Plot two lines on loglog

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Robert Calderon
Robert Calderon 2020 年 9 月 12 日
コメント済み: Robert Calderon 2020 年 9 月 12 日
I am using the function tic-toc to time how long it takes for my code to run. I want to be able to plot the iteration my code runs from 10^1-10^8, by using loglog. On the x-axis I want the iterations my code is ran for, and on the y-axis i want time it took for the code to run in each iteration. I'm pretty much trying to compare the time it takes for 2 equations, to run, that give me the same answer. I tried plotting it using loglog, but I can't seem to make it work. My professor told me to define time1 as a vector like I did for rtime1, but I don't really understand. Can someone steer me in the right direction please?
function Homework2
clc
a = [1,0,3];
b = [2,-2,-1];
c = 10;
fun1 = @(LJ1) term1(LJ1); %Calls function METHOD1, it will be timed by the following code
t1 = 1; %Iteration counter
tic;
for n = c^1:c^1:c^8
rtime1(t1) = n; %counts by the times 'n' runs (ITERATIONS)
stime1(t1) = tic;
t1 = t1+1; %each time code is run, counter increases by 1 for time
end
time1 = toc
fun2 = @(LJ2) term2(LJ2); %Calls function METHOD2
t2 = 1;
tic;
for n = c^1:c^1:c^8
rtime2(t2) = n; %counts by the times 'n' runs
stime2(t2) = tic;
t2 = t2+2;
end
time2 = toc
%loglog(rtime1,stime1)
plot(rtime2,stime2)
%loglog(rtime1,stime1,rtime2,stime2)
end
%Two separate POINTER functions are made for calculations
function y = term1(LJ1) %METHOD1
r = sqrt(sum((a-b).^2)); %Vectors are subtracted, then squared, then added together, and finally rooted
LJ1 = (1/r^6)+(1/r^12)
end
function y = term2(LJ2) %METHOD2
r2 = (sum((a-b).^2));
alpha = 1/(r2);
beta = alpha^3; %[1/r2^6] = (1/r2)(1/r2)(1/r2)
gamma = beta^2; %[1/r2^12] = (1/r2^6)(1/r2^6)
LJ2 = beta+gamma
end
  1 件のコメント
Robert Calderon
Robert Calderon 2020 年 9 月 12 日
Never mind guys, I figured it out! My mistake was increasing the count by 2 for t2, it was supposed to be one:
t2 = t2+2; -----> t2 = t2+1;
I plotted the graph without a problem afterwards.

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