Accumulating data in a loop while

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gravy
gravy 2024-1-29
评论： gravy 2024-1-29
Ran in:
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
gravy 2024-1-29
"b" this is part of the previous code
gravy
gravy 2024-1-29
编辑：Walter Roberson 2024-1-29
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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采纳的回答

Voss
Voss 2024-1-29
Ran in:
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
  4 个评论
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gravy
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
gravy 2024-1-29
I'm an idiot, your previous answer was correct. Thanks a lot

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