NodeMCU (Lua) Parallel Timers - timer

I'm quite a beginner and facing to an issue I can't solve for long time.
NodeMCU: 0.9.6 build 20150704
Lua: 5.1.4
********************* B1_up.lua ******************
Mpin = 8
gpio.mode(Mpin, gpio.OUTPUT)
gpio.write(Mpin, gpio.HIGH)
xdf = 0
tmr.alarm(4, 1000, 1, function()
if xdf < 5 then xdf = xdf+ 1
else
gpio.write(Mpin, gpio.LOW)
tmr.stop(4)
end
end)
********************* B2_up.lua ******************
Mpin = 0
gpio.mode(Mpin, gpio.OUTPUT)
gpio.write(Mpin, gpio.HIGH)
xkf = 0
tmr.alarm(5, 1000, 1, function()
if xkf < 5 then xkf = xkf+ 1
else
gpio.write(Mpin, gpio.LOW)
tmr.stop(5)
end
end)
Separately working fine each, BUT executing in sequence:
dofile("B1_up.lua")
dofile("B2_up.lua")
... only the last one (in this case B2_up.lua) executes the gpio.write(Mpin, gpio.LOW) command after 5 seconds. Why?

Related

Roblox Studio Lua if-statement in loop

I have a Roblox Game in this game the time changes using the code on the Roblox Developer site(robloxdev.com) I have been making a door with two unions called "open" and "closed". I want the door to be open between 10 in the morning and 5 in the evening. However the door won't open and it's not even bringing up the print open/close when it is the right time.
This is my current code Note: The script is in the same model (called: Door) as the two unions.
while true do
if game.Lighting.ClockTime > 10 and game.Lighting.ClockTime < 17 then
--Open the door
print("open")
script.Parent.Closed.Transparency = 1
script.Parent.Closed.CanCollide = false
script.Parent.Open.Transparency = 0
script.Parent.Open.CanCollide = true
else
--Close the door
print("close")
script.Parent.Closed.Transparency = 0
script.Parent.Closed.CanCollide = true
script.Parent.Open.Transparency = 1
script.Parent.Open.CanCollide = false
end
end
Thank's for any help.
You should add wait inside the while loop.
while true do
if game.Lighting.ClockTime > 10 and game.Lighting.ClockTime < 17 then
--Open the door
print("open")
script.Parent.Closed.Transparency = 1
script.Parent.Closed.CanCollide = false
script.Parent.Open.Transparency = 0
script.Parent.Open.CanCollide = true
else
--Close the door
print("close")
script.Parent.Closed.Transparency = 0
script.Parent.Closed.CanCollide = true
script.Parent.Open.Transparency = 1
script.Parent.Open.CanCollide = false
end
wait(1) -- change this to whatever you want
end

how to create multi line chart with single datatable in vb.net

I am working on vb.net application where i have to create a multi line chart. The table coming from the database using storing procedure is:-
TagName On Off Trip
P1 0 0 1
P2 0 1 1
P3 0 1 0
Q1 0 1 0
Q2 1 0 1
Q3 2 2 2
W1 4 2 1
W2 2 0 1
W3 1 1 0
W4 0 1 1
W5 2 1 1
And the code in vb.net i used to bind the chart named "chTrend" is:-
ds = ObjTags.GetTrendData()
If (ds.Tables(0).Rows.Count > 0) Then
dt = ds.Tables(0)
chTrend.DataSource = dt
chTrend.Series(0).XValueMember = "TagName"
chTrend.Series(0).YValueMembers = "On"
chTrend.Series(1).XValueMember = "TagName"
chTrend.Series(1).YValueMembers = "Off"
chTrend.Series(1).XValueMember = "TagName"
chTrend.Series(1).YValueMembers = "Trip"
For i = 0 To 2
chTrend.Series(i).ChartType = SeriesChartType.Line
chTrend.Series(i).IsVisibleInLegend = True
chTrend.Series(i).IsValueShownAsLabel = True
chTrend.Series(i).ToolTip = "Data Point Y Value #VALY{G}"
chTrend.Series(i).BorderWidth = 3
Next
End If
When i run the program a Error is coming as following
"Index was out of range. Must be non-negative and less than the size of the collection.Parameter name: index"
At the position
chTrend.Series(0).XValueMember = "TagName"
how to solve this?
Assuming that your DataTable only has the four DataColumns {TagName, On. Off. Trip}, then replace everything in your shown code below the line:
chTrend.DataSource = dt
with this.
chTrend.Series.Clear() ' clears any existing series
Dim s As Series
For columnIndex As Int32 = 1 To dt.Columns.Count - 1
Dim name As String = dt.Columns(columnIndex).ColumnName
s = chTrend.Series.Add(name)
s.XValueMember = dt.Columns(0).ColumnName
s.YValueMembers = name
s.ChartType = SeriesChartType.Line
s.IsVisibleInLegend = True
s.IsValueShownAsLabel = True
s.ToolTip = "Data Point Y Value #VALY{G}"
s.BorderWidth = 3
Next
chTrend.DataBind() ' loads the data from dt to the chart
The Chart databinding is a not a binding in the normal sense of binding. You need to tell the chart to copy the data from the source so that it has something to plot. That is what the DataBind method does.

MATLAB data indexing issue. What is going on here?

Let I be the identity, D an orthonormal projection, and p a vector.
I realized that several of my lines of code combined to be (I-(I-D))(p) and I could just simplify it to D(p). In replacing it, I computed the new method along-side the old to double check I was computing the same thing (Earlier in my code I had a line that was D = I - D. The D you see here is that D.) I wasn't getting the same answer, and traced it to an error in indexing D.
Here you can see I'm using the debugger and checking portions of D and getting the wrong data returned.
The values in the data explorer on the right are what I'd expect them to be. Sometimes I get what I'd expect from D(:,:,k,1), and sometimes I don't, even when I make the queries right after each other.
The vectors those red arrows are pointing to should be the same. Nothing else changed or was computed between those lines, and k = 2 when the first line was run. I've closed MATLAB and restarted it and get the same issue every time. (D depends on random input, but I'm not altering the seed, so I get the same thing every first run after newly opening MATLAB. The way D is computed, I do expect D(:,:,1,1) to be the identity matrix.)
What in the world is going on? Any help is appreciated.
I have wondered if MATLAB is messing with me on purpose. Sometimes when I open it, a pop-up dialog box says I need to update my student license. I click the update button, but nothing ever happens and the dialog box never closes, so I click cancel.
Edit:
K>> whos D P
Name Size Bytes Class Attributes
D 4-D 4608 double
P 4x1x6 192 double
K>> size(D)
ans =
4 4 6 6
I've been playing around with A and B a bit, and I get the same thing. Sometimes it computes correctly and sometimes it doesn't.
K>> B=permute(P,[1,3,2])
B =
0.4155 0.27554 0.52338 0.6991 -0.11346 0.20999
0.53573 -0.83781 0.53182 -0.022364 0.60291 -0.62601
-0.49246 -0.46111 -0.39168 0.45919 0.42377 0.47074
0.54574 0.097595 0.53835 -0.54763 0.66637 0.58516
K>> A=D
A(:,:,1,1) =
1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 1
A(:,:,2,1) =
0.99071 -0.091198 0.0020814 -0.029755
-0.091198 0.10503 0.020426 -0.292
0.0020814 0.020426 0.99953 0.0066643
-0.029755 -0.292 0.0066643 0.90473
A(:,:,3,1) =
0.46769 0.019281 -0.49725 0.036486
0.019281 0.9993 0.018011 -0.0013215
-0.49725 0.018011 0.53551 0.034083
0.036486 -0.0013215 0.034083 0.9975
A(:,:,4,1) =
0.96774 0.063488 -0.10826 0.12438
0.063488 0.87506 0.21304 -0.24477
-0.10826 0.21304 0.63673 0.41737
0.12438 -0.24477 0.41737 0.52047
A(:,:,5,1) =
0.7542 0.031217 0.42575 0.056052
0.031217 0.99604 -0.054071 -0.0071187
0.42575 -0.054071 0.26255 -0.097088
0.056052 -0.0071187 -0.097088 0.98722
A(:,:,6,1) =
0.9818 -0.10286 0.085279 0.0034902
-0.10286 0.41855 0.48208 0.01973
0.085279 0.48208 0.60031 -0.016358
0.0034902 0.01973 -0.016358 0.99933
A(:,:,1,2) =
0.99071 -0.091198 0.0020814 -0.029755
-0.091198 0.10503 0.020426 -0.292
0.0020814 0.020426 0.99953 0.0066643
-0.029755 -0.292 0.0066643 0.90473
A(:,:,2,2) =
1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 1
A(:,:,3,2) =
0.97125 -0.15889 -0.0080537 -0.051131
-0.15889 0.12194 -0.044507 -0.28256
-0.0080537 -0.044507 0.99774 -0.014323
-0.051131 -0.28256 -0.014323 0.90907
A(:,:,4,2) =
0.91488 -0.16388 -0.18495 0.12967
-0.16388 0.6845 -0.35607 0.24964
-0.18495 -0.35607 0.59815 0.28174
0.12967 0.24964 0.28174 0.80247
A(:,:,5,2) =
0.95461 0.16812 0.10326 0.066372
0.16812 0.37733 -0.38244 -0.24582
0.10326 -0.38244 0.76511 -0.15098
0.066372 -0.24582 -0.15098 0.90295
A(:,:,6,2) =
0.99628 0.012018 0.052874 0.027665
0.012018 0.96117 -0.17085 -0.089393
0.052874 -0.17085 0.24833 -0.39329
0.027665 -0.089393 -0.39329 0.79422
A(:,:,1,3) =
0.46769 0.019281 -0.49725 0.036486
0.019281 0.9993 0.018011 -0.0013215
-0.49725 0.018011 0.53551 0.034083
0.036486 -0.0013215 0.034083 0.9975
A(:,:,2,3) =
0.97125 -0.15889 -0.0080537 -0.051131
-0.15889 0.12194 -0.044507 -0.28256
-0.0080537 -0.044507 0.99774 -0.014323
-0.051131 -0.28256 -0.014323 0.90907
A(:,:,3,3) =
1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 1
A(:,:,4,3) =
0.98622 0.043449 -0.066709 0.085142
0.043449 0.86297 0.21038 -0.26852
-0.066709 0.21038 0.67698 0.41227
0.085142 -0.26852 0.41227 0.47382
A(:,:,5,3) =
0.62859 0.041458 0.47558 0.074661
0.041458 0.99537 -0.053085 -0.0083339
0.47558 -0.053085 0.39105 -0.0956
0.074661 -0.0083339 -0.0956 0.98499
A(:,:,6,3) =
0.95505 -0.16608 0.12371 0.0067153
-0.16608 0.38639 0.45705 0.02481
0.12371 0.45705 0.65956 -0.01848
0.0067153 0.02481 -0.01848 0.999
A(:,:,1,4) =
0.96774 0.063488 -0.10826 0.12438
0.063488 0.87506 0.21304 -0.24477
-0.10826 0.21304 0.63673 0.41737
0.12438 -0.24477 0.41737 0.52047
A(:,:,2,4) =
0.91488 -0.16388 -0.18495 0.12967
-0.16388 0.6845 -0.35607 0.24964
-0.18495 -0.35607 0.59815 0.28174
0.12967 0.24964 0.28174 0.80247
A(:,:,3,4) =
0.98622 0.043449 -0.066709 0.085142
0.043449 0.86297 0.21038 -0.26852
-0.066709 0.21038 0.67698 0.41227
0.085142 -0.26852 0.41227 0.47382
A(:,:,4,4) =
1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 1
A(:,:,5,4) =
0.73864 0.20112 -0.011394 0.39048
0.20112 0.84524 0.0087678 -0.30047
-0.011394 0.0087678 0.9995 0.017023
0.39048 -0.30047 0.017023 0.41662
A(:,:,6,4) =
0.87322 -0.15647 0.0029936 0.29363
-0.15647 0.80689 0.0036946 0.36238
0.0029936 0.0036946 0.99993 -0.0069332
0.29363 0.36238 -0.0069332 0.31996
A(:,:,1,5) =
0.7542 0.031217 0.42575 0.056052
0.031217 0.99604 -0.054071 -0.0071187
0.42575 -0.054071 0.26255 -0.097088
0.056052 -0.0071187 -0.097088 0.98722
A(:,:,2,5) =
0.95461 0.16812 0.10326 0.066372
0.16812 0.37733 -0.38244 -0.24582
0.10326 -0.38244 0.76511 -0.15098
0.066372 -0.24582 -0.15098 0.90295
A(:,:,3,5) =
0.62859 0.041458 0.47558 0.074661
0.041458 0.99537 -0.053085 -0.0083339
0.47558 -0.053085 0.39105 -0.0956
0.074661 -0.0083339 -0.0956 0.98499
A(:,:,4,5) =
0.73864 0.20112 -0.011394 0.39048
0.20112 0.84524 0.0087678 -0.30047
-0.011394 0.0087678 0.9995 0.017023
0.39048 -0.30047 0.017023 0.41662
A(:,:,5,5) =
1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 1
A(:,:,6,5) =
0.93556 0.24481 -0.0093576 0.016177
0.24481 0.069855 0.035553 -0.061461
-0.0093576 0.035553 0.99864 0.0023492
0.016177 -0.061461 0.0023492 0.99594
A(:,:,1,6) =
0.9818 -0.10286 0.085279 0.0034902
-0.10286 0.41855 0.48208 0.01973
0.085279 0.48208 0.60031 -0.016358
0.0034902 0.01973 -0.016358 0.99933
A(:,:,2,6) =
0.99628 0.012018 0.052874 0.027665
0.012018 0.96117 -0.17085 -0.089393
0.052874 -0.17085 0.24833 -0.39329
0.027665 -0.089393 -0.39329 0.79422
A(:,:,3,6) =
0.95505 -0.16608 0.12371 0.0067153
-0.16608 0.38639 0.45705 0.02481
0.12371 0.45705 0.65956 -0.01848
0.0067153 0.02481 -0.01848 0.999
A(:,:,4,6) =
0.87322 -0.15647 0.0029936 0.29363
-0.15647 0.80689 0.0036946 0.36238
0.0029936 0.0036946 0.99993 -0.0069332
0.29363 0.36238 -0.0069332 0.31996
A(:,:,5,6) =
0.93556 0.24481 -0.0093576 0.016177
0.24481 0.069855 0.035553 -0.061461
-0.0093576 0.035553 0.99864 0.0023492
0.016177 -0.061461 0.0023492 0.99594
A(:,:,6,6) =
1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 1
Edit 2:
Added relevant code. I've been pausing the code and getting the errors inside the for loops at the end. (I believe it's also giving errors in S, but I've been focusing on D trying to figure it out.)
mtimesx is from here.
n = 4;
M = 6;
P = Normalize(2*rand(n,1,M)-1);
%differences between p_i and p_j
%sum of p_i and p_j
d = Normalize(repmat(permute(P,[1,3,2]),[1,1,M]) - repmat(P,[1,M,1]));
s = Normalize(repmat(permute(P,[1,3,2]),[1,1,M]) + repmat(P,[1,M,1]));
d(isnan(d)) = 0;
%orthogonal projection onto d(:,i,j), i.e. outer product of differences
%orthogonal projection onto s(:,i,j), i.e. outer product of sums
D = mtimesx(permute(d,[1,4,2,3]), permute(d,[4,1,2,3]));
S = mtimesx(permute(s,[1,4,2,3]), permute(s,[4,1,2,3]));
D2 = D;
S2 = S;
%projection onto the complement of d(:,i,j)
%projection onto the complement of s(:,i,j)
D = repmat(eye(n),[1,1,M,M]) - D;
S = repmat(eye(n),[1,1,M,M]) - S;
%total distance to the nearest subspace
PDist = zeros([1,M]);
PDist2 = PDist;
for j = 1:M
for k = 1:M-1
for l = k:M
if j~=k && j~=l
PDist(j) = PDist(j) + min(norm(P(:,1,j) - mtimes(D(:,:,k,l),P(:,1,j))), norm(P(:,1,j) - mtimes(S(:,:,k,l),P(:,1,j))));
PDist2(j) = PDist2(j) + min(norm(D2(:,:,k,1)*P(:,1,j)),norm(S2(:,:,k,1)*P(:,1,j)));
end
end
end
end
PDist-PDist2
Normalize.m
%Normalize
%Accepts an array (of column vectors) and normalizes the columns
function B = Normalize(A)
B = A./repmat(sqrt(sum(A.*A)),size(A,1),1);
end
The problem is that you indexed the matrices using the constant 1 instead of the variable l (lowercase L), both in the first example and in the code for computing PDist2.
In general it is good to avoid using variable names that look similar to each other and/or similar to numbers.
This can be avoided by using an editor that highlights uses different colors for variables and constants (I don't know if this is possible in MATLAB). In fact, this is how I found the error in your code. As you can see, when indexing D2 for the computation of PDist2 the number 1 is colored red.

How to run a function during a limited time?

I've a function and would like to call here each 2 seconds during 3 seconds.
I tried timer.performwithDelay() but it doesn't answer to my question.
Here is the function I want to call each 2 secondes during 3 seconds :
function FuelManage(event)
if lives > 0 and pressed==true then
lifeBar[lives].isVisible=false
lives = lives - 1
-- print( lifeBar[lives].x )
livesValue.text = string.format("%d", lives)
end
end
How can I use timer.performwithDelay(2000, callback, 1) to call my function FuelManage(event) ?
So it looks like what you are actually after is to start a few check 2 seconds from "now", for a duration of 3 seconds. You can schedule registering and unregistering for the enterFrame events. Using this will call your FuelManage function every time step during the period of interest:
function cancelCheckFuel(event)
Runtime:removeListener('enterFrame', FuelManager)
end
function FuelManage(event)
if lives > 0 and pressed==true then
lifeBar[lives].isVisible=false
lives = lives - 1
-- print( lifeBar[lives].x )
livesValue.text = string.format("%d", lives)
end
end
-- fuel management:
local startFuelCheckMS = 2000 -- start checking for fuel in 2 seconds
local fuelCheckDurationMS = 3000 -- check for 3 seconds
local stopFuelCheckMS = startFuelCheckMS + fuelCheckDurationMS
timer.performWithDelay(
startFuelCheckMS,
function() Runtime:addEventListener('enterFrame', FuelManager) end,
1)
timer.performWithDelay(
stopFuelCheckMS,
function() Runtime:removeEventListener('enterFrame', FuelManager) end,
1)
If this is too high frequency, then you'll want to use a timer, and keep track of time:
local fuelCheckDurationMS = 3000 -- check for 3 seconds
local timeBetweenChecksMS = 200 -- check every 200 ms
local totalCheckTimeMS = 0
local startedChecking = false
function FuelManage(event)
if lives > 0 and pressed==true then
lifeBar[lives].isVisible=false
lives = lives - 1
-- print( lifeBar[lives].x )
livesValue.text = string.format("%d", lives)
end
if totalCheckTimeMS < 3000 then
timer.performWithDelay(timeBetweenChecksMS, FuelManage, 1)
if startedChecking then
totalCheckTimeMS = totalCheckTimeMS + timeBetweenChecksMS
end
startedChecking = true
end
end
-- fuel management:
local startFuelCheckMS = 2000 -- start checking for fuel in 2 seconds
timer.performWithDelay(startFuelCheckMS, FuelManage, 1)
Set a timer inside a timer like this:
function FuelManage(event)
if lives > 0 and pressed==true then
lifeBar[lives].isVisible=false
lives = lives - 1
-- print( lifeBar[lives].x )
livesValue.text = string.format("%d", lives)
end
end
-- Main timer, called every 2 seconds
timer.performwithDelay(2000, function()
-- Sub-timer, called every second for 3 seconds
timer.performwithDelay(1000, FuelManage, 3)
end, 1)
Be careful though because the way it's setup know you will have an infinite number of timer running very soon... Since the first timer has a lower lifetime than the second one. So you might think if you would like to secure the second timer by making sure it's cancelled first before calling it again, this kind of thing.

How can I open a file via autohotkey at certain time of the day?

I could not find a way for a simple script at autohotkey that can open a file at a specific time of the day. Any help?
hour = 09
min = 00
difhr := hour - A_hour difmin := min - A_min
if difhr > 0
sleepH := difhr * 3600000
else
There are 2 good ways to do this that I can think of.
SetTimer:
#Persistent
SetTimer, CheckTime, 5000
Return
CheckTime:
TheTime = %A_Hour%%A_Min%
If (TheTime = 1525) && !(Ran)
{
Run, C:\Test.txt
Ran := 1
}
Loop:
Loop
{
CurrTime = %A_Hour%%A_Min%
If (CurrTime = 1520) && !(Ran)
{
msgbox, Run File
Ran := 1
}
Sleep 5000
}

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