I have table with with multiple values and I want to print each of them.
To be like:
'value_1' 'value_2' etc..
table = {
{'value_1'},
{'value_2'},
{'value_3'},
{'value_4'},
}
I tried with for k, v but I failed:
for k, v in pairs(table) do
print(v)
end
The values of your table are tables themselves. So try this instead:
for k, v in pairs(table) do
print(v[1])
end
Or create a simpler table and use your original code:
table = {
'value_1',
'value_2',
'value_3',
'value_4',
}
I'm unsure if your example was meant to be production code or not, but there are a few optimizations (although small) you could make:
-Make the table a local variable (i.e): local table = {};
-Remove the unnecessary tables (i.e): {'value1'}; >> 'value1';
-Change the k,v loop to a generic for loop (I believe that would be more efficient?).
Final code (as I would put it):
local Table = {
"value_1";
"value_2";
"value_3";
"value_4";
};
for Key = 1, #Table, 1 do
print(Table[Key]);
end;
Feel free to ask any questions. Oh, and if you're planning on running this code many times, consider putting local print = print; above your code to define a local variable (they are faster).
You are working with multidimensional arrays when you have sub-tables. You can index a sub table like below.
local tab = {
{1, 2, 3},
{4, 5, 6},
{7, 8, 9}
}
for i, v in next, tab do
print(i, v)
for n, k in next, v do
print(">", n, k)
end
end
-- 1 table: 000001
-- > 1 1
-- > 2 2
-- > 3 3
-- 2 table: 000002
-- > 1 4
-- > 2 5
-- > 3 6
-- 3 table: 000003
-- > 1 7
-- > 2 8
-- > 3 9
To index the table above without for loops, you can use the []'s.
print(tab[1][1]) --> 1
print(tab[1][2]) --> 2
print(tab[2][1]) --> 4
print(tab[2][2]) --> 5
You are NOT restricted to number indices. You can use strings and a special way to index with them.
local tab = {
x = 5,
y = 10,
[3] = 15
}
print(tab.x, tab["y"], tab[3]) --> 5 10 15
Related
I'm doing some codewars and arr[index] keeps returning nil. I've done this a few different ways, and I'm sure the array exists, as well as the index. What's wrong here, is it syntax?
As I've mentioned in the title, I want to find the last digit of the array.
if arr[index] <= 0 then
return -1
end
Full Code:
local solution = {}
function solution.newAvg(arr, navg)
local currentAverage = 0
local index = 0
for i, v in pairs(arr) do
index = i
currentAverage = currentAverage + v
end
if arr[index] <= 0 then
return -1
end
return math.ceil(((index+1) * navg) - currentAverage)
end
return solution
I see two issues with your code:
Edge case: Empty array
If arr = {}, the loop for i, v in pairs(arr) do won't execute at all and index will remain at 0. Since arr is empty, arr[0] will be nil and arr[index] <= 0 will fail with an "attempt to compare a nil value" error.
Lack of ordering guarantee
You use pairs rather than ipairs to loop over what I assume is a list. This means keys & values might be traversed in any order. In practice pairs usually (but not always!) traverses the list part of a table in the same order as ipairs, but the reference manual clearly states that you can't rely on no specific order. I don't think CodeWars is this advanced but consider the possibility that pairs may be overridden to deliberately shuffle the order of traversal in order to check whether you're relying on the dreaded "undefined behavior". If this is the case, your "last index" might actually be any index that happens to be visited last, obviously breaking your algorithm.
Fixes
I'll assume arr is an "array", that is, it only contains keys from 1 to n and all values are non-nil (i.e. there are no holes). Then you can (and should!) use ipairs to loop over the "array":
for i, v in ipairs(arr) do ... end
I don't know the problem statement so it's hard to tell how an empty array should be handled. I'll assume that it should probably return 0. You could add a simply early return at the top of the function for that: if arr[1] == nil then return 0 end. Nonempty arrays will always have arr[1] ~= nil.
I want to find the last digit of the array.
If you mean the last integer (or entry/item) of the array:
local last = array[#array]
If you mean the last digit (for example array = {10, 75, 44, 62} and you want 2), then you can get the last item and then get the last digit using modulo 10:
local last = array[#array] % 10
for i, v in pairs(arr) do
index = i
currentAverage = currentAverage + v
end
Just a reminder:
#array returns the number of items in a table.
In Lua, arrays are implemented using integer-indexed tables.
There's a difference between pairs() and ipairs().
Regarding point 3 above, the following code:
local array = {
[1] = 12,
[2] = 32,
[3] = 41,
[4] = 30,
[5] = 14,
[6] = 50,
[7] = 62,
[8] = 57
}
for key, value in pairs(array) do
print(key, value)
end
produces the following output (note that the order of keys is not respected):
8 57
1 12
2 32
3 41
4 30
5 14
6 50
7 62
while the same code above with pairs() replaced with ipairs() gives:
1 12
2 32
3 41
4 30
5 14
6 50
7 62
8 57
So, this might be the cause of your problem.
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
In my Tabletop Simulator mod I have a bag, when something is dropped in the bag the emptyContents() function is called. For example I can drop 15 dice in the bag.
In the emptyContents() function I iterate over the objects in the bag. But as you can see I have to put in multiple if statements to catch the amount of dice put in because I want the dice to be spawned on different positions.
The contents variable is the amount of dice in the bag.
function emptyContents()
contents = self.getObjects()
for i, _ in ipairs(self.getObjects()) do
if i <= 6 then
self.takeObject(setPosition(5, -3))
elseif i <= 12 then
self.takeObject(setPosition(12.4,-5))
elseif i <= 18 then
self.takeObject(setPosition(19.8,-7))
end
end
end
How can I make the function less static? Because now I need to write if statements for each set of 6 dice.
maybe you can add a config like this:
local t = {
{6, 5, -3},
{12, 12.4, -5},
{18, 19.8, -7},
}
function emptyContents()
contents = self.getObjects()
for i, _ in ipairs(self.getObjects()) do
for _, v in ipairs(t) do
local l, p1, p2 = unpack(v)
if i <= l then
self.takeObject(setPosition(p1, p2))
break
end
end
end
end
I'm teaching myself Lua by reading Ierusalimschy's Programming in Lua (4th edition), and doing the exercises. Exercise 6.5 is
Write a function that takes an array and prints all combinations of the elements in the array.
After this succinct statement the book gives a hint that makes it clear that what one is expected to do is to write a function that prints all the C(n, m) combinations of m elements from an array of n elements.
I implemented the combinations function shown below:
function combinations (array, m)
local append = function (array, item)
local copy = {table.unpack(array)}
copy[#copy + 1] = item
return copy
end
local _combinations
_combinations = function (array, m, prefix)
local n = #array
if n < m then
return
elseif m == 0 then
print(table.unpack(prefix))
return
else
local deleted = {table.unpack(array, 2, #array)}
_combinations(deleted, m - 1, append(prefix, array[1]))
_combinations(deleted, m, prefix)
end
end
_combinations(array, m, {})
end
It works OK, but it is not tail-recursive.
Can someone show me a tail-recursive function that does the same thing as combinations above does?
(For what it's worth, I am using Lua 5.3.)
NB: I realize that the exercise does not require that the function be tail-recursive. This is a requirement I have added myself, out of curiosity.
EDIT: I simplified the function slightly, but removing a couple of nested functions that were not adding much.
There is a third option, one that doesn't have a snake eating it's tail. Although recursion with tail-calls don't lead to stack overflow, I avoid doing so out of personal preference. I use a while loop and a stack that holds the information for each iteration. Within the loop you pop the next task from the stack, do the work, then push next task onto the stack. I feel it looks cleaner and it's easier to visualize the nesting.
Here is how I would translate your code into the way I would write it:
function combinations(sequence, item)
local function append(array, item)
local copy = {table.unpack(array)}
copy[#copy + 1] = item
return copy
end
local stack = {}
local node = { sequence, item, {} }
while true do
local seq = node[ 1 ]
local itm = node[ 2 ]
local pre = node[ 3 ]
local n = #seq
if itm == 0 then
print(table.unpack(pre))
elseif n < itm then
-- do nothing
else
local reserve = {table.unpack(seq, 2, #seq)}
table.insert(stack, { reserve, itm, pre })
table.insert(stack, { reserve, itm-1, append(pre, seq[ 1 ]) })
end
if #stack > 0 then
node = stack[ #stack ] -- LIFO
stack[ #stack ] = nil
else
break
end
end
end
You can use this while-loop stack/node technique for just about any recursive method. Here is an example where it's applied to printing deeply nested tables: https://stackoverflow.com/a/42062321/5113346
My version, using your input example gives the same output:
1 2 3
1 2 4
1 2 5
1 3 4
1 3 5
1 4 5
2 3 4
2 3 5
2 4 5
3 4 5.
Forgive me if it doesn't work with other passed params because I didn't try to solve the answer to the exercise but rather just rewrite the code in your original post.
OK, I think I found one way to do this:
function combinations (array, m)
local dropfirst = function (array)
return {table.unpack(array, 2, #array)}
end
local append = function (array, item)
local copy = {table.unpack(array)}
copy[#copy + 1] = item
return copy
end
local _combinations
_combinations = function (sequence, m, prefix, queue)
local n = #sequence
local newqueue
if n >= m then
if m == 0 then
print(table.unpack(prefix))
else
local deleted = dropfirst(sequence)
if n > m then
newqueue = append(queue, {deleted, m, prefix})
else
newqueue = queue
end
return _combinations(deleted, m - 1,
append(prefix, sequence[1]),
newqueue)
end
end
if #queue > 0 then
newqueue = dropfirst(queue)
local newargs = append(queue[1], newqueue)
return _combinations(table.unpack(newargs))
end
end
_combinations(sequence, m, {}, {})
end
This version is, I think, tail-recursive. Unfortunately, it does not print out the results in as nice an order as did my original non-tail-recursive version (not to mention the added complexity of the code), but one can't have everything!
EDIT: Well, no, one can have everything! The version below is tail-recursive, and prints its results in the same order as does the original non-tail-recursive version:
function combinations (sequence, m, prefix, stack)
prefix, stack = prefix or {}, stack or {}
local n = #sequence
if n < m then return end
local newargs, newstack
if m == 0 then
print(table.unpack(prefix))
if #stack == 0 then return end
newstack = droplast(stack)
newargs = append(stack[#stack], newstack)
else
local deleted = dropfirst(sequence)
if n > m then
newstack = append(stack, {deleted, m, prefix})
else
newstack = stack
end
local newprefix = append(prefix, sequence[1])
newargs = {deleted, m - 1, newprefix, newstack}
end
return combinations(table.unpack(newargs)) -- tail call
end
It uses the following auxiliary functions:
function append (sequence, item)
local copy = {table.unpack(sequence)}
copy[#copy + 1] = item
return copy
end
function dropfirst (sequence)
return {table.unpack(sequence, 2, #sequence)}
end
function droplast (sequence)
return {table.unpack(sequence, 1, #sequence - 1)}
end
Example:
> combinations({1, 2, 3, 4, 5}, 3)
1 2 3
1 2 4
1 2 5
1 3 4
1 3 5
1 4 5
2 3 4
2 3 5
2 4 5
3 4 5
Ironically, this version achieves tail-recursion by implementing its own stack, so I am not sure it is ultimately any better than the non-tail-recursive version... Then again, I guess the function's stack actually lives in the heap (right?), because Lua's tables are passed around by reference (right?), so maybe this is an improvement, after all. (Please correct me if I'm wrong!)
The following script finds prime numbers in a range from 1 to 13.
When I explicitly iterate over the table that contains the results I can see that the script works as expected. However, if I use unpack() function on the table only the first 3 numbers get printed out.
From docs: unpack is "a special function with multiple returns. It receives an array and returns as results all elements from the array, starting from index 1".
Why is it not working in the script below?
t = {}
for i=1, 13 do t[i] = i end
primes = {}
for idx, n in ipairs(t) do
local isprime = true
for i=2, n-1 do
if n%i == 0 then
isprime = false
break
end
end
if isprime then
primes[idx] = n
end
end
print('loop printing:')
for i in pairs(primes) do
print(i)
end
print('unpack:')
print(unpack(primes))
Running
$ lua5.3 primes.lua
loop printing:
1
2
3
5
7
13
11
unpack:
1 2 3
Change
primes[idx] = n
to
primes[#primes+1] = n
The reason is that idx is not sequential as not every number is a prime.