Obtaining slope from fitlm results

With results from fitlm, how does one disentangle the linear regression's slope? I'm simply trying to plot the regression over the data. (I expect this'll be an easy answer; thank you in advance.)

Answer 1

[dpb]

Oh, it's a tangled web TMW has weaved, fer shure...what with all the different mechanisms to do the same thing in different toolboxen plus the base product, and nary the doubled twain shall meet...

Anyhoo, fitlm in Statistics Toolbox returns an object of the Linear Model class. There's a page documenting properties and methods for it you can get to by clicking on the Output Arguments section link mdl or the 'See Also' section for more details.

In short, the coefficients are obtainable by referencing the field Estimate in the table returned by Coefficients property. So, given

lm=fitlm(x,y,'poly1');             % a linear model
betahat=lm.Coefficients.Estimate;  % the coefficients

NB: These are case-sensitive names, btw...and must be spelled-out in their entirety, none of this "first n characters stuff" here (not that that's any different than for other dot addressing, just noting it).

BTW, the order of these are intercept first which is in direct conflict with the venerable polyfit/polyval pair.

To just evaluate the fit to draw the regression line, use predict method that will get the coefficients for you given just the model object and x vector.

[balsip]

dpb,

I appreciate a well-written response as much as the next guy, and you got me closer, but I'm still at a loss here.

I can drill into the fitlm results to get to the coefficient estimates. I see the y-intercept estimate in cell (1,1), so I've got myself one point right there (0,Y-int), but where is my second point?

The predictor variable's coefficient estimate (2,1) is—I assume—the x half of my second point, but then what is y? As I see it, I've got 3/4 of the numbers I need to get a slope.

On the other hand, is the coefficient in (2,1) an x-intercept?

Thanks for the time.

[dpb]

Well, I'm at a loss now, sorry... :)

I will admit I tend to the poetic and hyperbole isn't unknown in my reaction to much of what TMW has done with the "veritable plethora" of redundancies in toolboxes...sorry if that was the confusing factor, I thought I made the actual usage straightforward after the editorial content.

Show the code you used to obtain the linear model...the above for a linear regression (which I take to mean y=mx+b) is as given above from the LM object. I don't know where the cell references you're using come from, sorry. The LM return output is an object, not a cell array...

Carrying on with the above example, showing output--

>> lm.Coefficients
ans = 
                 Estimate       SE        tStat      pValue  
                 ________    ________    _______    _________
  (Intercept)    -0.13323     0.18049    -0.7382      0.50136
  x1              0.24822    0.046344     5.3561    0.0058618
> betahat=lm.Coefficients.Estimate  % save estimated coeff's
betahat =
 -0.1332
  0.2482
>>

where the first shows the table containing the coefficients and the various statistics besides, the second returns only the two estimated coefficients; as the table notes, the intercept is the first element of the array, the slope is the second.

For comparison to the base product polyfit, for the same data

>> polyfit(x,y,1)
ans =
  0.2482   -0.1332
>>

which are numerically the same but in opposite order as polyval expects the intercept term to be last.

If you want to set a regression line onto a plot, then

yhat=predict(lm,[x0; x1]);

For these values, the range of x was 1:6, those bounds would be good choice for the fit line--

plot(x,y,'*')
yhat=predict(lm,[1;6]);
hold on
plot([x(1);x(end)],yhat,'-')
eqn=sprintf('y=%0.2f*X + %0.2f',betahat);
text(3,1.2,eqn)

yielded

[dpb]

As noted earlier, peruse the <linearmodel class> doc for all the skinny on the properties and methods to retrieve data from the returned object.

[balsip]

Ahh! Now I'm cooking... that's all I needed to know: that the second Estimate coefficient is the slope.

[dpb]

Hmmm...I thought I said that a couple of times... :)

"BTW, the order of these are intercept first which is in direct conflict with the venerable polyfit/polyval pair."

Answer 2

[Image Analyst]

Like dpb, I was also wondering why you don't simply use polyfit() if you just want to fit a simply line. The first coefficient is the slope:

coefficients = polyfit(x, y, 1);
slope = coefficients(1);

If your linear model not a simple y=mx+b line? If not, what is your model and supply us your data.

[balsip]

That certainly works as well. I'm also using some of the other stats coming out of fitlm, so I'll stick with that output.

[Dan Hallatt]

I'm in the same boat. As far as I know, fit and polyfit both output confidence intervals of their coefficients but an advantage of fitlm is that there is direct output of SE's of the coefficients instead. Much easier to use them for things like pooled variance etc.. rather than trying to backtrack and solve for the SE or SD from CI's.

[Catherine Mauck]

My understanding is also that polyfit does not allow you to specify an intercept of 0 (or at least from my brief Googling trying to solve that, that led me to fitlm).

[dpb]

"My understanding is also that polyfit does not allow you to specify an intercept of 0..."

True; polyfit/polyval are a very simplistic toolset that was introduced in the very earliest years of MATLAB. It's useful for the simple case if all one cares about is the plain-vanilla results. Anything more than that is more easily obtained or can only be obtained by one or more of the later tools/functions or by reverting to base definitions and backslash for solution and then computing any desired statistics from first principles.