Error finding Linear Regression with polyfit and \

I am trying to plot the linear regression of my data throughout time however I have tried a few different methods but none of them having been working for me.

load data.mat;
y=data.MEAN_TEMPERATURE;
x=[1:numel(y)].';
p=polyfit(x,y,1)

b1=x\y
plot(x,y);

When I try using the plotfit or the \ method I get NaN as outputs. However when I plot it in a graph and then use the basic fitting tool I am able to get:

p1 = 0.00011312

p2 = 5.0801

for

y = p1*x + p2

So I am not sure what I am doing wrong here because clearly a linear regression can be found.

Answer 1

[Star Strider]

There is a NaN value in the temperature data.

Try this:

LD = load('data.mat');
data = LD.data;

x = data.LOCAL_DATE;
xdn = datenum(x);                                       % Copnvert To 'datenum' For Regression
y = data.MEAN_TEMPERATURE;
yv = ~ismissing(y);                                     % Find Missing Values

[p,S,mu] = polyfit(xdn(yv), y(yv), 1);                  % Use Scaling & Centring
yfit = polyval(p,xdn(yv),S,mu);                         % Use Scaling & Centring

figure
plot(xdn(yv), y(yv))
hold on
plot(xdn(yv), yfit, '-r')
hold off
grid
datetick('x', 'yyyy', 'keepticks','keeplimits')
xlabel('Date')
ylabel('Mean Temperature (°C)')

A linear regression is not going to tell much, other than that the mean termperatur is increasing slightly although not significantly over time.

This:

mdl = fitlm(xdn, y)

produces:

mdl = 


Linear regression model:
    y ~ 1 + x1

Estimated Coefficients:
                    Estimate         SE         tStat      pValue 
                   __________    __________    ________    _______

    (Intercept)       -76.185         124.8    -0.61044    0.54161
    x1             0.00011312    0.00017332     0.65267    0.51402


Number of observations: 3348, Error degrees of freedom: 3346
Root Mean Squared Error: 9.7
R-squared: 0.000127,  Adjusted R-Squared: -0.000172
F-statistic vs. constant model: 0.426, p-value = 0.514

The parameters that fitlm produces are not the same as those polyfit produces, since here polyfit uses centring and scaling (my choice, not absolutely necessary).

.

[JMG]

Okay great thank you for your help!

[Star Strider]

As always, my pleasure!

.

Answer 2

[Rik]

Removing the NaNs will do the trick. Although I'm not sure this data should have a linear fit.

fn=websave('data.mat','https://www.mathworks.com/matlabcentral/answers/uploaded_files/703197/data.mat');
S=load(fn);
data=S.data;

y=data.MEAN_TEMPERATURE;
y(isnan(y))=[];
x=[1:numel(y)].';
p=polyfit(x,y,1)

p = 1×2

0.0001 5.0795

b1=x\y

b1 = 0.0024

plot(x,y)

[JMG]

Thank you this also worked!