Extract regression coefficient values

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I have a regression model for some time series data investigating drug utilisation. The purpose is to fit a spline to a time series and work out 95% CI etc. The model goes as follows:

id <- ts(1:length(drug$Date))
a1 <- ts(drug$Rate)
a2 <- lag(a1-1)
tg <- ts.union(a1,id,a2)
mg <-lm (a1~a2+bs(id,df=df1),data=tg) 

The summary output of mg is:

Call:
lm(formula = a1 ~ a2 + bs(id, df = df1), data = tg)

Residuals:
     Min       1Q   Median       3Q      Max 
-0.31617 -0.11711 -0.02897  0.12330  0.40442 

Coefficients:
                  Estimate Std. Error t value Pr(>|t|)    
(Intercept)        0.77443    0.09011   8.594 1.10e-11 ***
a2                 0.13270    0.13593   0.976  0.33329    
bs(id, df = df1)1 -0.16349    0.23431  -0.698  0.48832    
bs(id, df = df1)2  0.63013    0.19362   3.254  0.00196 ** 
bs(id, df = df1)3  0.33859    0.14399   2.351  0.02238 *  
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 

I am using the Pr(>|t|) value of a2 to test if the data under investigation are autocorrelated.

Is it possible to extract this value of Pr(>|t|) (in this model 0.33329) and store it in a scalar to perform a logical test?

Alternatively, can it be worked out using another method?

Mar 26 in Machine Learning by Nandini
• 5,480 points
27 views

1 answer to this question.

0 votes

A quick rundown. These values are stored in a matrix named 'coefficients' by the lm object. As a result, you can get the value you're looking for by typing:

val <- summary(mg)$coefficients[2, 4]

Or, to put it another way, coef(summary(mg))["a2","Pr(>|t|)")].
It is suggested that you utilize those extractor functions since you can be sure that they will always return the proper information, even if the function authors decide to shuffle things about behind the scenes.

Maybe they add more information to one of the model list object's elements, which causes things to go out of order? All of your code will be broken.

answered Mar 30 by Dev
• 6,000 points

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