error of vector must be same length

Question

gnanaguru murugan 2021-6-19

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/860190-error-of-vector-must-be-same-length

评论： Image Analyst 2021-6-20

function voltot2 = recruitment(PIP,PEEP,opmin,opmax)
A = 0.0072;
B = 0.0072;
K = 0.15;
voltot2=[];
open=[];
voltot=0;
lung_level_units=9000;
 
for i=1:length(PIP)
    for pressure=PEEP:PIP
        for SP = -0.5:0.5:14.5
            for TOP = opmin:opmax
                if opmin == 0 || opmin == opmax
                    TOP = opmax;
                end
                
                if pressure > (SP+TOP)
                    volm = A-B*exp(-K*(PIP-SP));
                    vol = volm*lung_level_units/(1+opmax-opmin);
                else
                    vol = 0;
                end
                voltot = voltot+vol;
            end
        end
              voltot2(end+1)=voltot;
 
    end
end

phy_ode:

function dy = phy_ode_1(~,y,sp)
const=0.5;
top=0;
p_crit=sp+top;
dy=const*(y-p_crit);
end

call fn:

sp=-0.5:0.5:14.5;
PIP=35;
PEEP=0;
opmin=0;
opmax=0;
for i = 1:length(sp)
[t,y(:,i)]=ode45(@(t,y) phy_ode_1(t,y,sp(i)),0:0.5:10,35);
 p=y(:,i);
 vol=recruitment(PIP,PEEP,opmin,opmax);
 plot(p,vol')
end

error:

I got error of vector must be same length. could you help me?

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gnanaguru murugan 2021-6-20

here I attach the code in .mat file

Image Analyst 2021-6-20

@gnanaguru murugan, so did you post that just for completeness, because it's now working thanks to Walter's (as of yet unaccepted) Answer below? Or do you still have a problem even after trying his suggestions below?

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Answer 1

Walter Roberson 2021-6-19

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/860190-error-of-vector-must-be-same-length#answer_728295

在 MATLAB Online 中打开

PIP, PEEP, opmin ,opmax are all constants, so recruitment() invoked on those is always going to return the same thing.

p = y(:,i) on the other hand, is going to have one row for each time returned. You passed in a tspan that had more than 2 entries, so provided that the ode does not end early it should be the same length as the timespan, which is 21 entries. Let's test:

PIP=35;
PEEP=0;
opmin=0;
opmax=0;
vol=recruitment(PIP,PEEP,opmin,opmax);
size(vol)
ans = 1×2
     1    36

Nope, it returns 36 entries, which does not match the 21 you get from ode45() with that tspan.

... What is the point of calculating recruitment inside the loop when the inputs are not going to change?

function voltot2 = recruitment(PIP,PEEP,opmin,opmax)
    A = 0.0072;
    B = 0.0072;
    K = 0.15;
    voltot2=[];
    open=[];
    voltot=0;
    lung_level_units=9000;
     
    for i=1:length(PIP)
        for pressure=PEEP:PIP
            for SP = -0.5:0.5:14.5
                for TOP = opmin:opmax
                    if opmin == 0 || opmin == opmax
                        TOP = opmax;
                    end
                    
                    if pressure > (SP+TOP)
                        volm = A-B*exp(-K*(PIP-SP));
                        vol = volm*lung_level_units/(1+opmax-opmin);
                    else
                        vol = 0;
                    end
                    voltot = voltot+vol;
                end
            end
                  voltot2(end+1)=voltot;
     
        end
    end
end