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))
Related
I'm wanting to remove all of a possibly duplicated value in an array. At the moment I'm using the remove(x:T):Bool function in a while loop, but I'm wondering about the expression part.
I've started by using:
function removeAll(array:Array<String>, element:String):Void
while (array.remove(element)) {}
but I'm wondering if any of these lines would be more efficient:
while (array.remove(element)) continue;
while (array.remove(element)) true;
while (array.remove(element)) 0;
or if it makes any kind of difference.
I'm guessing that using continue is less efficient because it actually has to do something, true and 0 are slightly more efficient, but still do something, and {} would probably be most efficient.
Does anyone have any background information on this?
While other suggested filter, it will create a new instance of list/array which may cause your other code to lose reference.
If you loop array.remove, it is going to loop through all the elements in the front of the array every time, which is not so performant.
IMO a better approach is to use a reverse while loop:
var i = array.length;
while(--i >= 0)
if(array[i] == element) array.splice(i, 1);
It doesn't make any difference. In fact, there's not even any difference in the generated code for the {}, 0 and false cases: they all end up generating {}, at least on the JS target.
However, you could run into issues if you have a large array with many duplicates: in that case, remove() would be called many times, and it has to iterate over the array each time (until it finds a match, that is). In that case, it's probably more efficient to use filter():
function removeAll(array:Array<String>, element:String):Array<String>
return array.filter(function(e) return e != element);
Personally, I also find this to be a bit more elegant than your while-loop with an empty body. But again, it depends on the use case: this does create a new array, and thus causes an allocation. Usually, that's not worth worrying about, but if you for instance do it in the update loop of a game, you might want to avoid it.
In terms of the expression part of the while loop, it seems that it's just set to empty brases ({}) when compiled so it doesn't really matter what you do.
In terms of performance, a much better solution is the Method 2 from the following:
class Test
{
static function main()
{
var thing:Array<String> = new Array<String>();
for (index in 0...1000)
{
thing.push("0");
thing.push("1");
}
var copy1 = thing.copy();
var copy2 = thing.copy();
trace("epoch");
while (copy1.remove("0")) {}
trace("check");
// Method 2.
copy2 = [
for (item in Lambda.filter(copy2, function(v)
{return v != "0";}))
item
];
trace("check");
}
}
which can be seen [here](https://try.haxe.org/#D0468"Try Haxe example."). For 200,000 one-character elements in an Array<String>, Method 2 takes 0.017s while Method 1 takes 44.544s.
For large arrays it will be faster to use a temporary array and then assign that back after populating ( method3 in try )?
OR
If you don't want to use a temp you can assign back and splice ( method4 in try )?
https://try.haxe.org/#5f80c
Both are more verbose codewise as I setup vars, but on mac seems faster at runtime, summary of my method3 approach:
while( i < l ) { if( ( s = copy[ i++ ] ) != '0' ) arr[ j++ ] = s;
copy = arr;
am I missing something obvious against these approaches?
After working on my code for a while, optimizing the most obvious things, I've resulted in this:
function FindPath(start, finish, path)
--Define a table to hold the paths
local paths = {}
--Make a default argument
path = path or {start}
--Loop through connected nodes
for i,v in ipairs(start:GetConnectedParts()) do
--Determine if backtracking
local loop = false
for i,vv in ipairs(path) do
if v == vv then
loop = true
end
end
if not loop then
--Make a path clone
local npath = {unpack(path)}
npath[#npath+1] = v
if v == finish then
--If we reach the end add the path
return npath
else
--Otherwise add the shortest part extending from this node
paths[#paths+1] = FindPath(v, finish, npath) --NOTED HERE
end
end
end
--Find and return the shortest path
if #paths > 0 then
local lengths = {}
for i,v in ipairs(paths) do
lengths[#lengths+1] = #v
end
local least = math.min(unpack(lengths))
for i,v in ipairs(paths) do
if #v == least then
return v
end
end
end
end
The problem being, the line noted gets some sort of game script timeout error (which I believe is a because of mass recursion with no yielding). I also feel like once that problem is fixed, it'll probably be rather slow even on the scale of a pacman board. Is there a way I can further optimize it, or perhaps a better method I can look into similar to this?
UPDATE: I finally decided to trash my algorithm due to inefficiency, and implemented a Dijkstra algorithm for pathfinding. For anybody interested in the source code it can be found here: http://pastebin.com/Xivf9mwv
You know that Roblox provides you with the PathfindingService? It uses C-side A* pathing to calculate quite quickly. I'd recommend using it
http://wiki.roblox.com/index.php?title=API:Class/PathfindingService
Try to remodel your algorithm to make use of tail calls. This is a great mechanism available in Lua.
A tail call is a type of recursion where your function returns a function call as the last thing it does. Lua has proper tail calls implementation and it will dress this recursion as a 'goto' under the scenes, so your stack will never blow.
Passing 'paths' as one of the arguments of FindPath might help with that.
I saw your edit about ditching the code, but just to help others stumbling on this question:
ipairs is slower than pairs, which is slower than a numeric for-loop.
If performance matters, never use ipairs, but use a for i=1,#tab loop
If you want to clone a table, use a for-loop. Sometimes, you have to use unpack (returning dynamic amount of trailing nils), but this is not such a case. Unpack is also a slow function.
Replacing ipairs with pairs or numeric for-loops and using loops instead of unpack will increase the performance a lot.
If you want to get the lowest value in a table, use this code snippet:
local lowestValue = values[1]
for k,v in pairs(values) do
if v < lowestValue then
lowestValue = k,v
end
end
This could be rewritten for your path example as so:
local least = #path[1]
for k,v in pairs(path) do
if #v < least then
least = v
end
end
I have to admit, you're very inventive. Not a lot of people would use math.min(unpack(tab)) (not counting the fact it's bad)
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.
This is similar to a question I asked before, but is slightly different:
So I have a very large structure array in matlab. Suppose, for argument's sake, to simplify the situation, suppose I have something like:
structure(1).name, structure(2).name, structure(3).name structure(1).returns, structure(2).returns, structure(3).returns (in my real program I have 647 structures)
Suppose further that structure(i).returns is a vector (very large vector, approximately 2,000,000 entries) and that a condition comes along where I want to delete the jth entry from structure(i).returns for all i. How do you do this? or rather, how do you do this reasonably fast? I have tried some things, but they are all insanely slow (I will show them in a second) so I was wondering if the community knew of faster ways to do this.
I have parsed my data two different ways; the first way had everything saved as cell arrays, but because things hadn't been working well for me I parsed the data again and placed everything as vectors.
What I'm actually doing is trying to delete NaN data, as well as all data in the same corresponding row of my data file, and then doing the very same thing after applying the Hampel filter. The relevant part of my code in this attempt is:
for i=numStock+1:-1:1
for j=length(stock(i).return):-1:1
if(isnan(stock(i).return(j)))
for k=numStock+1:-1:1
stock(k).return(j) = [];
end
end
end
stock(i).return = sort(stock(i).return);
stock(i).returnLength = length(stock(i).return);
stock(i).medianReturn = median(stock(i).return);
stock(i).madReturn = mad(stock(i).return,1);
end;
for i=numStock:-1:1
for j = length(stock(i+1).volume):-1:1
if(isnan(stock(i+1).volume(j)))
for k=numStock:-1:1
stock(k+1).volume(j) = [];
end
end
end
stock(i+1).volume = sort(stock(i+1).volume);
stock(i+1).volumeLength = length(stock(i+1).volume);
stock(i+1).medianVolume = median(stock(i+1).volume);
stock(i+1).madVolume = mad(stock(i+1).volume,1);
end;
for i=numStock+1:-1:1
for j=stock(i).returnLength:-1:1
if (abs(stock(i).return(j) - stock(i).medianReturn) > 3*stock(i).madReturn)
for k=numStock+1:-1:1
stock(k).return(j) = [];
end
end;
end;
end;
for i=numStock:-1:1
for j=stock(i+1).volumeLength:-1:1
if (abs(stock(i+1).volume(j) - stock(i+1).medianVolume) > 3*stock(i+1).madVolume)
for k=numStock:-1:1
stock(k+1).volume(j) = [];
end
end;
end;
end;
However, this returns an error:
"Matrix index is out of range for deletion.
Error in Failure (line 110)
stock(k).return(j) = [];"
So instead I tried by parsing everything in as vectors. Then I decided to try and delete the appropriate entries in the vectors prior to building the structure array. This isn't returning an error, but it is very slow:
%% Delete bad data, Hampel Filter
% Delete bad entries
id=strcmp(returns,'');
returns(id)=[];
volume(id)=[];
date(id)=[];
ticker(id)=[];
name(id)=[];
permno(id)=[];
sp500(id) = [];
id=strcmp(returns,'C');
returns(id)=[];
volume(id)=[];
date(id)=[];
ticker(id)=[];
name(id)=[];
permno(id)=[];
sp500(id) = [];
% Convert returns from string to double
returns=cellfun(#str2double,returns);
sp500=cellfun(#str2double,sp500);
% Delete all data for which a return is not a number
nanid=isnan(returns);
returns(nanid)=[];
volume(nanid)=[];
date(nanid)=[];
ticker(nanid)=[];
name(nanid)=[];
permno(nanid)=[];
% Delete all data for which a volume is not a number
nanid=isnan(volume);
returns(nanid)=[];
volume(nanid)=[];
date(nanid)=[];
ticker(nanid)=[];
name(nanid)=[];
permno(nanid)=[];
% Apply the Hampel filter, and delete all data corresponding to
% observations deleted by the filter.
medianReturn = median(returns);
madReturn = mad(returns,1);
for i=length(returns):-1:1
if (abs(returns(i) - medianReturn) > 3*madReturn)
returns(i) = [];
volume(i)=[];
date(i)=[];
ticker(i)=[];
name(i)=[];
permno(i)=[];
end;
end
medianVolume = median(volume);
madVolume = mad(volume,1);
for i=length(volume):-1:1
if (abs(volume(i) - medianVolume) > 3*madVolume)
returns(i) = [];
volume(i)=[];
date(i)=[];
ticker(i)=[];
name(i)=[];
permno(i)=[];
end;
end
As I said, this is very slow, probably because I'm using a for loop on a very large data set; however, I'm not sure how else one would do this. Sorry for the gigantic post, but does anyone have a suggestion as to how I might go about doing what I'm asking in a reasonable way?
EDIT: I should add that getting the vector method to work is probably preferable, since my aim is to put all of the return vectors into a matrix and get all of the volume vectors into a matrix and perform PCA on them, and I'm not sure how I would do that using cell arrays (or even if princomp would work on cell arrays).
EDIT2: I have altered the code to match your suggestion (although I did decide to give up speed and keep with the for-loops to keep with the structure array, since reparsing this data will be way worse time-wise). The new code snipet is:
stock_return = zeros(numStock+1,length(stock(1).return));
for i=1:numStock+1
for j=1:length(stock(i).return)
stock_return(i,j) = stock(i).return(j);
end
end
stock_return = stock_return(~any(isnan(stock_return)), : );
This returns an Index exceeds matrix dimensions error, and I'm not sure why. Any suggestions?
I could not find a convenient way to handle structures, therefore I would restructure the code so that instead of structures it uses just arrays.
For example instead of stock(i).return(j) I would do stock_returns(i,j).
I show you on a part of your code how to get rid of for-loops.
Say we deal with this code:
for j=length(stock(i).return):-1:1
if(isnan(stock(i).return(j)))
for k=numStock+1:-1:1
stock(k).return(j) = [];
end
end
end
Now, the deletion of columns with any NaN data goes like this:
stock_return = stock_return(:, ~any(isnan(stock_return)) );
As for the absolute difference from medianVolume, you can write a similar code:
% stock_return_length is a scalar
% stock_median_return is a column vector (eg. [1;2;3])
% stock_mad_return is also a column vector.
median_return = repmat(stock_median_return, stock_return_length, 1);
is_bad = abs(stock_return - median_return) > 3.* stock_mad_return;
stock_return = stock_return(:, ~any(is_bad));
Using a scalar for stock_return_length means of course that the return lengths are the same, but you implicitly assume it in your original code anyway.
The important point in my answer is using any. Logical indexing is not sufficient in itself, since in your original code you delete all the values if any of them is bad.
Reference to any: http://www.mathworks.co.uk/help/matlab/ref/any.html.
If you want to preserve the original structure, so you stick to stock(i).return, you can speed-up your code using essentially the same scheme but you can only get rid of one less for-loop, meaning that your program will be substantially slower.
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