Alright, someone must know easier ways to do this than me.
I'm trying to write a random number generator using a fairly common formula.
--Random Number Generator
local X0=os.time()
local A1=710425941047
local B1=813633012810
local M1=711719770602
local X1=(((A1*X0)+B1)%M1)
local X2=(((A1*X1)+B1)%M1) --then I basically take the vaiable X1 and feed
--it back into itself.
print(X2)
local X3=(((A1*X2)+B1)%M1)
print(X3)
local X4=(((A1*X3)+B1)%M1)
print(X4)
local X5=(((A1*X4)+B1)%M1)
print(X5)
local X6=(((A1*X5)+B1)%M1)
print(X6)
local X7=(((A1*X6)+B1)%M1)
print(X7)
Etc Etc.
Does anybody know a faster way to do this?
I would love to be able to fit it into something along the lines of a:
for i=1,Number do
local X[loop count]=(((A1*X[Loop count-1])+B1)%M1)
math.randomseed(X[loop count])
local roll=math.random(1,20)
print("You rolled a "..roll)
end
io.read()
Type of string.
I'm using it to generate random numbers for pieces of track I'm making in a tabletop game.
Example hunk of code:
if trackclass == "S" then
for i=1,S do --Stated earlier that S=25
local roll=math.random(1,5)
local SP=math.random(1,3)
local Count= roll
if Count == 1 then
local Track = "Straightaway"
p(Track.." Of SP "..SP)
else
end
if Count == 2 then
local Track = "Curve"
p(Track.." of SP "..SP)
else
end
if Count == 3 then
local Track = "Hill"
p(Track.." of SP "..SP)
else
end
if Count == 4 then
local Track = "Water"
p(Track.." of SP "..SP)
else
end
if Count == 5 then
local Track = "Jump"
p(Track.." of SP "..SP)
else
end
end
end
Unfortunately this seems to generate a pretty poor set of random number distribution when I use it and I would really like it to work out better. Any possible assistance in fleshing out the variable loop cycle would be greatly appreciated.
Even something like a call so that every time math.random() is called, it adds one to the X[loop count]] so that every generated number is actually a better pseudo-random number distribution.
Please forgive my semi-rambling. My mind is not necessarily thinking in order right now.
Does anybody know a faster way to do this?
Each XN in the expression is always the previous X, so just restructure the code to use the previous X rather than creating new ones:
local X = os.time()
local A1 = 710425941047
local B1 = 813633012810
local M1 = 711719770602
function myrandomseed(val)
X = val
end
function myrandom()
X = (A1 * X + B1) % M1
return X
end
Now you can call myrandom to your heart's content:
for i=1,100 do
print(myrandom())
end
Another way of packaging it, to avoid static scope, would be generating random number generators as closures, which bind to their state variables:
function getrandom(seed)
local X = seed or os.time()
local A1 = 710425941047
local B1 = 813633012810
local M1 = 711719770602
return function()
X = (A1 * X + B1) % M1
return X
end
end
Now you call getrandom to get a random number generator for a given seed:
local rand = getrandom()
for i=1,100 do
print(rand())
end
I would love to be able to fit it into something along the lines of a:
math.randomseed(X[loop count])
local roll=math.random(1,20)
If you're calling randomseed every time you call random, you're not using Lua's (i.e. C's) random number generator at all. You can see why this is true by looking at myrandomseed above. Why are you funneling your numbers through Lua's random in the first place? Why not just use math.random and be done with it.
Just make to sure to call math.randomseed once rather than every time you call math.random and you'll be fine.
I'm using it to generate random numbers for pieces of track I'm making in a tabletop game. Example hunk of code:
When you see tons of nearly identical code, you should refactor it. When you see variables names like foo1, foo2, etc. you're either naming variables poorly or should be using a list. In your case you have a bunch of branches Count == 1, Count == 2, etc. when we could be using a list. For instance, this does the same thing as your code:
local trackTypes = { 'Straightaway', 'Curve', 'Hill', 'Water', 'Jump' }
for i=1,S do
local trackTypeIndex = math.random(1, #trackTypes)
local SP = math.random(1, 3)
p(trackTypes[trackTypeIndex]..' of SP '..SP)
end
Note that you can probably guess what trackTypes is just by reading the variable name. I have no idea what S and SP are. They are probably not good names.
Related
I have a structure called s in Matlab. This is a structure with two fields a and b. The structure size is 1 x 1,620,000.
It is a very large structure (that probably takes half of the ram of my machine). This is what the structure looks like:
I am looking for an efficient way to concatenate each of the fields a and b into two separate arrays that I can then export to csv. I built the code below, to do so, but even after 12 hours running it has not even reached a quarter of the loop. Any more efficient way of doing this?
a = [];
b =[];
total_n = size(s,2);
count = 1;
while size(s,2)>0
if size(s(1).a,1)
a = [a; s(1).a];
end
if size(s(1).b,1)
b = [b; s(1).b];
end
s(1) = []; %to save memory
if mod(count,1000) == 0
fprintf('Done %2f \n', [count/total_n])
end
count = count+1;
end
s(1) = []; %to save memory
ah, but such huge misunderstanding that comment is.
if size(s) is 1 x 1,620,000, you just suddenly forced the loop to do (under the hood, you dont see it)
snew=zeros(1,size(s,2)-1) # now you use double memory
snew=s(2:end) # now you force an unnecesary copy
So not only does that line make your code require double the memory, but also in each loop, you make an unnecesary copy of a large array.
Just replace your while for a normal for loop of for ii=1:size(s,2) and then index s!
Now, you can see hopefully then why the following is equally a big mistake (not only that, but any modern MATLAB version is currently telling you this is a bad idea in your editor)
a=[]
a=[a;s(1).a]
In here in each loop you are forcing MATLAB to make a new a that is 1 bigger than before, and copy the contents of the old a there.
instead, preallocate the size of a.
As you don't know what you are going to put there, I suggest using a cell array, as each s(ii).a has a different length.
You can then, after the loop, remove all empty (isempty) cells if you want.
Managed to do it efficiently:
s= struct2cell(s);
s= squeeze(s);
a = a(1,:);
a = a';
a = vertcat(a{:});
b = a(2,:);
b = b';
b = vertcat(b{:});
I've tried searching for help but I haven't found a solution yet, I'm trying to repeat math.random.
current code:
local ok = ""
for i = 0,10 do
local ok = ok..math.random(0,10)
end
print(ok)
no clue why it doesn't work, please help
Long answer
Even if the preferable answer is already given, just copying it will probably not lead to the solution you may expect or less future mistakes. So I decided to explain why your code fails and to fix it and also help better understand how DarkWiiPlayer's answer works (except for string.rep and string.gsub).
Issues
There are at least three issues in your code:
the math.random(m, n) function includes lower and the upper values
local declarations hide a same-name objects in outer scopes
math.random gives the same number sequence unless you set its seed with math.randomseed
See Detailed explanation section below for more.
Another point seems at least worth mentioning or suspicious to me, as I assume you might be puzzled by the result (it seems to me to reflect exactly the perspective of the C programmer, from which I also got to know Lua): the Lua for loop specifies start and end value, so both of these values are included.
Attempt to repair
Here I show how a version of your code that yields the same results as the answer you accepted: a sequence of 10 percent-encoded decimal digits.
-- this will change the seed value (but mind that its resolution is seconds)
math.randomseed(os.time())
-- initiate the (only) local variable we are working on later
local ok = ""
-- encode 10 random decimals (Lua's for-loop is one-based and inclusive)
for i = 1, 10 do
ok = ok ..
-- add fixed part
'%3' ..
-- concatenation operator implicitly converts number to string
math.random(0, 9) -- a random number in range [0..9]
end
print(ok)
Detailed explanation
This explanation makes heavily use of the assert function instead of adding print calls or comment what the output should be. In my opinion assert is the superior choice for illustrating expected behavior: The function guides us from one true statement - assert(true) - to the next, at the first miss - assert(false) - the program is exited.
Random ranges
The math library in Lua provides actually three random functions depending on the count of arguments you pass to it. Without arguments, the result is in the interval [0,1):
assert(math.random() >= 0)
assert(math.random() < 1)
the one-argument version returns a value between 1 and the argument:
assert(math.random(1) == 1)
assert(math.random(10) >= 1)
assert(math.random(10) <= 10)
the two-argument version explicitly specifies min and max values:
assert(math.random(2,2) == 2)
assert(math.random(0, 9) >= 0)
assert(math.random(0, 9) <= 9)
Hidden outer variable
In this example, we have two variables x of different type, the outer x is not accessible from the inner scope.
local x = ''
assert(type(x) == 'string')
do
local x = 0
assert(type(x) == 'number')
-- inner x changes type
x = x .. x
assert(x == '00')
end
assert(type(x) == 'string')
Predictable randomness
The first call to math.random() in a Lua program will return always the same number because the pseudorandom number generator (PRNG) starts at seed 1. So if you call math.randomseed(1), you'll reset the PRNG to its initial state.
r0 = math.random()
math.randomseed(1)
r1 = math.random()
assert(r0 == r1)
After calling math.randomseed(os.time()) calls to math.random() will return different sequences presuming that subsequent program starts differ at least by one second. See question Current time in milliseconds and its answers for more information about the resolutions of several Lua functions.
string.rep(".", 10):gsub(".", function() return "%3" .. math.random(0, 9) end)
That should give you what you want
I have a function that takes a variable amount of ints as arguments.
thisFunction(1,1,1,2,2,2,2,3,4,4,7,4,2)
this function was given in a framework and I'd rather not change the code of the function or the .lua it is from. So I want a function that repeats a number for me a certain amount of times so this is less repetitive. Something that could work like this and achieve what was done above
thisFunction(repeatNum(1,3),repeatNum(2,4),3,repeatNum(4,2),7,4,2)
is this possible in Lua? I'm even comfortable with something like this:
thisFunction(repeatNum(1,3,2,4,3,1,4,2,7,1,4,1,2,1))
I think you're stuck with something along the lines of your second proposed solution, i.e.
thisFunction(repeatNum(1,3,2,4,3,1,4,2,7,1,4,1,2,1))
because if you use a function that returns multiple values in the middle of a list, it's adjusted so that it only returns one value. However, at the end of a list, the function does not have its return values adjusted.
You can code repeatNum as follows. It's not optimized and there's no error-checking. This works in Lua 5.1. If you're using 5.2, you'll need to make adjustments.
function repeatNum(...)
local results = {}
local n = #{...}
for i = 1,n,2 do
local val = select(i, ...)
local reps = select(i+1, ...)
for j = 1,reps do
table.insert(results, val)
end
end
return unpack(results)
end
I don't have 5.2 installed on this computer, but I believe the only change you need is to replace unpack with table.unpack.
I realise this question has been answered, but I wondered from a readability point of view if using tables to mark the repeats would be clearer, of course it's probably far less efficient.
function repeatnum(...)
local i = 0
local t = {...}
local tblO = {}
for j,v in ipairs(t) do
if type(v) == 'table' then
for k = 1,v[2] do
i = i + 1
tblO[i] = v[1]
end
else
i = i + 1
tblO[i] = v
end
end
return unpack(tblO)
end
print(repeatnum({1,3},{2,4},3,{4,2},7,4,2))
So I'm trying to create a little something and I have looked all over the place looking for ways of generating a random number. However no matter where I test my code, it results in a non-random number. Here is an example I wrote up.
local lowdrops = {"Wooden Sword","Wooden Bow","Ion Thruster Machine Gun Blaster"}
local meddrops = {}
local highdrops = {}
function randomLoot(lootCategory)
if lootCategory == low then
print(lowdrops[math.random(3)])
end
if lootCategory == medium then
end
if lootCategory == high then
end
end
randomLoot(low)
Wherever I test my code I get the same result. For example when I test the code here http://www.lua.org/cgi-bin/demo it always ends up with the "Ion Thruster Machine Gun Blaster" and doesen't randomize. For that matter testing simply
random = math.random (10)
print(random)
gives me 9, is there something i'm missing?
You need to run math.randomseed() once before using math.random(), like this:
math.randomseed(os.time())
One possible problem is that the first number may not be so "randomized" in some platforms. So a better solution is to pop some random number before using them for real:
math.randomseed(os.time())
math.random(); math.random(); math.random()
Reference: Lua Math Library
Ok here's a basic for loop
local a = {"first","second","third","fourth"}
for i=1,#a do
print(i.."th iteration")
a = {"first"}
end
As it is now, the loop executes all 4 iterations.
Shouldn't the for-loop-limit be calculated on the go? If it is calculated dynamically, #a would be 1 at the end of the first iteration and the for loop would break....
Surely that would make more sense?
Or is there any particular reason as to why that is not the case?
The main reason why numerical for loops limits are computed only once is most certainly for performance.
With the current behavior, you can place arbitrary complex expressions in for loops limits without a performance penalty, including function calls. For example:
local prod = 1
for i = computeStartLoop(), computeEndLoop(), computeStep() do
prod = prod * i
end
The above code would be really slow if computeEndLoop and computeStep required to be called at each iteration.
If the standard Lua interpreter and most notably LuaJIT are so fast compared to other scripting languages, it is because a number of Lua features have been designed with performance in mind.
In the rare cases where the single evaluation behavior is undesirable, it is easy to replace the for loop with a generic loop using while end or repeat until.
local prod = 1
local i = computeStartLoop()
while i <= computeEndLoop() do
prod = prod * i
i = i + computeStep()
end
The length is computed once, at the time the for loop is initialized. It is not re-computed each time through the loop - a for loop is for iterating from a starting value to an ending value. If you want the 'loop' to terminate early if the array is re-assigned to, you could write your own looping code:
local a = {"first", "second", "third", "fourth"}
function process_array (fn)
local inner_fn
inner_fn =
function (ii)
if ii <= #a then
fn(ii,a)
inner_fn(1 + ii)
end
end
inner_fn(1, a)
end
process_array(function (ii)
print(ii.."th iteration: "..a[ii])
a = {"first"}
end)
Performance is a good answer but I think it also makes the code easier to understand and less error-prone. Also, that way you can (almost) be sure that a for loop always terminates.
Think about what would happen if you wrote that instead:
local a = {"first","second","third","fourth"}
for i=1,#a do
print(i.."th iteration")
if i > 1 then a = {"first"} end
end
How do you understand for i=1,#a? Is it an equality comparison (stop when i==#a) or an inequality comparison (stop when i>=#a). What would be the result in each case?
You should see the Lua for loop as iteration over a sequence, like the Python idiom using (x)range:
a = ["first", "second", "third", "fourth"]
for i in range(1,len(a)+1):
print(str(i) + "th iteration")
a = ["first"]
If you want to evaluate the condition every time you just use while:
local a = {"first","second","third","fourth"}
local i = 1
while i <= #a do
print(i.."th iteration")
a = {"first"}
i = i + 1
end