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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)

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.

There are other ways of selecting models, such as using `lasso`

, `lassoglm`

, `sequentialfs`

, or `plsregress`

.

Name | 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. |

`'poly` | Model is a polynomial with all terms up to degree in
the first predictor, degree `i` in the second
predictor, etc. Use numerals `j` `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)

`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 dsa = 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.

dsa = 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 character vector of the form

`'`

,* Y* ~

`terms`

is the response name.`Y`

contains`terms`

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

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 and make adjustments.

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

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