Azzera filtri
Azzera filtri

Sum of Maple to MATLAB

3 visualizzazioni (ultimi 30 giorni)
Bibha
Bibha il 14 Mag 2011
Hi All,
There is a saying "A pee to an ant is flooding".... I guess my question appears simple but being new to MATLAB, couldn't find my way. Seeking some help...
Lm = [1 2 3]';
Um = [4 5 6]';
Maple:
sum(Lm[p,1] * sum(Lm[q,1] * (Um[p,1] - Um[q,1]),q=1..3), p=1..3)
What will be the equivalent MATLAB for the above?
Thank you.
  1 Commento
Matt Fig
Matt Fig il 14 Mag 2011
I would have to know what that Maple command did first! Since this is a MATLAB site, it is possible that nobody on here knows what that Maple command does...

Accedi per commentare.

Risposta accettata

Walter Roberson
Walter Roberson il 14 Mag 2011
The most direct equivalent would be
Lm(1,1) * (Lm(2,1) * (Um(1,1) - Um(2,1)) + Lm(3,1) * (Um(1,1) - Um(3,1))) + Lm(2,1) * (Lm(1,1) * (Um(2,1) - Um(1,1)) + Lm(3,1) * (Um(2,1) - Um(3,1))) + Lm(3,1) * (Lm(1,1) * (Um(3,1) - Um(1,1)) + Lm(2,1) * (Um(3,1) - Um(2,1)))
Which works out to be 0
If we generalize to
sum(Lm[p,1] * sum(Lm[q,1] * (Um[p,1] - Um[q,1]),q=1..N), p=1..N) %MAPLE
where N is the length of Um and Lm, then:
breaking it down,
T(p) := sum(Lm[q,1] * (Um[p,1] - Um[q,1]),q=1..N) %Maple
would be
T(p) = sum(Lm .* (Um(p)-Um)) %MATLAB
and when Lm and Um are column vectors, that could be written
T(p) = Lm.' * (Um(p)-Um) %MATLAB
Then,
sum(Lm[p,1] * T[p,1],p=1..N) %Maple
would be
T = zeros(length(Lm),1); %MATLAB
for p=1:T
T(p) = Lm.' * (Um(p)-Um)
end
sum(Lm .* T)
but the sum(Lm .* T) could also be written as Lm.' * T leading to
T = zeros(length(Lm),1); %MATLAB
for p=1:T
T(p) = Lm.' * (Um(p)-Um)
end
Lm.' * T
In turn, in MATLAB that could be written as
Lm.' * arrayfun(@(p) Lm.' * (Um(p)-Um),1:length(Lm)).'
Note here that arrayfun is returning a row vector instead of a column vector, so the .' after the arrayfun is needed to switch back to a column vector for the array multiplication.

Più risposte (3)

Andrei Bobrov
Andrei Bobrov il 14 Mag 2011
more variant
Lm.'*bsxfun(@minus,Um,Um.')*Lm
  2 Commenti
Walter Roberson
Walter Roberson il 14 Mag 2011
Nice compaction!
Bibha
Bibha il 14 Mag 2011
Thank you again!
Just phenomenal.

Accedi per commentare.


Matt Fig
Matt Fig il 14 Mag 2011
It would help, for those of us who do not have Maple and consequently do not know what the result of that Maple command is, if you would show the expected output. Even if you have to make a simple example and create the output by hand, show inputs and expected outputs.
The best I can guess would be:
S = 0; % This is the result you want at the end.
for p = 1:3
tmp = 0;
for q = 1:3
tmp = tmp + Lm(q) * (Um(p)-Um(q));
end
S = S + tmp*Lm(p);
end
S % Look at the final result...

Bibha
Bibha il 14 Mag 2011
Thanks for all the answers!
I wish I could select all the answers but I am selecting the one where an educated person is trying to educate other rather than giving the answer right away.
Well, no doubt, I learned more functions from all of you guys but I am giving credit to the hard worker.
Quite amazed by the one line of code that did the trick! Thank you andrei.

Categorie

Scopri di più su MATLAB in Help Center e File Exchange

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by