这个拟合的曲线,为什​么第一条好第二条在图​例中是一样.。

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sepepo
sepepo am 25 Nov. 2022
Beantwortet: rimomov am 25 Nov. 2022
这个拟合的曲线,为什么第一条好第二条在图例中是一样的呢?
d=[500 1000 1500 2000 2500];
t1=[0.299 0.566 0.841 1.117 1.393];
d.^2;
t1.^2;
scatter(d.^2,t1.^2)
hold on
p1=polyfit(d.^2,t1.^2,1);
f1=polyval(p1,d.^2);
plot(d.^2,t1.^2,'r-*')
t2=[0.364 0.517 0.701 0.897 1.099];
t2.^2;
scatter(d.^2,t2.^2)
p2=polyfit(d.^2,t2.^2,1);
f2=polyval(p2,d.^2);
plot(d.^2,f2,'b-s')
t3=[0.592 0.638 0.708 0.799 0.896];
t3.^2;
scatter(d.^2,t3.^2,[0.25 0.25 0.25])
p3=polyfit(d.^2,t3.^2,1);
f3=polyval(p3,d.^2);
plot(d.^2,t3.^2,'g-d')
legend('第一条','第二条','第三条',2)

Akzeptierte Antwort

rimomov
rimomov am 25 Nov. 2022
d=[500 1000 1500 2000 2500];
t1=[0.299 0.566 0.841 1.117 1.393];
d.^2;
t1.^2;
scatter(d.^2,t1.^2,'ro')
hold on
p1=polyfit(d.^2,t1.^2,1);
f1=polyval(p1,d.^2);
plot(d.^2,t1.^2,'r-*')
t2=[0.364 0.517 0.701 0.897 1.099];
t2.^2;
scatter(d.^2,t2.^2,'+')
p2=polyfit(d.^2,t2.^2,1);
f2=polyval(p2,d.^2);
plot(d.^2,f2,'b-s')
t3=[0.592 0.638 0.708 0.799 0.896];
t3.^2;
scatter(d.^2,t3.^2,'gv')
p3=polyfit(d.^2,t3.^2,1);
f3=polyval(p3,d.^2);
plot(d.^2,t3.^2,'g-^')
legend('第一条 (data)','第一条 (fit)','第二条 (data)','第二条 (fit)','第三条 (data)','第三条 (fit)',2)

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