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Choose a Model or Range of Models Examine Quality and Adjust the Fitted Model |
To begin fitting a regression, put your data into a form that fitting functions expect. All regression techniques begin with input data in an array X and response data in a separate vector y, or input data in a table or dataset array tbl and response data as a column in tbl. Each row of the input data represents one observation. Each column represents one predictor (variable).
For a table or dataset array tbl, indicate the response variable with the 'ResponseVar' name-value pair:
mdl = fitlm(tbl,'ResponseVar','BloodPressure'); % or mdl = fitglm(tbl,'ResponseVar','BloodPressure');
The response variable is the last column by default.
You can use numeric categorical predictors. A categorical predictor is one that takes values from a fixed set of possibilities.
For a numeric array X, indicate the categorical predictors using the 'Categorical' name-value pair. For example, to indicate that predictors 2 and 3 out of six are categorical:
mdl = fitlm(X,y,'Categorical',[2,3]); % or mdl = fitglm(X,y,'Categorical',[2,3]); % or equivalently mdl = fitlm(X,y,'Categorical',logical([0 1 1 0 0 0]));
For a table or dataset array tbl, fitting functions assume that these data types are categorical:
Logical
Categorical (nominal or ordinal)
String or character array
If you want to indicate that a numeric predictor is categorical, use the 'Categorical' name-value pair.
Represent missing numeric data as NaN. To represent missing data for other data types, see Missing Group Values.
To create a dataset array from an Excel^{®} spreadsheet:
ds = dataset('XLSFile','hospital.xls',... 'ReadObsNames',true);
To create a dataset array from workspace variables:
load carsmall
ds = dataset(MPG,Weight);
ds.Year = ordinal(Model_Year);
To create a table from an Excel spreadsheet:
tbl = readtable('hospital.xls',... 'ReadRowNames',true);
To create a table from workspace variables:
load carsmall
tbl = table(MPG,Weight);
tbl.Year = ordinal(Model_Year);
For example, to create numeric arrays from workspace variables:
load carsmall
X = [Weight Horsepower Cylinders Model_Year];
y = MPG;
To create numeric arrays from an Excel spreadsheet:
[X Xnames] = xlsread('hospital.xls'); y = X(:,4); % response y is systolic pressure X(:,4) = []; % remove y from the X matrix
Notice that the nonnumeric entries, such as sex, do not appear in X.
There are three ways to fit a model to data:
Use fitlm to construct a least-squares fit of a model to the data. This method is best when you are reasonably certain of the model's form, and mainly need to find its parameters. This method is also useful when you want to explore a few models. The method requires you to examine the data manually to discard outliers, though there are techniques to help (see Residuals — Model Quality for Training Data).
Use fitlm with the RobustOpts name-value pair to create a model that is little affected by outliers. Robust fitting saves you the trouble of manually discarding outliers. However, step does not work with robust fitting. This means that when you use robust fitting, you cannot search stepwise for a good model.
Use stepwiselm to find a model, and fit parameters to the model. stepwiselm starts from one model, such as a constant, and adds or subtracts terms one at a time, choosing an optimal term each time in a greedy fashion, until it cannot improve further. Use stepwise fitting to find a good model, which is one that has only relevant terms.
The result depends on the starting model. Usually, starting with a constant model leads to a small model. Starting with more terms can lead to a more complex model, but one that has lower mean squared error. See Compare large and small stepwise models.
You cannot use robust options along with stepwise fitting. So after a stepwise fit, examine your model for outliers (see Residuals — Model Quality for Training Data).
There are several ways of specifying a model for linear regression. Use whichever you find most convenient.
For fitlm, the model specification you give is the model that is fit. If you do not give a model specification, the default is 'linear'.
For stepwiselm, the model specification you give is the starting model, which the stepwise procedure tries to improve. If you do not give a model specification, the default starting model is 'constant', and the default upper bounding model is 'interactions'. Change the upper bounding model using the Upper name-value pair.
Note: There are other ways of selecting models, such as using lasso, lassoglm, sequentialfs, or plsregress. |
String | Model Type |
---|---|
'constant' | Model contains only a constant (intercept) term. |
'linear' | Model contains an intercept and linear terms for each predictor. |
'interactions' | Model contains an intercept, linear terms, and all products of pairs of distinct predictors (no squared terms). |
'purequadratic' | Model contains an intercept, linear terms, and squared terms. |
'quadratic' | Model contains an intercept, linear terms, interactions, and squared terms. |
'polyijk' | Model is a polynomial with all terms up to degree i in the first predictor, degree j in the second predictor, etc. Use numerals 0 through 9. For example, 'poly2111' has a constant plus all linear and product terms, and also contains terms with predictor 1 squared. |
For example, to specify an interaction model using fitlm with matrix predictors:
mdl = fitlm(X,y,'interactions');
To specify a model using stepwiselm and a table or dataset array tbl of predictors, suppose you want to start from a constant and have a linear model upper bound. Assume the response variable in tbl is in the third column.
mdl2 = stepwiselm(tbl,'constant',... 'Upper','linear','ResponseVar',3);
A terms matrix is a t-by-(p + 1) matrix specifying terms in a model, where t is the number of terms, p is the number of predictor variables, and plus one is for the response variable.
The value of T(i,j) is the exponent of variable j in term i. Suppose there are three predictor variables A, B, and C:
[0 0 0 0] % Constant term or intercept [0 1 0 0] % B; equivalently, A^0 * B^1 * C^0 [1 0 1 0] % A*C [2 0 0 0] % A^2 [0 1 2 0] % B*(C^2)
The 0 at the end of each term represents the response variable. In general,
If you have the variables in a table or dataset array, then 0 must represent the response variable depending on the position of the response variable. The following example illustrates this.
Load the sample data and define the dataset array.
load hospital ds = dataset(hospital.Sex,hospital.BloodPressure(:,1),hospital.Age,... hospital.Smoker,'VarNames',{'Sex','BloodPressure','Age','Smoker'});
Represent the linear model 'BloodPressure ~ 1 + Sex + Age + Smoker' in a terms matrix. The response variable is in the second column of the dataset array, so there must be a column of 0s for the response variable in the second column of the terms matrix.
T = [0 0 0 0;1 0 0 0;0 0 1 0;0 0 0 1]
T = 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 1
Redefine the dataset array.
ds = dataset(hospital.BloodPressure(:,1),hospital.Sex,hospital.Age,... hospital.Smoker,'VarNames',{'BloodPressure','Sex','Age','Smoker'});
Now, the response variable is the first term in the dataset array. Specify the same linear model, 'BloodPressure ~ 1 + Sex + Age + Smoker', using a terms matrix.
T = [0 0 0 0;0 1 0 0;0 0 1 0;0 0 0 1]
T = 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1
If you have the predictor and response variables in a matrix and column vector, then you must include 0 for the response variable at the end of each term. The following example illustrates this.
Load the sample data and define the matrix of predictors.
load carsmall
X = [Acceleration,Weight];
Specify the model 'MPG ~ Acceleration + Weight + Acceleration:Weight + Weight^2' using a term matrix and fit the model to the data. This model includes the main effect and two-way interaction terms for the variables, Acceleration and Weight, and a second-order term for the variable, Weight.
T = [0 0 0;1 0 0;0 1 0;1 1 0;0 2 0]
T = 0 0 0 1 0 0 0 1 0 1 1 0 0 2 0
Fit a linear model.
mdl = fitlm(X,MPG,T)
mdl = Linear regression model: y ~ 1 + x1*x2 + x2^2 Estimated Coefficients: Estimate SE tStat pValue (Intercept) 48.906 12.589 3.8847 0.00019665 x1 0.54418 0.57125 0.95261 0.34337 x2 -0.012781 0.0060312 -2.1192 0.036857 x1:x2 -0.00010892 0.00017925 -0.6076 0.545 x2^2 9.7518e-07 7.5389e-07 1.2935 0.19917 Number of observations: 94, Error degrees of freedom: 89 Root Mean Squared Error: 4.1 R-squared: 0.751, Adjusted R-Squared 0.739 F-statistic vs. constant model: 67, p-value = 4.99e-26
Only the intercept and x2 term, which correspond to the Weight variable, are significant at the 5% significance level.
Now, perform a stepwise regression with a constant model as the starting model and a linear model with interactions as the upper model.
T = [0 0 0;1 0 0;0 1 0;1 1 0];
mdl = stepwiselm(X,MPG,[0 0 0],'upper',T)
1. Adding x2, FStat = 259.3087, pValue = 1.643351e-28 mdl = Linear regression model: y ~ 1 + x2 Estimated Coefficients: Estimate SE tStat pValue (Intercept) 49.238 1.6411 30.002 2.7015e-49 x2 -0.0086119 0.0005348 -16.103 1.6434e-28 Number of observations: 94, Error degrees of freedom: 92 Root Mean Squared Error: 4.13 R-squared: 0.738, Adjusted R-Squared 0.735 F-statistic vs. constant model: 259, p-value = 1.64e-28
The results of the stepwise regression are consistent with the results of fitlm in the previous step.
A formula for a model specification is a string of the form
'Y ~ terms',
Y is the response name.
terms contains
Variable names
+ to include the next variable
- to exclude the next variable
: to define an interaction, a product of terms
* to define an interaction and all lower-order terms
^ to raise the predictor to a power, exactly as in * repeated, so ^ includes lower order terms as well
() to group terms
Tip Formulas include a constant (intercept) term by default. To exclude a constant term from the model, include -1 in the formula. |
Examples:
'Y ~ A + B + C' is a three-variable
linear model with intercept.
'Y ~ A + B +
C - 1' is a three-variable linear model without intercept.
'Y ~ A + B + C + B^2' is a three-variable
model with intercept and a B^2 term.
'Y
~ A + B^2 + C' is the same as the previous example, since B^2 includes
a B term.
'Y ~ A + B +
C + A:B' includes an A*B term.
'Y
~ A*B + C' is the same as the previous example, since A*B
= A + B + A:B.
'Y ~ A*B*C - A:B:C' has
all interactions among A, B,
and C, except the three-way interaction.
'Y
~ A*(B + C + D)' has all linear terms, plus products of A with
each of the other variables.
For example, to specify an interaction model using fitlm with matrix predictors:
mdl = fitlm(X,y,'y ~ x1*x2*x3 - x1:x2:x3');
To specify a model using stepwiselm and a table or dataset array tbl of predictors, suppose you want to start from a constant and have a linear model upper bound. Assume the response variable in tbl is named 'y', and the predictor variables are named 'x1', 'x2', and 'x3'.
mdl2 = stepwiselm(tbl,'y ~ 1','Upper','y ~ x1 + x2 + x3');
The most common optional arguments for fitting:
For robust regression in fitlm, set the 'RobustOpts' name-value pair to 'on'.
Specify an appropriate upper bound model in stepwiselm, such as set 'Upper' to 'linear'.
Indicate which variables are categorical using the 'CategoricalVars' name-value pair. Provide a vector with column numbers, such as [1 6] to specify that predictors 1 and 6 are categorical. Alternatively, give a logical vector the same length as the data columns, with a 1 entry indicating that variable is categorical. If there are seven predictors, and predictors 1 and 6 are categorical, specify logical([1,0,0,0,0,1,0]).
For a table or dataset array, specify the response variable using the 'ResponseVar' name-value pair. The default is the last column in the array.
For example,
mdl = fitlm(X,y,'linear',... 'RobustOpts','on','CategoricalVars',3); mdl2 = stepwiselm(tbl,'constant',... 'ResponseVar','MPG','Upper','quadratic');
After fitting a model, examine the result.
A linear regression model shows several diagnostics when you enter its name or enter disp(mdl). This display gives some of the basic information to check whether the fitted model represents the data adequately.
For example, fit a linear model to data constructed with two out of five predictors not present and with no intercept term:
X = randn(100,5); y = X*[1;0;3;0;-1]+randn(100,1); mdl = fitlm(X,y)
mdl = Linear regression model: y ~ 1 + x1 + x2 + x3 + x4 + x5 Estimated Coefficients: Estimate SE tStat pValue (Intercept) 0.038164 0.099458 0.38372 0.70205 x1 0.92794 0.087307 10.628 8.5494e-18 x2 -0.075593 0.10044 -0.75264 0.45355 x3 2.8965 0.099879 29 1.1117e-48 x4 0.045311 0.10832 0.41831 0.67667 x5 -0.99708 0.11799 -8.4504 3.593e-13 Number of observations: 100, Error degrees of freedom: 94 Root Mean Squared Error: 0.972 R-squared: 0.93, Adjusted R-Squared 0.926 F-statistic vs. constant model: 248, p-value = 1.5e-52
Notice that:
The display contains the estimated values of each coefficient in the Estimate column. These values are reasonably near the true values [0;1;0;3;0;-1].
There is a standard error column for the coefficient estimates.
The reported pValue (which are derived from the t statistics under the assumption of normal errors) for predictors 1, 3, and 5 are extremely small. These are the three predictors that were used to create the response data y.
The pValue for (Intercept), x2 and x4 are much larger than 0.01. These three predictors were not used to create the response data y.
The display contains R^{2}, adjusted R^{2}, and F statistics.
To examine the quality of the fitted model, consult an ANOVA table. For example, use anova on a linear model with five predictors:
X = randn(100,5); y = X*[1;0;3;0;-1]+randn(100,1); mdl = fitlm(X,y); tbl = anova(mdl)
tbl = SumSq DF MeanSq F pValue x1 106.62 1 106.62 112.96 8.5494e-18 x2 0.53464 1 0.53464 0.56646 0.45355 x3 793.74 1 793.74 840.98 1.1117e-48 x4 0.16515 1 0.16515 0.17498 0.67667 x5 67.398 1 67.398 71.41 3.593e-13 Error 88.719 94 0.94382
This table gives somewhat different results than the default display (see Model Display). The table clearly shows that the effects of x2 and x4 are not significant. Depending on your goals, consider removing x2 and x4 from the model.
Diagnostic plots help you identify outliers, and see other problems in your model or fit. For example, load the carsmall data, and make a model of MPG as a function of Cylinders (nominal) and Weight:
load carsmall tbl = table(Weight,MPG,Cylinders); tbl.Cylinders = ordinal(tbl.Cylinders); mdl = fitlm(tbl,'MPG ~ Cylinders*Weight + Weight^2');ˋ
Make a leverage plot of the data and model.
plotDiagnostics(mdl)
There are a few points with high leverage. But this plot does not reveal whether the high-leverage points are outliers.
Look for points with large Cook's distance.
plotDiagnostics(mdl,'cookd')
There is one point with large Cook's distance. Identify it and remove it from the model. You can use the Data Cursor to click the outlier and identify it, or identify it programmatically:
[~,larg] = max(mdl.Diagnostics.CooksDistance); mdl2 = fitlm(tbl,'MPG ~ Cylinders*Weight + Weight^2',... 'Exclude',larg);
There are several residual plots to help you discover errors, outliers, or correlations in the model or data. The simplest residual plots are the default histogram plot, which shows the range of the residuals and their frequencies, and the probability plot, which shows how the distribution of the residuals compares to a normal distribution with matched variance.
Load the carsmall data, and make a model of MPG as a function of Cylinders (nominal) and Weight:
load carsmall tbl = table(Weight,MPG,Cylinders); tbl.Cylinders = ordinal(tbl.Cylinders); mdl = fitlm(tbl,'MPG ~ Cylinders*Weight + Weight^2');
Examine the residuals:
plotResiduals(mdl)
The observations above 12 are potential outliers.
plotResiduals(mdl,'probability')
The two potential outliers appear on this plot as well. Otherwise, the probability plot seems reasonably straight, meaning a reasonable fit to normally distributed residuals.
You can identify the two outliers and remove them from the data:
outl = find(mdl.Residuals.Raw > 12)
outl = 90 97
To remove the outliers, use the Exclude name-value pair:
mdl2 = fitlm(tbl,'MPG ~ Cylinders*Weight + Weight^2',... 'Exclude',outl);
Examine a residuals plot of mdl2:
plotResiduals(mdl2)
The new residuals plot looks fairly symmetric, without obvious problems. However, there might be some serial correlation among the residuals. Create a new plot to see if such an effect exists.
plotResiduals(mdl2,'lagged')
The scatter plot shows many more crosses in the upper-right and lower-left quadrants than in the other two quadrants, indicating positive serial correlation among the residuals.
Another potential issue is when residuals are large for large observations. See if the current model has this issue.
plotResiduals(mdl2,'fitted')
There is some tendency for larger fitted values to have larger residuals. Perhaps the model errors are proportional to the measured values.
This example shows how to understand the effect each predictor has on a regression model using a variety of available plots.
Create a model of mileage from some predictors in the carsmall data.
load carsmall tbl = table(Weight,MPG,Cylinders); tbl.Cylinders = ordinal(tbl.Cylinders); mdl = fitlm(tbl,'MPG ~ Cylinders*Weight + Weight^2');
Examine a slice plot of the responses. This displays the effect of each predictor separately.
plotSlice(mdl)
You can drag the individual predictor values, which are represented by dashed blue vertical lines. You can also choose between simultaneous and non-simultaneous confidence bounds, which are represented by dashed red curves.
Use an effects plot to show another view of the effect of predictors on the response.
plotEffects(mdl)
This plot shows that changing Weight from about 2500 to 4732 lowers MPG by about 30 (the location of the upper blue circle). It also shows that changing the number of cylinders from 8 to 4 raises MPG by about 10 (the lower blue circle). The horizontal blue lines represent confidence intervals for these predictions. The predictions come from averaging over one predictor as the other is changed. In cases such as this, where the two predictors are correlated, be careful when interpreting the results.
Instead of viewing the effect of averaging over a predictor as the other is changed, examine the joint interaction in an interaction plot.
plotInteraction(mdl,'Weight','Cylinders')
The interaction plot shows the effect of changing one predictor with the other held fixed. In this case, the plot is much more informative. It shows, for example, that lowering the number of cylinders in a relatively light car (Weight = 1795) leads to an increase in mileage, but lowering the number of cylinders in a relatively heavy car (Weight = 4732) leads to a decrease in mileage.
For an even more detailed look at the interactions, look at an interaction plot with predictions. This plot holds one predictor fixed while varying the other, and plots the effect as a curve. Look at the interactions for various fixed numbers of cylinders.
plotInteraction(mdl,'Cylinders','Weight','predictions')
Now look at the interactions with various fixed levels of weight.
plotInteraction(mdl,'Weight','Cylinders','predictions')
This example shows how to understand the effect of each term in a regression model using a variety of available plots.
Create a model of mileage from some predictors in the carsmall data.
load carsmall tbl = table(Weight,MPG,Cylinders); tbl.Cylinders = ordinal(tbl.Cylinders); mdl = fitlm(tbl,'MPG ~ Cylinders*Weight + Weight^2');
Create an added variable plot with Weight^2 as the added variable.
plotAdded(mdl,'Weight^2')
This plot shows the results of fitting both Weight^2 and MPG to the terms other than Weight^2. The reason to use plotAdded is to understand what additional improvement in the model you get by adding Weight^2. The coefficient of a line fit to these points is the coefficient of Weight^2 in the full model. The Weight^2 predictor is just over the edge of significance (pValue < 0.05) as you can see in the coefficients table display. You can see that in the plot as well. The confidence bounds look like they could not contain a horizontal line (constant y), so a zero-slope model is not consistent with the data.
Create an added variable plot for the model as a whole.
plotAdded(mdl)
The model as a whole is very significant, so the bounds don't come close to containing a horizontal line. The slope of the line is the slope of a fit to the predictors projected onto their best-fitting direction, or in other words, the norm of the coefficient vector.
There are two ways to change a model:
step — Add or subtract terms one at a time, where step chooses the most important term to add or remove.
addTerms and removeTerms — Add or remove specified terms. Give the terms in any of the forms described in Choose a Model or Range of Models.
If you created a model using stepwiselm, step can have an effect only if you give different upper or lower models. step does not work when you fit a model using RobustOpts.
For example, start with a linear model of mileage from the carbig data:
load carbig tbl = table(Acceleration,Displacement,Horsepower,Weight,MPG); mdl = fitlm(tbl,'linear','ResponseVar','MPG')
mdl = Linear regression model: MPG ~ 1 + Acceleration + Displacement + Horsepower + Weight Estimated Coefficients: Estimate SE tStat pValue (Intercept) 45.251 2.456 18.424 7.0721e-55 Acceleration -0.023148 0.1256 -0.1843 0.85388 Displacement -0.0060009 0.0067093 -0.89441 0.37166 Horsepower -0.043608 0.016573 -2.6312 0.008849 Weight -0.0052805 0.00081085 -6.5123 2.3025e-10 Number of observations: 392, Error degrees of freedom: 387 Root Mean Squared Error: 4.25 R-squared: 0.707, Adjusted R-Squared 0.704 F-statistic vs. constant model: 233, p-value = 9.63e-102
Try to improve the model using step for up to 10 steps:
mdl1 = step(mdl,'NSteps',10)
1. Adding Displacement:Horsepower, FStat = 87.4802, pValue = 7.05273e-19 mdl1 = Linear regression model: MPG ~ 1 + Acceleration + Weight + Displacement*Horsepower Estimated Coefficients: Estimate SE tStat pValue (Intercept) 61.285 2.8052 21.847 1.8593e-69 Acceleration -0.34401 0.11862 -2.9 0.0039445 Displacement -0.081198 0.010071 -8.0623 9.5014e-15 Horsepower -0.24313 0.026068 -9.3265 8.6556e-19 Weight -0.0014367 0.00084041 -1.7095 0.088166 Displacement:Horsepower 0.00054236 5.7987e-05 9.3531 7.0527e-19 Number of observations: 392, Error degrees of freedom: 386 Root Mean Squared Error: 3.84 R-squared: 0.761, Adjusted R-Squared 0.758 F-statistic vs. constant model: 246, p-value = 1.32e-117
step stopped after just one change.
To try to simplify the model, remove the Acceleration and Weight terms from mdl1:
mdl2 = removeTerms(mdl1,'Acceleration + Weight')
mdl2 = Linear regression model: MPG ~ 1 + Displacement*Horsepower Estimated Coefficients: Estimate SE tStat pValue (Intercept) 53.051 1.526 34.765 3.0201e-121 Displacement -0.098046 0.0066817 -14.674 4.3203e-39 Horsepower -0.23434 0.019593 -11.96 2.8024e-28 Displacement:Horsepower 0.00058278 5.193e-05 11.222 1.6816e-25 Number of observations: 392, Error degrees of freedom: 388 Root Mean Squared Error: 3.94 R-squared: 0.747, Adjusted R-Squared 0.745 F-statistic vs. constant model: 381, p-value = 3e-115
mdl2 uses just Displacement and Horsepower, and has nearly as good a fit to the data as mdl1 in the Adjusted R-Squared metric.
There are three ways to use a linear model to predict or simulate the response to new data:
This example shows how to predict and obtain confidence intervals on the predictions using the predict method.
Load the carbig data and make a default linear model of the response MPG to the Acceleration, Displacement, Horsepower, and Weight predictors.
load carbig
X = [Acceleration,Displacement,Horsepower,Weight];
mdl = fitlm(X,MPG);
Create a three-row array of predictors from the minimal, mean, and maximal values. There are some NaN values, so use functions that ignore NaN values.
Xnew = [nanmin(X);nanmean(X);nanmax(X)]; % new data
Find the predicted model responses and confidence intervals on the predictions.
[NewMPG NewMPGCI] = predict(mdl,Xnew)
NewMPG = 34.1345 23.4078 4.7751 NewMPGCI = 31.6115 36.6575 22.9859 23.8298 0.6134 8.9367
The confidence bound on the mean response is narrower than those for the minimum or maximum responses, which is quite sensible.
When you construct a model from a table or dataset array, feval is often more convenient for predicting mean responses than predict. However, feval does not provide confidence bounds.
This example shows how to predict mean responses using the feval method.
Load the carbig data and make a default linear model of the response MPG to the Acceleration, Displacement, Horsepower, and Weight predictors.
load carbig tbl = table(Acceleration,Displacement,Horsepower,Weight,MPG); mdl = fitlm(tbl,'linear','ResponseVar','MPG');
Create a three-row array of predictors from the minimal, mean, and maximal values. There are some NaN values, so use functions that ignore NaN values.
X = [Acceleration,Displacement,Horsepower,Weight];
Xnew = [nanmin(X);nanmean(X);nanmax(X)]; % new data
The Xnew array has the correct number of columns for prediction, so feval can use it for predictions.
Find the predicted model responses.
NewMPG = feval(mdl,Xnew)
NewMPG = 34.1345 23.4078 4.7751
The random method simulates new random response values, equal to the mean prediction plus a random disturbance with the same variance as the training data.
This example shows how to simulate responses using the random method.
Load the carbig data and make a default linear model of the response MPG to the Acceleration, Displacement, Horsepower, and Weight predictors.
load carbig
X = [Acceleration,Displacement,Horsepower,Weight];
mdl = fitlm(X,MPG);
Create a three-row array of predictors from the minimal, mean, and maximal values. There are some NaN values, so use functions that ignore NaN values.
Xnew = [nanmin(X);nanmean(X);nanmax(X)]; % new data
Generate new predicted model responses including some randomness.
rng('default') % for reproducibility NewMPG = random(mdl,Xnew)
NewMPG = 36.4178 31.1958 -4.8176
Because a negative value of MPG does not seem sensible, try predicting two more times.
NewMPG = random(mdl,Xnew)
NewMPG = 37.7959 24.7615 -0.7783
NewMPG = random(mdl,Xnew)
NewMPG = 32.2931 24.8628 19.9715
Clearly, the predictions for the third (maximal) row of Xnew are not reliable.
Suppose you have a linear regression model, such as mdl from the following commands:
load carbig tbl = table(Acceleration,Displacement,Horsepower,Weight,MPG); mdl = fitlm(tbl,'linear','ResponseVar','MPG');
To share the model with other people, you can:
Provide the model display.
mdl
mdl = Linear regression model: MPG ~ 1 + Acceleration + Displacement + Horsepower + Weight Estimated Coefficients: Estimate SE tStat pValue (Intercept) 45.251 2.456 18.424 7.0721e-55 Acceleration -0.023148 0.1256 -0.1843 0.85388 Displacement -0.0060009 0.0067093 -0.89441 0.37166 Horsepower -0.043608 0.016573 -2.6312 0.008849 Weight -0.0052805 0.00081085 -6.5123 2.3025e-10 Number of observations: 392, Error degrees of freedom: 387 Root Mean Squared Error: 4.25 R-squared: 0.707, Adjusted R-Squared 0.704 F-statistic vs. constant model: 233, p-value = 9.63e-102
Provide just the model definition and coefficients.
mdl.CoefficientNames
ans = '(Intercept)' 'Acceleration' 'Displacement' 'Horsepower' 'Weight'
mdl.Coefficients.Estimate
ans = 45.2511 -0.0231 -0.0060 -0.0436 -0.0053
mdl.Formula
ans = MPG ~ 1 + Acceleration + Displacement + Horsepower + Weight
This example shows how to fit a linear regression model. A typical workflow involves the following: import data, fit a regression, test its quality, modify it to improve the quality, and share it.
Step 1. Import the data into a dataset array.
hospital.xls is an Excel® spreadsheet containing patient names, sex, age, weight, blood pressure, and dates of treatment in an experimental protocol. First read the data into a table.
patients = readtable('hospital.xls',... 'ReadRowNames',true);
Examine the first row of data.
patients(1,:)
ans = name sex age wgt smoke sys dia trial1 trial2 trial3 trial4 _______ ___ ___ ___ _____ ___ ___ ______ ______ ______ ______ YPL-320 'SMITH' 'm' 38 176 1 124 93 18 -99 -99 -99
The sex and smoke fields seem to have two choices each. So change these fields to nominal.
patients.smoke = nominal(patients.smoke,{'No','Yes'}); patients.sex = nominal(patients.sex);
Step 2. Create a fitted model.
Your goal is to model the systolic pressure as a function of a patient's age, weight, sex, and smoking status. Create a linear formula for 'sys' as a function of 'age', 'wgt', 'sex', and 'smoke' .
modelspec = 'sys ~ age + wgt + sex + smoke';
mdl = fitlm(patients,modelspec)
mdl = Linear regression model: sys ~ 1 + sex + age + wgt + smoke Estimated Coefficients: Estimate SE tStat pValue _________ ________ ________ __________ (Intercept) 118.28 7.6291 15.504 9.1557e-28 sex_m 0.88162 2.9473 0.29913 0.76549 age 0.08602 0.06731 1.278 0.20438 wgt -0.016685 0.055714 -0.29947 0.76524 smoke_Yes 9.884 1.0406 9.498 1.9546e-15 Number of observations: 100, Error degrees of freedom: 95 Root Mean Squared Error: 4.81 R-squared: 0.508, Adjusted R-Squared 0.487 F-statistic vs. constant model: 24.5, p-value = 5.99e-14
The sex, age, and weight predictors have rather high -values, indicating that some of these predictors might be unnecessary.
Step 3. Locate and remove outliers.
See if there are outliers in the data that should be excluded from the fit. Plot the residuals.
plotResiduals(mdl)
There is one possible outlier, with a value greater than 12. This is probably not truly an outlier. For demonstration, here is how to find and remove it.
Find the outlier.
outlier = mdl.Residuals.Raw > 12; find(outlier)
ans = 84
Remove the outlier.
mdl = fitlm(patients,modelspec,... 'Exclude',84); mdl.ObservationInfo(84,:)
ans = Weights Excluded Missing Subset _______ ________ _______ ______ WXM-486 1 true false false
Observation 84 is no longer in the model.
Try to obtain a simpler model, one with fewer predictors but the same predictive accuracy. step looks for a better model by adding or removing one term at a time. Allow step take up to 10 steps.
mdl1 = step(mdl,'NSteps',10)
1. Removing wgt, FStat = 4.6001e-05, pValue = 0.9946 2. Removing sex, FStat = 0.063241, pValue = 0.80199 mdl1 = Linear regression model: sys ~ 1 + age + smoke Estimated Coefficients: Estimate SE tStat pValue ________ ________ ______ __________ (Intercept) 115.11 2.5364 45.383 1.1407e-66 age 0.10782 0.064844 1.6628 0.09962 smoke_Yes 10.054 0.97696 10.291 3.5276e-17 Number of observations: 99, Error degrees of freedom: 96 Root Mean Squared Error: 4.61 R-squared: 0.536, Adjusted R-Squared 0.526 F-statistic vs. constant model: 55.4, p-value = 1.02e-16
step took two steps. This means it could not improve the model further by adding or subtracting a single term.
Plot the effectiveness of the simpler model on the training data.
plotResiduals(mdl1)
The residuals look about as small as those of the original model.
Step 5. Predict responses to new data.
Suppose you have four new people, aged 25, 30, 40, and 65, and the first and third smoke. Predict their systolic pressure using mdl1.
ages = [25;30;40;65]; smoker = {'Yes';'No';'Yes';'No'}; systolicnew = feval(mdl1,ages,smoker)
systolicnew = 127.8561 118.3412 129.4734 122.1149
To make predictions, you need only the variables that mdl1 uses.
You might want others to be able to use your model for prediction. Access the terms in the linear model.
coefnames = mdl1.CoefficientNames
coefnames = '(Intercept)' 'age' 'smoke_Yes'
View the model formula.
mdl1.Formula
ans = sys ~ 1 + age + smoke
Access the coefficients of the terms.
coefvals = mdl1.Coefficients(:,1); % table
coefvals = table2array(coefvals)
coefvals = 115.1066 0.1078 10.0540
The model is sys = 115.1066 + 0.1078*age + 10.0540*smoke, where smoke is 1 for a smoker, and 0 otherwise.