Linear regression coefficient information as a data frame or matrix
I am trying to create a script to optimize linear regression analysis and I would really like to work on the output of the model, in particular the Pr value (> | t |). Unfortunately I don't know how to get the output of the model into a matrix or data table.
Here's an example: In the code below, I create seven columns of data and fit the seventh using the other six. When I receive a model summary, it is clear that three parameters are much more important than the other three. If I could access the output of the coefficient numerically, I could create a script to remove the least significant parameter and rerun the analysis ... however, anyway, I do it manually.
What's the best way to do this?
Thanks in advance for your help.
q = matrix(
c(2,14,-4,1,10,9,41,8,13,2,0,20,3,27,1,10,-1,0,
10,-6,23,6,13,-8,1,15,-7,55,7,14,10,0,20,-3,6,4,20,
-1,5,19,-2,48,10,19,8,8,10,-2,24,8,13,9,8,14,5,7,7,
12,1,0,16,7,27,7,10,-1,1,15,7,31,2,20,-5,10,12,3,57,
0,19,-8,8,11,-4,63,5,11,7,8,10,-7,6,9,10,-7,2,19,8,
51,2,18,3,3,14,4,30), nrow=15, ncol=7, byrow = TRUE)
#
colnames(q) <- c("A","B","C","D","E","F","Z")
#
q <- as.data.frame(q)
#
qmodel <- lm(Z~.,data=q)
#
summary(qmodel)
#
Output:
Call:
lm(formula = Z ~ ., data = q)
Residuals:
Min 1Q Median 3Q Max
-1.25098 -0.52655 -0.02931 0.62350 1.26649
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) -2.09303 1.51627 -1.380 0.205
A 0.91161 0.11719 7.779 5.34e-05 ***
B 1.99503 0.09539 20.914 2.87e-08 ***
C -2.98252 0.04789 -62.283 4.91e-12 ***
D 0.13458 0.10377 1.297 0.231
E 0.15191 0.09397 1.617 0.145
F 0.01417 0.04716 0.300 0.772
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 0.9439 on 8 degrees of freedom
Multiple R-squared: 0.9986, Adjusted R-squared: 0.9975
F-statistic: 928.9 on 6 and 8 DF, p-value: 6.317e-11
Now, here's what I would like to see:
> coeffs
Estimate Std. Error t value Pr(>|t|)
(Intercept) -2.09303 1.51627 -1.380 2.05e-01
A 0.91161 0.11719 7.779 5.34e-05
B 1.99503 0.09539 20.914 2.87e-08
C -2.98252 0.04789 -62.283 4.91e-12
D 0.13458 0.10377 1.297 2.31e-01
E 0.15191 0.09397 1.617 1.45e-01
F 0.01417 0.04716 0.300 7.72e-01
Anyway, I got it this way ... didn't automate at all ...
coeffs = matrix(
c(-2.09303,1.51627,-1.38,0.205,0.91161,0.11719,
7.779,0.0000534,1.99503,0.09539,20.914,0.0000000287,
-2.98252,0.04789,-62.283,0.00000000000491,0.13458,
0.10377,1.297,0.231,0.15191,0.09397,1.617,0.145,
0.01417,0.04716,0.3,0.772), nrow=7, ncol=4, byrow = TRUE)
#
rownames(coeffs) <- c("(Intercept)","A","B","C","D","E","F")
colnames(coeffs) <- c("Estimate","Std. Error","t value","Pr(>|t|)")
#
coeffs <- as.data.frame(coeffs)
#
coeffs
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What you want is a coefficients
pivot object component .
m <- lm(Z~.,data=q)
summary(m)$coefficients
Some additional comments:
- Use
step
to select a step variable, not to encode it; - The stepwise selection of variables has poor statistical properties; consider something like
glmnet
(in a package of the same name) to create an ordered build of the model instead.
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If I understand correctly, you want the matrix returned by the summary. This is pretty straight forward:
fit <- lm( formula, data=yourData)
coeffs <- summary(fit)$coefficients
After that, you can select records from coeffs
that match your conditions, just like with any matrix. Example:
coeffs[coeffs[4,] < 1e-12,]
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