count)])
local repeating = true
print(Count)
while repeating == true do
if CurrentVal ~= nil then
Count = Count + 1
CurrentVal = CueNames[tonumber(Count)]
else
repeating = false
TableSize = Count - 1
end
end
print(TableSize)
end
seems when the elements of the table is added by insert method, getn will return correctly. Otherwise, we have to count all elements
mytable = {}
element1 = {version = 1.1}
element2 = {version = 1.2}
table.insert(mytable, element1)
table.insert(mytable, element2)
print(table.getn(mytable))
It will print 2 correctly

How to get number of entries in a Lua table?

Sounds like a "let me google it for you" question, but somehow I can't find an answer. The Lua # operator only counts entries with integer keys, and so does table.getn:
tbl = {}
tbl["test"] = 47
tbl[1] = 48
print(#tbl, table.getn(tbl)) -- prints "1 1"
count = 0
for _ in pairs(tbl) do count = count + 1 end
print(count) -- prints "2"
How do I get the number of all entries without counting them?
You already have the solution in the question -- the only way is to iterate the whole table with pairs(..).
function tablelength(T)
local count = 0
for _ in pairs(T) do count = count + 1 end
return count
end
Also, notice that the "#" operator's definition is a bit more complicated than that. Let me illustrate that by taking this table:
t = {1,2,3}
t[5] = 1
t[9] = 1
According to the manual, any of 3, 5 and 9 are valid results for #t. The only sane way to use it is with arrays of one contiguous part without nil values.
You can set up a meta-table to track the number of entries, this may be faster than iteration if this information is a needed frequently.
The easiest way that I know of to get the number of entries in a table is with '#'. #tableName gets the number of entries as long as they are numbered:
tbl={
[1]
[2]
[3]
[4]
[5]
}
print(#tbl)--prints the highest number in the table: 5
Sadly, if they are not numbered, it won't work.
There's one way, but it might be disappointing: use an additional variable (or one of the table's field) for storing the count, and increase it every time you make an insertion.
count = 0
tbl = {}
tbl["test"] = 47
count = count + 1
tbl[1] = 48
count = count + 1
print(count) -- prints "2"
There's no other way, the # operator will only work on array-like tables with consecutive keys.
function GetTableLng(tbl)
local getN = 0
for n in pairs(tbl) do
getN = getN + 1
end
return getN
end
You're right. There are no other way to get length of table
You could use penlight library. This has a function size which gives the actual size of the table.
It has implemented many of the function that we may need while programming and missing in Lua.
Here is the sample for using it.
> tablex = require "pl.tablex"
> a = {}
> a[2] = 2
> a[3] = 3
> a['blah'] = 24
> #a
0
> tablex.size(a)
3
local function CountedTable(x)
assert(type(x) == 'table', 'bad parameter #1: must be table')
local new_t = {}
local mt = {}
-- `all` will represent the number of both
local all = 0
for k, v in pairs(x) do
all = all + 1
end
mt.__newindex = function(t, k, v)
if v == nil then
if rawget(x, k) ~= nil then
all = all - 1
end
else
if rawget(x, k) == nil then
all = all + 1
end
end
rawset(x, k, v)
end
mt.__index = function(t, k)
if k == 'totalCount' then return all
else return rawget(x, k) end
end
return setmetatable(new_t, mt)
end
local bar = CountedTable { x = 23, y = 43, z = 334, [true] = true }
assert(bar.totalCount == 4)
assert(bar.x == 23)
bar.x = nil
assert(bar.totalCount == 3)
bar.x = nil
assert(bar.totalCount == 3)
bar.x = 24
bar.x = 25
assert(bar.x == 25)
assert(bar.totalCount == 4)
I stumbled upon this thread and want to post another option. I'm using Luad generated from a block controller, but it essentially works by checking values in the table, then incrementing which value is being checked by 1. Eventually, the table will run out, and the value at that index will be Nil.
So subtract 1 from the index that returned a nil, and that's the size of the table.
I have a global Variable for TableSize that is set to the result of this count.
function Check_Table_Size()
local Count = 1
local CurrentVal = (CueNames[tonumber(Count)])
local repeating = true
print(Count)
while repeating == true do
if CurrentVal ~= nil then
Count = Count + 1
CurrentVal = CueNames[tonumber(Count)]
else
repeating = false
TableSize = Count - 1
end
end
print(TableSize)
end
seems when the elements of the table is added by insert method, getn will return correctly. Otherwise, we have to count all elements
mytable = {}
element1 = {version = 1.1}
element2 = {version = 1.2}
table.insert(mytable, element1)
table.insert(mytable, element2)
print(table.getn(mytable))
It will print 2 correctly

How do I get the number of keys in a hash table in Lua?

myTable = {}
myTable["foo"] = 12
myTable["bar"] = "blah"
print(#myTable) -- this prints 0
Do I actually have to iterate through the items in the table to get the number of keys?
numItems = 0
for k,v in pairs(myTable) do
numItems = numItems + 1
end
print(numItems) -- this prints 2
I experimented with both the # operator and table.getn(). I thought table.getn() would do what you wanted but as it turns out it's returning the same value as #, namely 0. It appears that dictionaries insert nil placeholders as necessary.
Looping over the keys and counting them seems like the only way to get the dictionary size.
The length operator:
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 t[1] 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).
so only way to get length is iterate over it.
Aside from iterating through the keys manually, it is simple to automatically keep track of it via metamethods. Considering you probably don't want to keep track of every table you make, you can just write a function that will allow you to convert any table into a key-countable object. The following isn't perfect, but I think it would illustrate the point:
function CountedTable(x)
assert(type(x) == 'table', 'bad parameter #1: must be table')
local mt = {}
-- `keys` will represent the number of non integral indexes
-- `indxs` will represent the number of integral indexes
-- `all` will represent the number of both
local keys, indxs, all = 0, 0, 0
-- Do an initial count of current assets in table.
for k, v in pairs(x) do
if (type(k) == 'number') and (k == math.floor(k)) then indxs = indxs + 1
else keys = keys + 1 end
all = all + 1
end
-- By using `__nexindex`, any time a new key is added, it will automatically be
-- tracked.
mt.__newindex = function(t, k, v)
if (type(k) == 'number') and (k == math.floor(k)) then indxs = indxs + 1
else keys = keys + 1 end
all = all + 1
t[k] = v
end
-- This allows us to have fields to access these datacounts, but won't count as
-- actual keys or indexes.
mt.__index = function(t, k)
if k == 'keyCount' then return keys
elseif k == 'indexCount' then return indxs
elseif k == 'totalCount' then return all end
end
return setmetatable(x, mt)
end
Examples of using this would include:
-- Note `36.35433` would NOT be counted as an integral index.
local foo = CountedTable { 1, 2, 3, 4, [36.35433] = 36.35433, [54] = 54 }
local bar = CountedTable { x = 23, y = 43, z = 334, [true] = true }
local foobar = CountedTable { 1, 2, 3, x = 'x', [true] = true, [64] = 64 }
print(foo.indexCount) --> 5
print(bar.keyCount) --> 4
print(foobar.totalCount) --> 6
Live Working Example
Hope this helped! :)
Lua stores table as two separates parts : a hash part and an array part, the len operator only deal with the array part, meaning value indexed by a number value, plus using rules mentioned below, so you don't have any choice for counting "hash" value you need to iterate over the table with pairs() function.

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