Vector with if condition
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Hi everyone,
I got stuck while trying to create a nx1 vector from a nxm matrix.
The matrix is created by two inputs, T01(n,1) and RH(1,m). The result is Td(n,m).
When an element of each row is bigger than an other parameter (Ts2), I have to pick up the value of RH.
I created this code, but it isn't working. (I'm writing here the entire code and I know it's a little bit long)
It returns a vector with all hundrends or zeros .
Could anyone help me please ?
Thank you !
function [c] = cond_humidity1 (T01, RH, c2)
n=length(T01); %size of the martixes
m=length(RH);
rho=zeros(n,1); % Allocation of some vectors
Ps2=zeros(n,1);
Ts2=zeros(n,1);
Pv_sat=zeros(n,1);
Pv=zeros(n,m);
Pv2=zeros(n,m);
Td=zeros(n,m);
T_wall=zeros(n,1); %This is the vector who's values has to be compared with Td(i,j) values
for i=1:n
rho(i) = 28.96/(0.0821*(T01(i) +273.16));
Ps2(i) = 101300 - ((c2^2)/2)*rho(i);
Ts2(i) = T01(i)-(c2^2)/2060;
T_wall(i) = Ts2(i) + 0.8*(T01(i)-Ts2(i));
Pv_sat(i) = 6.1 *10^((7.4*T01(i))/(T01(i)+240.73));
for j=1:m
Pv(i,j) = Pv_sat(i) *(RH(j)/100);
Pv2(i,j) = (Ps2(i)/101300)*Pv(i,j);
Td(i,j) = 230.73/((7.4/log10(Pv2(i,j)/6.1))-1); %Till here everything works well
if Td(i,j)>=T_wall(i) % and I checked that there exists values
c(i)=RH(j); % Td(i,j) higher than T_wall
end
end
end
end
It returns a vector with all values 100 or 0
4 个评论
Geoff Hayes
2020-3-27
Klaudio - can you provide a set of inputs to the above function that generates the problem that you are observing?
Klaudio Myrtaj
2020-3-27
编辑：Klaudio Myrtaj
2020-3-27
darova
2020-3-27
I tried this case
n = 10;
m = 15;
T01 = -rand(1,n)*350;
RH = rand(1,m);
c2 = 1;
and get
c'
ans =
Columns 1 through 5
0 0 -202.1897 -164.6581 -230.7300
Columns 6 through 10
-230.7300 -214.6403 0 0 0
Klaudio Myrtaj
2020-3-27
编辑：Klaudio Myrtaj
2020-3-27
采纳的回答
Geoff Hayes
2020-3-27
Klaudio - so with your sample inputs and changing the last line assignment to c(i) = RH(j); then I see (like you) that all elements of c are set to 100. In this code
for i=1:n
% snip!
for j=1:m
% snip
Td(i,j) = 230.73/((7.4/log10(Pv2(i,j)/6.1))-1);
if Td(i,j)>=T_wall(i)
c(i)=RH(j);
end
end
end
note how we may update c(i) with RH(j) for different j. Is this correct? Because it seems that once we find one j that satisfies Td(i,j) >= T_wall(i), then all subsequent j satisify this too...and so we continue to update c(i) with different RH(j) until we reach the last element of RH which is 100. So do you want to do this? Or do you want jump or break out of this inner for loop upon the first update to c(i) like
if Td(i,j)>=T_wall(i)
c(i)=RH(j);
break;
end
?
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