Documentation

## Linear Regression

### Prepare Data

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');`

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 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 vector

• Categorical vector

• Character array

• String 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.

#### Dataset Array for Input and Response Data

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 = categorical(Model_Year);```

#### Table for Input and Response Data

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 = categorical(Model_Year);```

#### Numeric Matrix for Input Data, Numeric Vector for Response

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`.

### Choose a Fitting Method

There are three ways to fit a model to data:

#### Least-Squares Fit

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 Examine Quality and Adjust Fitted Model).

#### Robust Fit

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.

#### Stepwise Fit

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 Examine Quality and Adjust Fitted Model).

### Choose a Model or Range of Models

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`.

#### Brief Name

NameModel 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);```

#### Terms Matrix

A terms matrix `T` 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  +1 accounts for the response variable. The value of `T(i,j)` is the exponent of variable `j` in term `i`.

For example, suppose that an input includes three predictor variables `A`, `B`, and `C` and the response variable `Y` in the order `A`, `B`, `C`, and `Y`. Each row of `T` represents one term:

• `[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, a column vector of zeros in a terms matrix represents the position of the response variable. If you have the predictor and response variables in a matrix and column vector, then you must include `0` for the response variable in the last column of each row.

#### Formula

A formula for a model specification is a character vector or string scalar 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');`

### Fit Model to Data

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');```

### Examine Quality and Adjust Fitted Model

After fitting a model, examine the result and make adjustments.

Model Display

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 (`tStat`) 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 ${\mathit{R}}^{2}$, adjusted ${\mathit{R}}^{2}$, and F statistics.

ANOVA

To examine the quality of the fitted model, consult an ANOVA table. For example, use `anova` on a linear model with five predictors:

`tbl = anova(mdl)`
```tbl=6×5 table 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 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

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` (categorical) and `Weight`:

```load carsmall tbl = table(Weight,MPG,Cylinders); tbl.Cylinders = categorical(tbl.Cylinders); mdl = fitlm(tbl,'MPG ~ Cylinders*Weight + Weight^2');```

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);```

Residuals — Model Quality for Training Data

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.

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 = 2×1 90 97 ```

To remove the outliers, use the `Exclude` name-value pair:

`mdl3 = fitlm(tbl,'MPG ~ Cylinders*Weight + Weight^2','Exclude',outl);`

Examine a residuals plot of mdl2:

`plotResiduals(mdl3)`

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(mdl3,'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(mdl3,'fitted')`

There is some tendency for larger fitted values to have larger residuals. Perhaps the model errors are proportional to the measured values.

Plots to Understand Predictor Effects

This example shows how to understand the effect each predictor has on a regression model using a variety of available plots.

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')`

Plots to Understand Terms Effects

This example shows how to understand the effect of each term in a regression model using a variety of available plots.

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.

Change Models

There are two ways to change a model:

If you created a model using `stepwiselm`, then `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.

### Predict or Simulate Responses to New Data

A `LinearModel` object offers three functions to predict or simulate the response to new data: `predict`, `feval`, and `random`.

`predict`

Use the `predict` function to predict and obtain confidence intervals on the predictions.

Load the `carbig` data and create 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. `X` contains some `NaN` values, so use functions that ignore `NaN` values.

`Xnew = [nanmin(X);nanmean(X);nanmax(X)];`

Find the predicted model responses and confidence intervals on the predictions.

`[NewMPG, NewMPGCI] = predict(mdl,Xnew)`
```NewMPG = 3×1 34.1345 23.4078 4.7751 ```
```NewMPGCI = 3×2 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.

`feval`

Use the `feval` function to predict responses. When you create a model from a table or dataset array, `feval` is often more convenient than `predict` for predicting responses. When you have new predictor data, you can pass it to `feval` without creating a table or matrix. However, `feval` does not provide confidence bounds.

Load the `carbig` data set and create a default linear model of the response `MPG` to the predictors `Acceleration`, `Displacement`, `Horsepower`, and `Weight`.

```load carbig tbl = table(Acceleration,Displacement,Horsepower,Weight,MPG); mdl = fitlm(tbl,'linear','ResponseVar','MPG');```

Predict the model response for the mean values of the predictors.

`NewMPG = feval(mdl,nanmean(Acceleration),nanmean(Displacement),nanmean(Horsepower),nanmean(Weight))`
```NewMPG = 23.4078 ```

`random`

Use the `random` function to simulate responses. The `random` function simulates new random response values, equal to the mean prediction plus a random disturbance with the same variance as the training data.

Load the `carbig` data and create 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.

`Xnew = [nanmin(X);nanmean(X);nanmax(X)];`

Generate new predicted model responses including some randomness.

```rng('default') % for reproducibility NewMPG = random(mdl,Xnew)```
```NewMPG = 3×1 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 = 3×1 37.7959 24.7615 -0.7783 ```
`NewMPG = random(mdl,Xnew)`
```NewMPG = 3×1 32.2931 24.8628 19.9715 ```

Clearly, the predictions for the third (maximal) row of `Xnew` are not reliable.

### Share Fitted Models

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 the model definition and coefficients.

`mdl.Formula`
```ans = MPG ~ 1 + Acceleration + Displacement + Horsepower + Weight ```
`mdl.CoefficientNames`
```ans = 1x5 cell array Columns 1 through 4 {'(Intercept)'} {'Acceleration'} {'Displacement'} {'Horsepower'} Column 5 {'Weight'} ```
`mdl.Coefficients.Estimate`
```ans = 5×1 45.2511 -0.0231 -0.0060 -0.0436 -0.0053 ```