Is it possible to extract the variable from constant in lua [duplicate] - lua

I have this table in lua:
local values={"a", "b", "c"}
is there a way to return the index of the table if a variable equals one the table entries?
say
local onevalue = "a"
how can I get the index of "a" or onevalue in the table without iterating all values?

There is no way to do that without iterating.
If you find yourself needing to do this frequently, consider building an inverse index:
local index={}
for k,v in pairs(values) do
index[v]=k
end
return index["a"]

The accepted answer works, but there is room for improvement:
Why not exit the loop once the element is found? And why bother copying the entire source table into a new throwaway table?
Usually, this sort of function returns the first array index with that value, not an arbitrary array index with that value.
For arrays:
-- Return the first index with the given value (or nil if not found).
function indexOf(array, value)
for i, v in ipairs(array) do
if v == value then
return i
end
end
return nil
end
print(indexOf({'b', 'a', 'a'}, 'a')) -- 2
For hash tables:
-- Return a key with the given value (or nil if not found). If there are
-- multiple keys with that value, the particular key returned is arbitrary.
function keyOf(tbl, value)
for k, v in pairs(tbl) do
if v == value then
return k
end
end
return nil
end
print(keyOf({ a = 1, b = 2 }, 2)) -- 'b'

If you use Lua for Roblox development, you can use the table.find method:
print(table.find({'a', 'b', 'c'}, 'b'))

Related

lua - how to compare different array

im trying to compare 2 arrays but i dont know how
for example:
local array1 = { 'friend', 'work', 'privat' }
local array2 = { 'apple', 'juice', 'privat' }
if both arrays have the same value it should do a print.
i know i need to work with something like this
for x in ipairs(array1) do
if x == array2 then
print ("Hi")
end
end
but ofcourse it didnt work.
so how can i check if the array1 value contains a values from array2?
Think of it this way: You have to check each element in the first array to its counterpart in the second. If any element is not equal, you know right away that the arrays aren't equal. If every element checks out as equal, the arrays are equal.
local function arrayEqual(a1, a2)
-- Check length, or else the loop isn't valid.
if #a1 ~= #a2 then
return false
end
-- Check each element.
for i, v in ipairs(a1) do
if v ~= a2[i] then
return false
end
end
-- We've checked everything.
return true
end
how can i check if the array1 value contains a values from array2?
#luther's answer will not always work for your question..
If the arrays are different sizes, it completely fails.
If you have an array where similar element are not in the exact same index, it can return a false negative.
for example a = {'one', 'two'}; b = {'two', 'one'} will return false
Using table.sort to solve this would be a band-aid solution without fixing the real problem.
The function below will work with arrays of different sizes containing elements in any order
function array_compare(a, b)
for ia, va in ipairs(a) do
for ib, vb in ipairs(b) do
if va == vb then
print("matching:",va)
end
end
end
end
In array_compare we go through all the combinations of elements in table a and table b, compare them, and print if they are equal.
ipairs(table) uses index, value (instead of just value)
For example
local array1 = { 'friend', 'work', 'privat' }
local array2 = { 'apple', 'juice', 'privat' }
array_compare(array1, array2)
will print
matching: privat
(I'm writing a second answer to account for another possible interpretation of the question.)
If you want to see if array1 contains any value that's also in array2, you can do the following:
Convert array1 to a set. A set is a new table where the array's values become keys whose values are true.
Iterate through array2 to see if any of its values are a key in the set.
local set = {}
for _, v in ipairs(array1) do
set[v] = true
end
for _, v in ipairs(array2) do
if set[v] then
print'Hi'
-- Use a break statement if you only want to say hi once.
end
end
If the arrays are large, this algorithm should be faster than a nested loop that compares every value in array1 to every value in array2.

how to represent nil in a table

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.

get key list in descending order in lua scipt

I have one list in redis key store. it's containing date as keyname like this.
key
===
20160429
20160430
20160501
20160502
Now I want to key last 2 keys, for this I am doing following in my lua script.
local data = {};
local keyslist = redis.call('keys', 'stats:day:*');
local key, users, redisData;
-- keyslist = #keyslist.sort(#keyslist, function(a, b) return a[2] > b[2] end);
-- keyslist = #keyslist.sort(#keyslist, function(a,b) if a>b then return true; else return false; end end);
for iCtr = 1, #keyslist do
key = string.gsub(keyslist[iCtr], 'stats:day:','');
redisData = redis.call('hmget', keyslist[iCtr], 'image','video');
table.insert(data, {date=key, imgctr=redisData[1], vidctr=redisData[2]});
if iCtr == 2 then break end
end
but this is returning first 2 records, I need last 2 records (e.g. following keys)
20160501
20160502
How Do I get descending key list?
If I understand you right, you might want to do the following:
local count = 0
for iCtr = #keyslist-1,#keyslist do
count=count+1
--do your stuff
if count == 2 then break end
--or
if iCtr == #keyslist then break end
end
This will start at the penultimate item in the keyslist and then count upwards.
Note, I did not test the code, but it should work..
Sample code for sorting a Lua table:
keylist = {1,2,5,8,3, 5}
-- after the following line keylist will be sorted ascending (default)
table.sort(keylist)
-- this line is equivalent:
table.sort(keylist, function (a,b) return a < b end)
the second parameter is to table.sort is a function that takes two table values and returns true if the first one is smaller than the second one.
To sort a table decending you simply call
table.sort(keylist, function(a,b)return a > b end)
Please keep in mind that you can only use this to sort the table values, not their keys. But as you are using keys in a different context this should solve your problem.

LUA: Loop in 2d table to display first key name

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.

How to pop/remove the next item (any) in a key-value-pair table in Lua?

In Lua how would one pop/remove the next item (any order) in a key-value-pair table?
Is this possible without having to iterate using pairs?
There is a primitive function next, you can call next(t,k), where k is a key of the table t, returns a next key in the table, in an arbitrary order, and the value associated with this key.
If k is nil, next(t,k) returns the first element if there is one. So you can iterate the table from calling next(t,nil) and end when the next key is nil.
This is an simple example to demonstrate the use of next:
local t = {a = "va", b = "vb", c = "vc"}
local k,v = next(t,nil)
print(k,v)
k,v = next(t,k)
print(k,v)
k,v = next(t,k)
print(k,v)
k,v = next(t,k)
print(k,v)
Output:
a va
c vc
b vb
nil nil
The global function next is useful here. The docs explain it pretty well in general. To use it iteratively, this is 'key':
You may ... modify existing fields. In particular, you may clear
existing fields.
A simple pop function:
-- removes and returns an arbitrary key-value pair from a table, otherwise nil
local pop = function (t)
local key, value = next(t)
if key ~= nil then
t[key] = nil
end
return key, value
end
Demo:
local test = { "one", c = "see", "two", a = "ayy", b = "bee", "three" }
assert(next(test), "Table was expected to be non-empty")
local key, value = pop(test)
while key do
print(key, value)
key, value = pop(test)
end
assert(not next(test), "Table was expected to be empty")
If you run the demo multiple times, you might see the arbitrariness of the table sequence.

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