Can't modify loop-variable in lua [duplicate] - lua

This question already has answers here:
Lua for loop reduce i? Weird behavior [duplicate]
(3 answers)
Closed 7 years ago.
im trying this in lua:
for i = 1, 10,1 do
print(i)
i = i+2
end
I would expect the following output:
1,4,7,10
However, it seems like i is getting not affected, so it gives me:
1,2,3,4,5,6,7,8,9,10
Can someone tell my a bit about the background concept and what is the right way to modify the counter variable?

As Colonel Thirty Two said, there is no way to modify a loop variable in Lua. Or rather more to the point, the loop counter in Lua is hidden from you. The variable i in your case is merely a copy of the counter's current value. So changing it does nothing; it will be overwritten by the actual hidden counter every time the loop cycles.
When you write a for loop in Lua, it always means exactly what it says. This is good, since it makes it abundantly clear when you're doing looping over a fixed sequence (whether a count or a set of data) and when you're doing something more complicated.
for is for fixed loops; if you want dynamic looping, you must use a while loop. That way, the reader of the code is aware that looping is not fixed; that it's under your control.

When using a Numeric for loop, you can change the increment by the third value, in your example you set it to 1.
To see what I mean:
for i = 1,10,3 do
print(i)
end
However this isn't always a practical solution, because often times you'll only want to modify the loop variable under specific conditions. When you wish to do this, you can use a while loop (or if you want your code to run at least once, a repeat loop):
local i = 1
while i < 10 do
print(i)
i = i + 1
end
Using a while loop you have full control over the condition, and any variables (be they global or upvalues).

All answers / comments so far only suggested while loops; here's two more ways of working around this problem:
If you always have the same step size, which just isn't 1, you can explicitly give the step size as in for i =start,end,stepdo … end, e.g. for i = 1, 10, 3 do … or for i = 10, 1, -1 do …. If you need varying step sizes, that won't work.
A "problem" with while-loops is that you always have to manually increment your counter and forgetting this in a sub-branch easily leads to infinite loops. I've seen the following pattern a few times:
local diff = 0
for i = 1, n do
i = i+diff
if i > n then break end
-- code here
-- and to change i for the next round, do something like
if some_condition then
diff = diff + 1 -- skip 1 forward
end
end
This way, you cannot forget incrementing i, and you still have the adjusted i available in your code. The deltas are also kept in a separate variable, so scanning this for bugs is relatively easy. (i autoincrements so must work, any assignment to i below the loop body's first line is an error, check whether you are/n't assigning diff, check branches, …)

Related

(Lua) How do I increment a variable every time one of three if statements runs?

I'm building a moving and sensing bot in CoppelliaSim for school. CopelliaSim uses Lua for scripting. Basically every time the bot's forward sensors hit something, one of three if statements will run. I want it to count how many times any of these if statements runs, and once that count gets to a certain amount (like 20), I'll run something else, which will do the same thing (find collisions, add to the count, reach an amount, and switch back to the first).
i=0
result=sim.readProximitySensor(noseSensor) -- Read the proximity sensor
-- If we detected something, we set the backward mode:
if (result>0) then backUntilTime=sim.getSimulationTime()+3
print("Collision Detected")
i=i+1
print(i)
end
result=sim.readProximitySensor(noseSensor0) -- Read the proximity sensor
-- If we detected something, we set the backward mode:
if (result>0) then backUntilTime=sim.getSimulationTime()+3
print("Collision Detected")
i=i+1
print(i)
end
result=sim.readProximitySensor(noseSensor1) -- Read the proximity sensor
-- If we detected something, we set the backward mode:
if (result>0) then backUntilTime=sim.getSimulationTime()+3
print("Collision Detected")
i=i+1
print(i)
end
Above is the start of a function and one of the three If statements. I'm printing just to see if it actually increments. It is printing, but it is not incrementing (just 1 over and over). This bot has 3 sensors on it (an if statement for each sensor) and it adds 1 to i for the first collision and ignores the rest, even if it's from the same sensor. I feel like my problem is just some simple syntax issue with Lua that I don't know and can't find how to properly fix.
I'm happy to provide more code if this little snippet was not sufficient to answer this question.
Assuming that you have a looping function such as sysCall_actuation which is being executed per simulation step. As Joseph Sible-Reinstate Monica has already stated, you are setting your variable i back to zero every time a simulation step is executed. To achieve your goal, you would have to set your variable to 0 outside your function. There are two appropriate approaches to achieve that:
Define the variable outside any function, in the beginning of your file (or before you define any function that uses your variable e.g right before the definition of sysCall_actuation).
-- Beginning of the file.
local i = 0
..
function sysCall_actuation()
..
i = i + 1
..
end
Define your variable in sysCall_init function, which is the appropriate approach in CoppeliaSim.
function sysCall_init()
i = 0
..
end
Finally, you can use your variable in your sysCall_actuation function with basic comparison operations:
function sysCall_actuation()
..
if i > 20 then
i = 0 -- Reset 'i' so this function will not be running every step again and again.
-- Do something here.
..
end
As a side note, practice using local variables whenever you can, to keep the memory clean and avoid having ambiguous variables.

Is there a reason why this code is not outputting a worth

Problem
Hello, StackOverflow community! I am working on this Lua game, and I was testing to see if it would change the text on my TextLabel to the Bitcoins current worth, I was utterly disappointed when nothing showed up.
I have tried to do research on Google, and my code seems to be just right.
Code
Change = false
updated = false
while Change[true] do --While change = true do
worth = math.random(1,4500) --Pick random number
print('Working!') --Say its working
Updated = true --Change the updated local var.
end --Ending while loop
script.Parent.TextLabel.Text.Text = 'Bitcoin is currently worth: ' .. worth
--Going to the Text, and changing in to a New worth.
while Updated[false] do --While updated = false do
wait(180) --Wait
Change = true --After waits 3 minutes it makes an event trigger
end -- Ending while loop
wait(180) --Wait
Updated = false --Reseting Script.
I expect the output on the Label to be a random number.
I can't really speak to roblox, but there are a couple of obvious problems with your code:
Case
You have confusion between capitalized ("Updated", "Change") and lowercase ("updated", "change" [in commented while statement]), which will fail. See, for example:
bj#bj-lt:~$ lua
Lua 5.2.4 Copyright (C) 1994-2015 Lua.org, PUC-Rio
> Updated = true
> print(Updated)
true
> print(updated)
nil
So be super-careful about what identifiers you capitalize. In general, most programmers leave variables like that in all-lowercase (or sometimes things like camelCase). I suppose there might be some oddball lua runtime out there that is case-insensitive, but I don't know of one.
Type misuse.
Updated is a boolean (a true/false value), so the syntax:
while Change[true] do
...is invalid. See:
> if Updated[true] then
>> print("foo")
>> end
stdin:1: attempt to index global 'Updated' (a boolean value)
stack traceback:
stdin:1: in main chunk
[C]: in ?
Note also that the "While change == true do" is also wrong because of case ("While" is not valid lua, but "while" is).
Lastly:
Lack of threading.
You have basically two different things that you're trying to do at once, namely randomly change the "worth" variable as fast as possible (it's in a loop) and see a set a label to match it (it looks like you probably want it to change constantly). This requires two threads of operation (one to change worth and another to read it and stick it on the label). You've written this like you're assuming you have a spreadsheet or something and that. What your code is actually doing is:
Setting some variables
Updating worth indefinitely, printing 'Working!' a bunch, and...
Never stopping
The rest of the code never runs, because the rest of the code isn't in a background thread (basically the first bit monopolizes the runtime and never yields to everything else).
Lastly, even if the top code was running in the background, you only set the Text label one-time to exactly "Bitcoin is currently worth: 3456" (or some similar number) one time. The way this is written there won't be any updates thereafter (and, if it runs once before the other thread has warmed up, it might not be set to anything useful at all).
My guess is that your runtime is spitting out errors left and right due to the identifier problems and/or is running in a tight infinite loop and never actually getting to the label refresh logic.
BJ Black has given an excellent description of the issues with the syntax, so I'll try to cover the Roblox piece of this. In order for this kind of thing to work properly in a Roblox game, here are some assumptions to double check :
Since we are working with a TextLabel, is it inside a ScreenGui? Or a SurfaceGui?
If it's in a ScreenGui, make sure that ScreenGui is in StarterGui, and is this code in a LocalScript
If it's in a SurfaceGui, make sure that SurfaceGui is adorning a Part and this code
is in a Script
After you checked all those pieces, maybe this is closer to what you were thinking :
-- define the variables we're working with
local textLabel = script.Parent.TextLabel
local worth = 0
-- create an infinite loop
spawn(function()
while true do
--Pick random number
worth = math.random(1,4500)
-- update the text of the label with the new worth
textLabel.Text = string.format("Bitcoin is currently worth: %d", worth)
-- wait for 3 minutes, then loop
wait(180)
end
end)
I removed Updated and Changed because all they were doing was deciding whether or not to change the value. The flow of your loop was:
do nothing and display an undefined number. Wait 3 minutes
update the number, display it, wait 6 minutes
repeat 1 and 2.
So hopefully this is a little clearer and closer to what you were thinking.

Some lines ending with ";1" in Rails legacy code

Hey Stack Overflow fellows,
I need your help on this one. For a few months now I'm dealing with Ruby on Rails application that is mostly Legacy. Today, I noticed the weirdest thing about the codebase. Some files, not many but the significant ones, contain a few lines of code that would end up with ; 1. Like for example: Users.find(id); 1. Occurrence of those suffixes does not create any form of the pattern. Sometimes is ; 1 appears after puts or after expression that will always return value e.g. nil || 'default_value'; 1.
Does it make any sense to use the suffix? Is there any reason behind this? Maybe there used to be a tool that worked with Ruby code and ; 1 was form of annotation. I would gladly remove the suffix but I want to make sure that it's 100% safe.
Here is a code sample from the project added in the same commit:
times = events.map{|x| [x.time, x.time_from_impression_id]};1
times = times.map{|x| (x.first - x.last) / 1.day}.sort;1
time_to_event_success = times[(times.length.to_f * 0.95).to_i]
events = events.select{|e| e.time_from_impression_id < time_to_event_success.days.ago};1
Semi-colons in ruby terminate statements in the same manner as a line break. The ; 1 really isn't doing anything useful.
Logically the code you posted is equivalent to:
times = events.map{|x| [x.time, x.time_from_impression_id]}
1
times = times.map{|x| (x.first - x.last) / 1.day}.sort
1
time_to_event_success = times[(times.length.to_f * 0.95).to_i]
events = events.select{|e| e.time_from_impression_id <
time_to_event_success.days.ago}
1
The only thing I can think of is that if someone was testing the code out in IRB adding the ; 1 to the end of a line would prevent the return value of the previous statement from echoing. That or they didn't quite understand how implicit return and truthy and falsy values work in ruby.

Initialising hist function in Julia for use in a loop

When I use a loop, to access the variables outside of the loop they need to be initialised before you enter the loop. For example:
Y = Array{Int}()
for i = 1:end
Y = i
end
Since I have initialised Y before entering the loop, I can access it later by typing
Y
If I had not initialised it before entering the loop, typing Y would not have returned anything.
I want to extend this functionality to the output of the 'hist' function. I don't know how to set up the empty hist output before the loop. The only work around I have found is below.
yHistData = [hist(DataSet[1],Bins)]
for j = 2:NumberOfLayers
yHistData = [yHistData;hist(DataSet[j],Bins)]
end
Now when I access this later on by simply typing
yHistData
I get the correct values returned to me.
How can I initialise this hist data before entering the loop without defining it using the first value of the list I'm iterating over?
This can be done with a loop like follows:
yHistData = []
for j = 1:NumberOfLayers
push!(yHistData, hist(DataSet[j], Bins))
end
push! modifies the array by adding the specified element to the end. This increases code speed because we do not need to create copies of the array all the time. This code is nice and simple, and runs faster than yours. The return type, however, is now Array{Any, 1}, which can be improved.
Here I have typed the array so that the performance when using this array in the future is better. Without typing the array, the performance is sometimes better and sometimes worse than your code, depending on NumberOfLayers.
yHistData = Tuple{FloatRange{Float64},Array{Int64,1}}[]
for j = 1:NumberOfLayers
push!(yHistData, hist(DataSet[j], Bins))
end
Assuming length(DataSet) == NumberOfLayers, we can use anonymous functions to simplify the code even further:
yHistData = map(data -> hist(data, Bins), DataSet)
This solution is short, easy to read, and very fast on Julia 0.5. However, this version is not yet released. On 0.4, the currently released version, the performance of this version will be slower.

matlab indexing into nameless matrix [duplicate]

For example, if I want to read the middle value from magic(5), I can do so like this:
M = magic(5);
value = M(3,3);
to get value == 13. I'd like to be able to do something like one of these:
value = magic(5)(3,3);
value = (magic(5))(3,3);
to dispense with the intermediate variable. However, MATLAB complains about Unbalanced or unexpected parenthesis or bracket on the first parenthesis before the 3.
Is it possible to read values from an array/matrix without first assigning it to a variable?
It actually is possible to do what you want, but you have to use the functional form of the indexing operator. When you perform an indexing operation using (), you are actually making a call to the subsref function. So, even though you can't do this:
value = magic(5)(3, 3);
You can do this:
value = subsref(magic(5), struct('type', '()', 'subs', {{3, 3}}));
Ugly, but possible. ;)
In general, you just have to change the indexing step to a function call so you don't have two sets of parentheses immediately following one another. Another way to do this would be to define your own anonymous function to do the subscripted indexing. For example:
subindex = #(A, r, c) A(r, c); % An anonymous function for 2-D indexing
value = subindex(magic(5), 3, 3); % Use the function to index the matrix
However, when all is said and done the temporary local variable solution is much more readable, and definitely what I would suggest.
There was just good blog post on Loren on the Art of Matlab a couple days ago with a couple gems that might help. In particular, using helper functions like:
paren = #(x, varargin) x(varargin{:});
curly = #(x, varargin) x{varargin{:}};
where paren() can be used like
paren(magic(5), 3, 3);
would return
ans = 16
I would also surmise that this will be faster than gnovice's answer, but I haven't checked (Use the profiler!!!). That being said, you also have to include these function definitions somewhere. I personally have made them independent functions in my path, because they are super useful.
These functions and others are now available in the Functional Programming Constructs add-on which is available through the MATLAB Add-On Explorer or on the File Exchange.
How do you feel about using undocumented features:
>> builtin('_paren', magic(5), 3, 3) %# M(3,3)
ans =
13
or for cell arrays:
>> builtin('_brace', num2cell(magic(5)), 3, 3) %# C{3,3}
ans =
13
Just like magic :)
UPDATE:
Bad news, the above hack doesn't work anymore in R2015b! That's fine, it was undocumented functionality and we cannot rely on it as a supported feature :)
For those wondering where to find this type of thing, look in the folder fullfile(matlabroot,'bin','registry'). There's a bunch of XML files there that list all kinds of goodies. Be warned that calling some of these functions directly can easily crash your MATLAB session.
At least in MATLAB 2013a you can use getfield like:
a=rand(5);
getfield(a,{1,2}) % etc
to get the element at (1,2)
unfortunately syntax like magic(5)(3,3) is not supported by matlab. you need to use temporary intermediate variables. you can free up the memory after use, e.g.
tmp = magic(3);
myVar = tmp(3,3);
clear tmp
Note that if you compare running times with the standard way (asign the result and then access entries), they are exactly the same.
subs=#(M,i,j) M(i,j);
>> for nit=1:10;tic;subs(magic(100),1:10,1:10);tlap(nit)=toc;end;mean(tlap)
ans =
0.0103
>> for nit=1:10,tic;M=magic(100); M(1:10,1:10);tlap(nit)=toc;end;mean(tlap)
ans =
0.0101
To my opinion, the bottom line is : MATLAB does not have pointers, you have to live with it.
It could be more simple if you make a new function:
function [ element ] = getElem( matrix, index1, index2 )
element = matrix(index1, index2);
end
and then use it:
value = getElem(magic(5), 3, 3);
Your initial notation is the most concise way to do this:
M = magic(5); %create
value = M(3,3); % extract useful data
clear M; %free memory
If you are doing this in a loop you can just reassign M every time and ignore the clear statement as well.
To complement Amro's answer, you can use feval instead of builtin. There is no difference, really, unless you try to overload the operator function:
BUILTIN(...) is the same as FEVAL(...) except that it will call the
original built-in version of the function even if an overloaded one
exists (for this to work, you must never overload
BUILTIN).
>> feval('_paren', magic(5), 3, 3) % M(3,3)
ans =
13
>> feval('_brace', num2cell(magic(5)), 3, 3) % C{3,3}
ans =
13
What's interesting is that feval seems to be just a tiny bit quicker than builtin (by ~3.5%), at least in Matlab 2013b, which is weird given that feval needs to check if the function is overloaded, unlike builtin:
>> tic; for i=1:1e6, feval('_paren', magic(5), 3, 3); end; toc;
Elapsed time is 49.904117 seconds.
>> tic; for i=1:1e6, builtin('_paren', magic(5), 3, 3); end; toc;
Elapsed time is 51.485339 seconds.

Resources