On Nov 27, 2008, at 9:49 AM, David Winsemius wrote:
Have you looked at results of str on a regression object? I would
not think that there would be a single p.value associated with such
a beast, but that there might be if you examined individual
coefficients.
? coefficients
?coef
That wasn't on as on point as I thought. Take a look at this screen
dialog:
> x <- 1:5; coef(lm(c(1:3,7,6) ~ x))
(Intercept) x
-0.7 1.5
> str(coef(lm(c(1:3,7,6) ~ x)))
Named num [1:2] -0.7 1.5
- attr(*, "names")= chr [1:2] "(Intercept)" "x"
> anova(lm(c(1:3,7,6) ~ x))
Analysis of Variance Table
Response: c(1:3, 7, 6)
Df Sum Sq Mean Sq F value Pr(>F)
x 1 22.5000 22.5000 15.698 0.02872 *
Residuals 3 4.3000 1.4333
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
> str(anova(lm(c(1:3,7,6) ~ x)))
Classes ‘anova’ and 'data.frame': 2 obs. of 5 variables:
$ Df : int 1 3
$ Sum Sq : num 22.5 4.3
$ Mean Sq: num 22.5 1.43
$ F value: num 15.7 NA
$ Pr(>F) : num 0.0287 NA
- attr(*, "heading")= chr "Analysis of Variance Table\n" "Response:
c(1:3, 7, 6)"
> anova(lm(c(1:3,7,6) ~ x))$"Pr(>F)"
[1] 0.02871561 NA
> anova(lm(c(1:3,7,6) ~ x))$"Pr(>F)"[1]
[1] 0.02871561
--
David Winsemius
On Nov 27, 2008, at 4:03 AM, ales grill wrote:
Dear all,
I have wrote a code for a linear regression. I want to
write a loop for so, that I can get estimate for pavlues for six
predictors.
But I am getting for estmate for only last one. How can I get
pvalues for
all my predictors in a loop??
Anticipating your help
Thanks
Ales
mat<-matrix(rnorm(36),nrow=6)
mat
[,1] [,2] [,3] [,4]
[,5] [,6]
[1,] 1.10536338 -0.7613770 -1.7100569 -1.8762241 -0.36579280
0.6465219
[2,] -1.34836804 -0.2174270 -0.1153477 -0.1727683 -1.88406206
1.7484955
[3,] 0.96814418 -2.1483727 0.5839668 -1.2361659 0.04592844
1.9937995
[4,] 0.01960219 -1.2339691 0.8290761 0.1002795 -0.15952881
0.3969251
[5,] 1.62343073 1.3741222 -1.2045854 0.4180127 -0.09898615
1.3575119
[6,] -0.95260509 -0.1522824 -1.4257526 1.0057412 -1.20068336
-0.4306761
res<-rnorm(6)
res
[1] 0.2045252 -0.9824761 0.7727004 0.6439993 1.8005737 1.0167214
pval<-NULL
for(i in c(1:6))
+ {
+ reg<-lm(res~mat[,i])
+ reg
+ pval[i]<-reg$p.value
+ }
pval
NULL
reg
Call:
lm(formula = res ~ mat[, i])
Coefficients:
(Intercept) mat[, i]
0.8195 -0.2557
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and provide commented, minimal, self-contained, reproducible code.
______________________________________________
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
