Accumulating data in a loop while

Question

gravy 2024-1-29

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2075881-accumulating-data-in-a-loop-while

评论： gravy 2024-1-29

Good evening, I need to collect data into an array with accumulation.

For example, there are 9 time values t=9, the array should look like this (0 t1 t1+t2 t1+t2+t3 and...) I want to do this in a loop so that it would be possible to regulate the number of t values.

So far I've thought of this, but it doesn't work correctly:

index=1;
n=10;
c=[];
while index<10
    t = b(index)+b(index+1)
    if t>b(index) && index<10
        t1=0;
        t1 = t + b(index)
        index=index+1;
    end
    c=[c t1];
end
Unrecognized function or variable 'b'.
c

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gravy 2024-1-29

"b" this is part of the previous code

gravy 2024-1-29

编辑：Walter Roberson 2024-1-29

在 MATLAB Online 中打开

This is what it looks like in simple form

%t0=0;

%t1=delt0;

%t2=delt0+delt1;

%t3=delt0+delt1+delt2;

%t4=delt0+delt1+delt2+delt3;

%t5=delt0+delt1+delt2+delt3+delt4;

%t6=delt0+delt1+delt2+delt3+delt4+delt5;

%t7=delt0+delt1+delt2+delt3+delt4+delt5+delt6;

%t8=delt0+delt1+delt2+delt3+delt4+delt5+delt6+delt7;

%t9=delt0+delt1+delt2+delt3+delt4+delt5+delt6+delt7+delt8;

%t10=delt0+delt1+delt2+delt3+delt4+delt5+delt6+delt7+delt8+delt9;

%t = [t0 t1 t2 t3 t4 t5 t6 t7 t8 t9 t10];

I want to automate this process and not write new variables myself every time.

Here is the full code, maybe it will become clearer:

X = linspace(0,100,12);
for x = X    
    if (0<=x) && (x<5.5)
        y1=((x-0)/(5.5-0))*(7.74-9.19)+9.19;
        
    end
    
    if (5.5<=x) && (x<14.5)
        y2=((x-5.5)/(14.5-5.5))*(6.45-7.74)+7.74;
        
    end
    if (14.5<=x) && (x<22.5)
        y3=((x-14.5)/(22.5-14.5))*(5.56-6.45)+6.45;
        
    end
    if (22.5<=x) && (x<30)
        y4=((x-22.5)/(30-22.5))*(4.79-5.56)+5.56;
        
    end
    if (30<=x) && (x<43.5)
        y5=((x-30)/(43.5-30))*(3.43-4.79)+4.79;
        
    end
    if (43.5<=x) && (x<49)
        y6=((x-43.5)/(49-43.5)).*(3.04-3.43)+3.43;
        
    end
    if (49<=x) && (x<56)
        y7=((x-49)/(56-49))*(2.39-3.04)+3.04;
        
    end
    if (56<=x) && (x<64)
        y8=((x-56)/(60-56)).*(0.8-2.39)+2.39;
        
    end
    if (64<=x) && (x<73)
        y9=((x-60)/(70-60))*(-1.98-0.8)+0.8;
        
    end
    if (73<=x) && (x<83)
        y10=((x-70)/(80-70))*(-3.88+1.98)-1.98;
        
    end
    if (83<=x) && (x<93)
        y11=((x-80)/(90-80))*(-5.45+3.88)-3.88;
        
    end
    if (93<=x) && (x<=100)
        y12=((x-90)/(100-90))*(-6.74+5.45)-5.45;
        
    end
end
Y=[y1 y2 y3 y4 y5 y6 y7 y8 y9 y10 y11 y12];
X = linspace(0,100,12);
v = [ 0:10:100 ];
delv=10;
dzeta = 120;
w0=0;
v1=0;
a=[];
while v1<100 %&& w0<3
    fycp = abs(((interp1(X,Y,v1)-w0)+((interp1(X,Y,v1+delv))-w0)))/2;
    v1=v1+10;
    %  w0=w0+1;
    a=[a fycp];
    
end
b=[];
index=1;
while index<11
    delt = delv/(dzeta*a(index));
    index=index+1;
    b=[b delt];
end
index=1;
n=10;
c=[];
while index<10
    t = b(index)+b(index+1)
    if t>b(index) && index<10
        t1=0;
        t1 = t + b(index)
        index=index+1;
    end
    c=[c t1];
end

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Answer 1

Voss 2024-1-29

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2075881-accumulating-data-in-a-loop-while#answer_1399241

在 MATLAB Online 中打开

t = rand(1,9) % 9 random t values
t = 1×9
    0.7787    0.5803    0.4362    0.3903    0.0329    0.8313    0.1897    0.7956    0.7963
c = cumsum([0 t]) % c = [0, t(1), t(1)+t(2), t(1)+t(2)+t(3), etc.]
c = 1×10
         0    0.7787    1.3590    1.7951    2.1854    2.2183    3.0496    3.2394    4.0349    4.8312

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gravy 2024-1-29

编辑：gravy 2024-1-29

still not the same, it should look like you wrote before:

t = rand(1.9) % 9 random t values

c = cumsum([0 t]) % c = [0, t(1), t(1)+t(2), t(1)+t(2)+t(3), etc.]

But the summation in the array should be done automatically, I want to change the value of t, but not rewrite the array every time. There will be a graph with "b" and "c" and their number in the array should converge.

I tried to set some kind of condition, but I couldn’t figure it out yet.

gravy 2024-1-29