Re: [Scilab-users] automatic traverse question
as I did thank you -- View this message in context: http://mailinglists.scilab.org/automatic-traverse-question-tp4036327p4036383.html Sent from the Scilab users - Mailing Lists Archives mailing list archive at Nabble.com. ___ users mailing list users@lists.scilab.org http://lists.scilab.org/mailman/listinfo/users
Re: [Scilab-users] automatic traverse question
Hello Erhy, >I studied your solution and think there will be a very huge array with >kronecker operator and generated the planes with a for loop. There is a common compromise between the used amount of memory and the algorithmic speed. There we are. >And now I have the problem with the max() function for all planes, >at which in the resulting plane each pixel should have the max. values of >the according pixels of the planes mentioned. > >max() works of a known number of planes: >resu=max(monos (:,:,1),monos (:,:,2),monos (:,:,3)) > >What is the smartest way to code, if the number of planes can differ? I am afraid that i don't clearly catch your point. If what i guess is not too far from what you mean, here is a way: MAX = monos(:,:,1); for i = 2:N MAX = max(MAX, monos(:,:,i)) end Samuel ___ users mailing list users@lists.scilab.org http://lists.scilab.org/mailman/listinfo/users
Re: [Scilab-users] linking (jumping) from line of code to an other
Jumps are generally bad style; I don't know if Scilab even supports
them. If it does I advise you not to use them for something this
trivial (error handling is the only place I've seen them seriously
recommended, and even there they can be very problematical -- it's why
people invented exceptions).
For something that small the cost in making your code into spaghetti is
larger than the cost of just doing it.
The two ways that software professionals would deal with this are to
either get rid of 100 as a "magic number" and assign its value to a
variable (i.e., before the "for" statement assign PowerIncrement =
100), or rearrange the conditional so that the P = P + PowerIncrement
line is only encountered once.
Or do both.
Certainly if you use the existing structure and if the two lines that
increment P are supposed to do the same thing always, then using a
named variable is a Good Thing -- that means that at some later date,
when some poor ignorant fool reads your code they don't have to wonder
if the "P+100" in one spot means the same thing as the "P+100" in
another, AND they don't run the risk of changing the (hopefully proper)
increment in one place but not another.
And note, based on my experience writing software for the last 40
years, that often the poor ignorant fool who has to maintain your
software is you, months or years down the road -- so it pays to make
your code readable. Nothing makes you feel dumber than not being able
to understand your own code, because no matter which way the blame
flies, it always lands squarely on you.
On Fri, 2017-05-12 at 11:15 +0200, Frieder Nikolaisen wrote:
> Hello,
> another question to solve the locomotive stuff, giving you a example
> with the not implented code.
> I don't want to code the same stuff twice. Thats why I want to jump
> between lines of code. Is this possible and how?
> P = 200;
> DM = 1;
>
> for n = 1:10
> if DM == 1 then
> if P > 100 then DM = 1
> P = P - 60
> disp('P bigger 100. DM = ' + string(DM))
> elseDM = 0
> P = P + 100 //instead of writing P = P +
> 100, I would like ...
> disp('P smaller 100. DM = ' + string(DM))
> end
> else // D == 0
> if P > 100 then DM = 1
> disp('P bigger 100. DM = ' +
> string(DM))
> P = P - 30
> elseDM = 0 // ... to continue here
> disp('P smaller 100. DM = ' +
> string(DM))
> P = P + 100 // to use this P = P + 100
> end
> end
> disp(string(n) + ' ' + string(P))
> end
>
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www.wescottdesign.com
Control & Communications systems, circuit & software design.
Phone: 503.631.7815
Cell: 503.349.8432
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Re: [Scilab-users] linking (jumping) from line of code to an other
The Solution is: Using whil:
i = 1
a = 1
while i < 5
disp(i)
a
= a + 1
i = i +1
if a == 3 then
i = i - 1
end
end
Am 2017-05-12
11:15, schrieb Frieder Nikolaisen:
> Hello,
>
> another question to
solve the locomotive stuff, giving you a example with the not implented
code.
>
> I don't want to code the same stuff twice. Thats why I want
to jump between lines of code. Is this possible and how?
>
> P =
200;
> DM = 1;
>
> for n = 1:10
> if DM == 1 then
> if P > 100 then DM
= 1
> P = P - 60
> disp('P bigger 100. DM = ' + string(DM))
> else DM =
0
> P = P + 100 //instead of writing P = P + 100, I would like ...
>
disp('P smaller 100. DM = ' + string(DM))
> end
> else // D == 0
> if
P > 100 then DM = 1
> disp('P bigger 100. DM = ' + string(DM))
> P = P
- 30
> else DM = 0 // ... to continue here
> disp('P smaller 100. DM =
' + string(DM))
> P = P + 100 // to use this P = P + 100
> end
> end
>
disp(string(n) + ' ' + string(P))
> end
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Re: [Scilab-users] automatic traverse question
I studied your solution and think there will be a very huge array with kronecker operator and generated the planes with a for loop. And now I have the problem with the max() function for all planes, at which in the resulting plane each pixel should have the max. values of the according pixels of the planes mentioned. max() works of a known number of planes: resu=max(monos (:,:,1),monos (:,:,2),monos (:,:,3)) What is the smartest way to code, if the number of planes can differ? -- View this message in context: http://mailinglists.scilab.org/automatic-traverse-question-tp4036327p4036377.html Sent from the Scilab users - Mailing Lists Archives mailing list archive at Nabble.com. ___ users mailing list users@lists.scilab.org http://lists.scilab.org/mailman/listinfo/users
[Scilab-users] linking (jumping) from line of code to an other
Hello,
another question to solve the locomotive stuff, giving you
a example with the not implented code.
I don't want to code the same
stuff twice. Thats why I want to jump between lines of code. Is this
possible and how?
P = 200;
DM = 1;
for n = 1:10
if DM == 1 then
if
P > 100 then DM = 1
P = P - 60
disp('P bigger 100. DM = ' +
string(DM))
else DM = 0
P = P + 100 //instead of writing P = P + 100,
I would like ...
disp('P smaller 100. DM = ' + string(DM))
end
else
// D == 0
if P > 100 then DM = 1
disp('P bigger 100. DM = ' +
string(DM))
P = P - 30
else DM = 0 // ... to continue here
disp('P
smaller 100. DM = ' + string(DM))
P = P + 100 // to use this P = P +
100
end
end
disp(string(n) + ' ' + string(P))
end
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Re: [Scilab-users] Avoiding a loop
Hello Tim,
Yes. batt is the is the State of Charge of the battery.
P(n,2) (kN) is the power taken rom battery, engine or both. I will
calculate the battery in As, so P(n,2) is only an in alternate value -
but thats fare to complicated for the example code.
"Proving that it's
correct will be a pain." - prooving the for-loop and getting the same
result with your way - that might be a prooving.
As I have to code the
loop anyway, I will consider your if the loop is very slow.
Best
regards
Frieder
Am 2017-05-11 18:48, schrieb Tim Wescott:
>
Depending on how often you switch between battery and generator, and
>
how icky-picky you're willing to be, there may be a way to reduce
>
computation.
>
> It looks like the term P(n,2) * (P(n+1,1) - P(n,1)) is
always there,
> and you're either adding it to 'gen' (is it energy
production?) or
> subtracting it from 'batt'.
>
> If you really want to
go there, you can vectorize "if" statements by
> using boolean
expressions on vectors and the "find" function, which
> returns indexes
of true results. Then you can use "cumsum" on your
> P(n,2) * (P(n+1,1)
- P(n,1)) term to find where the battery state of
> charge (I assume
that's what 'batt' is) hits 800.
>
> It'll be complicated. It'll be
prone to error. Proving that it's
> correct will be a pain. But when you
get it working, it'll be
> considerably faster.
>
> On Thu, 2017-05-11
at 09:17 +0200, Frieder Nikolaisen wrote:
>
>> Thanks for all the
answers. I feared that there is no way around a loop. During the process
batt (Battery) is charged and discharged. In my example, it is only
discharged. I will code the entire problem with a loop, maybe somebody
knows something to speed up the process with the full problem. (Tim: I
am not a programming pro, a C-function might not be a solution. ) Why do
I try avoidng a loop? I do have txt-document with 50 000 to 100 000
lines about a (hybrid-)locomotive shunting process. I do need to
optimize the energy managment. Because I am not mathemtic student, I
have to solve the problem empirical (try and error). The programm has to
run a few hundred times. With a matrix thats no problem, but with
matrixes only, I can only calculate the diesel usage without any battery
energy storage. Thanks for the checking my code anyway. Am 10.05.2017 um
20:53 schrieb Amanda Osvaldo:
>>
>>> What it's the equation you need
to compute ? Perhaps I can help. I think it's possible to compute with
something in this way: map = find (P(:,2) > 100 ); if batt > 800 then
batt = batt - P(map,2) * (P(map+1,1) - P(map,1)); end On Wed, 2017-05-10
at 17:23 +0200, Frieder Nikolaisen wrote:
>>>
Hello, I did write
an example code, but I do not like the time consuming way I solved the
problem. With 50 000 lines in the matrix, it wouldn't be fun. How can I
avoid using the for-loop? 10, 80; 11, 200 15, 0]; batt = 1000; gen = 0;
n = 1 for n=1:5 if P(n,2) > 100 then if batt > 800 then batt = batt -
P(n,2) * (P(n+1,1) - P(n,1)) else gen = gen + P(n,2) * (P(n+1,1) -
P(n,1)) end else batt = batt - P(n,2) * (P(n+1,1) - P(n,1)) end disp('n
' + string(n)) disp('batt ' + string(batt)) disp('gen ' + string(gen))
end Thanks alot! Best regards Frieder
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