perform calculation using for loop

1 vue (au cours des 30 derniers jours)
Adam Kevin Francis Baker
Adam Kevin Francis Baker le 17 Fév 2019
Commenté : Walter Roberson le 18 Fév 2019
So I have to use a for loop because I'm a student and the teacher said so. I've been trying to figure this one out and I'm sure theres a simple solution but because we are using matlab there aren't a whole lot of tutorials on using for loops when array operations are so much better.
I need to calculate the specific humidity using a for loop.
The equation is: q = (E*e)/P
where E = 0.622, e is a 96x144 matrix and, P represents p1 in my code which is another 96x144 matrix
I've coded something but I feel like the for loop is too simple.
Also the for loop and array operation return different values...is this normal?
% use for loop to calculate specific humidity
for q = 1:numel(p1)
q_for = (0.622.*e)./(p1(q));
end
% calculate specific humidity using array operations
q_array = (0.622.*e)./p1;
  2 commentaires
Walter Roberson
Walter Roberson le 17 Fév 2019
Your output in your for loop should probably be going into a variable indexed at something.
Adam Kevin Francis Baker
Adam Kevin Francis Baker le 17 Fév 2019
q_for is the variable and i'm not sure what you mean by indexed at something...I thought thats what p1(q) was doing.

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Adam Kevin Francis Baker
Adam Kevin Francis Baker le 18 Fév 2019
Ok I had to do a double for loop to make it work because I was dealing with a matrix so I ended up writing this code to solve my problem:
for q = 1:length(p1)
for p = 1:min(size(p1))
q_for(p,q) = (0.622.*e(p,q))./(p1(p,q));
end
end
The array operation was simple:
q_array = (e.*0.622)./p1;

Plus de réponses (1)

Adam Kevin Francis Baker
Adam Kevin Francis Baker le 18 Fév 2019
This is what I did...
for q = 1:numel(p1)
q_for(q) = (0.622.*e(q))./(p1(q));
end
and for the array operation
q_array = (e.*0.622)./p1;
  2 commentaires
Adam Kevin Francis Baker
Adam Kevin Francis Baker le 18 Fév 2019
Still not quite right...the for loop is producing a 1x13824 array instead of a 96x144 array.
Walter Roberson
Walter Roberson le 18 Fév 2019
That would have been okay under one of two circumstances:
  1. You pre-initialized q_for to be size(p1); or
  2. You reshaped vector q_for to be size(p1)

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