Having some Lua trouble with a a modification of Fisher-Yates shuffle in place. For example, let's say I have a 16 item table (sequence). I want to shuffle integers 1-4 then apply the shuffled pattern in the table to 1-4, 5-8, 9-12, 13-16. So:
{ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16 }
with a 4 item shuffling pattern of 4,2,3,1 would become:
{ 4, 2, 3, 1, 8, 6, 7, 5, 12, 10, 11, 9, 16, 14, 15, 13 }
The code here is from context and includes the "rising edge" input I am using to reshuffle. If you look at the test pic below you can see that yes, it shuffles each section in place, but it reshuffles each section -- I want the shuffled pattern to repeat.
t = {0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15}
range = 4
local function ShuffleInPlace(t)
for i = #t, 2, -1 do
local j = math.random(1, range)
local k = (math.floor(i/(range+.001)))*range + j
t[i], t[j] = t[j], t[i]
end
end
-- initialize new table for shuffling
if s == nil then s = {} end
-- use gate rising edge to shuffle
if prev == nil then prev = 0 end
if gate > 0 and prev <= 0 then
s = t
ShuffleInPlace(s)
end
prev = gate
Test pic:
LMD, thank you, your helpful reply is uncovering a solution (by creating the shuffled "pattern" sequence first, outside the iterator). (Still some issues with the first value I'm working out. And I might be looking at some biproducts of the not-so-great math.random function, but that's another story). I'm a novice so any suggestions are appreciated!
-- range input is 0 to 1
seqRange = math.floor(range*(#t*.99))
local function ShuffleRange(x)
if rdm == nil then rdm = {} end
for m = 1, x do rdm[m] = m end
for m = #rdm, 2, -1 do
local j = math.random(m)
rdm[m], rdm[j] = rdm[j], rdm[m]
return rdm[m]
end
end
local function ShuffleInPlace(t)
y = ShuffleRange(seqRange)
for i = #t, 2, -1 do
local j = (math.floor(i/(seqRange*1.001)))*seqRange + y
t[i], t[j] = t[j], t[i]
end
end
Here's how I would do it, implementing the simple approach of first generating a series of swaps and then applying that to the sublists of length n:
math.randomseed(os.time()) -- seed the random
local t = {}; for i = 1, 16 do t[i] = i end -- build table
local n = 4 -- size of subtables
local swaps = {} -- list of swaps of offsets (0-based)
for i = 0, n - 1 do
-- Insert swap into list of swaps to carry out
local j = math.random(i, n - 1)
table.insert(swaps, {i, j})
end
-- Apply swaps to every subtable from i to i + n
for i = 1, #t, n do
for _, swap in ipairs(swaps) do
-- Swap: First add offsets swap[1] & swap[2] respectively
local a, b = i + swap[1], i + swap[2]
t[a], t[b] = t[b], t[a]
end
end
print(table.concat(t, ", "))
Example output: 4, 2, 1, 3, 8, 6, 5, 7, 12, 10, 9, 11, 16, 14, 13, 15
Related
i have a problem with my Script if i try to loop thought my list the output is completly random shuffled
minimal Code:
list = {
numbers = {
number1 = 1,
number2 = 2,
number3 = 3,
number4 = 4,
number5 = 5,
number6 = 6,
number7 = 7,
}
}
for k, numbers in pairs(list) do
for k, number in pairs(numbers) do
print(number)
end
end
output:
5
7
2
3
4
6
1
the only fix i figured out is to remove the variables number1 to number7
and just enter the numbers
Lua tables do not have an order.
In addition to that you're using pairs which internally uses next.
From the Lua manual:
The order in which the indices are enumerated is not specified, even
for numeric indices. (To traverse a table in numerical order, use a
numerical for.)
In your case the keys have a numeric component so you could simply create them in a numeric loop.
local numbers = {
number1 = 1,
number2 = 2,
number3 = 3,
number4 = 4,
number5 = 5,
number6 = 6,
number7 = 7,
}
for i = 1, 7 do
print(numbers["number"..i])
end
For other non-numeric keys you would have to use a second table that lists the keys in an ordered sequence:
local numbers = { bob = 1, bill = 3, john = 2}
local orderedKeys = { "bob", "john", "bill"}
for k,v in ipairs(orderedKeys) do
print(numbers[v])
end
A numeric loop will always work for any integer keys.
local numbers = {
[0] = 0,
[5] = 5,
[3] = 3,
[1] = 0,
}
for i = 0, 5 do
if numbers[i] then
print(numbers[i])
end
end
Read through this carefully:
A table with exactly one border is called a sequence. For instance,
the table {10, 20, 30, 40, 50} is a sequence, as it has only one
border (5). The table {10, 20, 30, nil, 50} has two borders (3 and 5),
and therefore it is not a sequence. (The nil at index 4 is called a
hole.) The table {nil, 20, 30, nil, nil, 60, nil} has three borders
(0, 3, and 6) and three holes (at indices 1, 4, and 5), so it is not a
sequence, too. The table {} is a sequence with border 0. Note that
non-natural keys do not interfere with whether a table is a sequence.
Things like ipairs, the length operator #, table.sort, table.concat and others only work with sequences.
Keys that do not contribute to the sequence are ignored by those functions. You can only loop over all keys of a table with next or pairs respectively. But then order is not guaranteed.
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
I'm making a lua script for custom encrypting a string, but I can't find a loop that substitutes the letter to a number correspondent to the alphabet table.
It must return something like this:
"hello"
> 8, 4, 12, 12, 15
That'll be the first part of the encryption proccess but I can't program such a loop, any help? (Script below:)
local text = "what"
local key = math.random(100000, 500000)
local pie = math.pi
local n = 26
local alpha = { a = 1, b = 2, c = 3, d = 4, e = 5, f = 6 , g = 7, h = 8, i = 9, j = 10,
k = 11, l = 12, m = 13, n = 14, o = 15, p = 16, q = 17, r = 18, s = 19, t = 20, u = 21, v = 22, w = 23, x = 24, y = 25, z = 26}
function enumerate(str)
return str:gsub("%l", function(c) return c.char(c:byte()-96) end):byte(1,-1)
end
UPDATE: New question, how do I return the numbers as:
46, 68, 46, 32, 4, 12, 18, 15, 23, 32, 15, 23, 5, 8
Instead of a string or with whitespaces:
"46 68 46 32 4 12 18 15 23 32 15 23 5 8"
You just have to collect each char on a table with explode + encode (see below) and then call unpack (or table.unpack) to get a tuple.
First explode:
function explode(str)
local ret = {}
for c in str:gmatch(".") do
table.insert(ret, c)
end
return ret
end
The encoding part can be done this way:
local BASE_CHAR = ("a"):byte()
function encode(c)
return c:byte() - BASE_CHAR + 1
end
Putting everything together:
unpack = unpack or table.unpack -- Handling lua 5.1 or higher
function enumerate(str) -- Keeping the original name
local exploded = {}
for i, c in ipairs(explode(str)) do
exploded[i] = encode(c)
end
return unpack(exploded)
end
And a test:
print(enumerate("test"))
Which yields the following result:
20 5 19 20
Note:
I prefer to keep results in table and only unpack if necessary. This way you can manipulate the table the way you want and it's more convenient than dealing with tuples:
function enumerate(str) -- Keeping the original name
local exploded = {}
for i, c in ipairs(explode(str)) do
exploded[i] = encode(c)
end
return exploded
end
print(unpack(enumerate("test")))
And if you want a string:
print(table.concat(enumerate("test"), ", "))
Hope this solves the problem
I've been asking questions on random numbers, and I decide the Fisher-Yates shuffle would be the best option. I make a table 't'
t = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}
Now, how would I even shuffle these and be able to use them individually, for example generate the results in another table u
u = {}
For those that find this answer later, this will shuffle in place without making a new table:
local function ShuffleInPlace(t)
for i = #t, 2, -1 do
local j = math.random(i)
t[i], t[j] = t[j], t[i]
end
end
And this one that returns a shuffled table without touching the original (unlike the current answer, which both shuffles in-place and returns a copy):
local function Shuffle(t)
local s = {}
for i = 1, #t do s[i] = t[i] end
for i = #t, 2, -1 do
local j = math.random(i)
s[i], s[j] = s[j], s[i]
end
return s
end
Usage:
local t = {"a", "b", "c", "d", "e", "f"}
print(table.concat(t)) --> abcdef
local s = Shuffle(t)
print(table.concat(t)) --> abcdef (unchanged)
print(table.concat(s)) --> fbcade (shuffled)
ShuffleInPlace(t)
print(table.concat(t)) --> dcbfea (shuffled)
And a quick sanity check that they're uniform:
local t = {"a", "b", "c"}
local results = {abc = 0,acb = 0,bac = 0,bca = 0,cab = 0,cba = 0}
for i = 1, 10000000 do
ShuffleInPlace(t)
local r = table.concat(t)
results[r] = results[r] + 1
end
for k, v in pairs(results) do print(k, v) end
--[[
cba 1667473
cab 1666235
bca 1665672
bac 1666782
acb 1666447
abc 1667391
--]]
NOTE: Check the other answer https://stackoverflow.com/a/68486276/1190388 which fixes an issue in the code snippet below as well as providing other alternatives
If you do not have holes in your table:
math.randomseed(os.time()) -- so that the results are always different
function FYShuffle( tInput )
local tReturn = {}
for i = #tInput, 1, -1 do
local j = math.random(i)
tInput[i], tInput[j] = tInput[j], tInput[i]
table.insert(tReturn, tInput[i])
end
return tReturn
end
I have the following array of numbers.
arr = {3412323450, 8912745671, 3212367894}
I want to convert it into a simple two-dimensional matrix.
mat = {
{3, 4, 1, 2, 3, 2, 3, 4, 5, 0},
{8, 9, 1, 2, 7, 4, 5, 6, 7, 1},
{3, 2, 1, 2, 3, 6, 7, 8, 9, 4}
}
Initially, I would iterate over arr, convert it into a string, then split the string, iterate over each string char and convert it back to number storing every row and number in mat accordingly. This would be really ugly.
Is there a more conventional method to convert an array into a matrix in
Lua?
Is there a luarock package that people use frequently to convert an array to a matrix?
Personally I think converting to a string and grabbing all of the digits is far prettier than the alternatives (massively dividing by 10, or any other elaborate means you can think of). This is especially true if you wrap the operations up in functions, so your conversions are not constantly appearing throughout your code.
function Digits(n)
local digits = {}
for d in tostring(n):gmatch('%d') do
digits[#digits+1] = tonumber(d)
end
return digits
end
function ArrayToMatrix(array)
local matrix = {}
for i,v in ipairs(array) do
matrix[i] = Digits(v)
end
return matrix
end
Ok, here is my try.
arr = {3412323450, 8912745671, 3212367894}
function arr2matrix(arr)
local mat = {}
for i, row in ipairs(arr) do
mat[i] = {}
local j = 0
row_str = string.gsub(row, '%d', '%0 ')
for c in string.gmatch(row_str, '%S') do
j = j + 1
mat[i][j] = tonumber(c)
end
end
return mat
end
-- checking the result
m = arr2matrix(arr)
for i=1, #m do
for j=1, #m[i] do
io.write(m[i][j]..',')
end
io.write('\n')
end
Running the above gives:
3,4,1,2,3,2,3,4,5,0,
8,9,1,2,7,4,5,6,7,1,
3,2,1,2,3,6,7,8,9,4,