Finding slope for the polyfit line

70 vues (au cours des 30 derniers jours)
Michelangelo Cannistraro
Michelangelo Cannistraro le 2 Déc 2022
Commenté : Torsten le 2 Déc 2022
load data_matrix.mat
data=SECTION_L;
data(:,1:3)=[];
row=9;
cleanup=data([1 2 3 row],:);
% year range
y1 = 1975;
y2 = 2016;
% init
x = y1:y2;
nx = numel(x);
mo_av = zeros(nx,12);
mo_max = zeros(nx,12);
mo_min = zeros(nx,12);
max_day = zeros(nx,12);
min_day = zeros(nx,12);
k = 0;
for k = 1:nx % year loop
col_dat_yr=find(cleanup(1,:)== x(k)); % find data corresponding to year = x(k)
data_yr=cleanup(:,col_dat_yr);
for mmonth = 1:12
col_dat_mo=find(data_yr(2,:)==mmonth);
data_mo=data_yr(:,col_dat_mo);
mo_av(k,mmonth)=mean(data_mo(4,:));
mo_max(k,mmonth)=max(data_mo(4,:));
mo_min(k,mmonth)=min(data_mo(4,:));
max_day(k,mmonth)=find(data_mo(4,:)==mo_max(k,mmonth));
min_day(k,mmonth)=find(data_mo(4,:)==mo_min(k,mmonth));
end
end
% plot montly data
str_mo = {'Jan' 'Feb' 'Mar' 'Apr' 'May' 'Jun' 'Jul' 'Aug' 'Sep' 'Oct' 'Nov' 'Dec'};
figure
sgtitle('Monthly Data : Avg')
for ck = 1:12
coefs1=polyfit(x,mo_av(:,ck),1);
curve1=polyval(coefs1,x);
subplot(3,4,ck)
plot(x,mo_av(:,ck),'-',x,curve1);
xlim([1970,2020])
ylim([-25,5])
xlabel('Years')
ylabel('Temperature C')
title(str_mo(ck));
end
figure
sgtitle('Monthly Data : Max')
for ck = 1:12
coefs2=polyfit(x,mo_max(:,ck),1);
curve2=polyval(coefs2,x);
subplot(3,4,ck)
plot(x,mo_max(:,ck),'-',x,curve2);
xlim([1970,2020])
ylim([0,25])
xlabel('Years')
ylabel('Temperature C')
title(str_mo(ck));
end
figure
sgtitle('Monthly Data : Min')
for ck = 1:12
coefs3=polyfit(x,mo_min(:,ck),1);
curve3=polyval(coefs3,x);
subplot(3,4,ck)
plot(x,mo_min(:,ck),'-',x,curve3);
xlim([1970,2020])
ylim([-30,10])
xlabel('Years')
ylabel('Temperature C')
title(str_mo(ck));
end
I have made the plots with 12 graphs for three figures and have put a line through the 36 graphs. How would you find the slope of the polyfit and polyval lines for every graph.

Réponse acceptée

Voss
Voss le 2 Déc 2022
polyfit returns the coefficients in a vector, with the highest-degree coefficient first. Therefore, the slopes are the values of coefs1(1) (similarly, coefs2(1) and coefs3(1)) in each iteration.
  2 commentaires
Michelangelo Cannistraro
Michelangelo Cannistraro le 2 Déc 2022
but i need to get 36 slopes and my variable curve1 etc only give one slope
Torsten
Torsten le 2 Déc 2022
Replace
coefs1=polyfit(x,mo_av(:,ck),1);
by
coefs1=polyfit(x,mo_av(:,ck),1);
slope1(ck) = coefs1(1);
(same for the remaining two loops).
Then you have 36 slopes (slope1(1:12),slope2(1:12) and slope3(1:12)).

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