On Aug 20, 2009, at 3:42 PM, g...@ucalgary.ca wrote:
Thanks!
On Aug 20, 2009, at 1:46 PM, g...@ucalgary.ca wrote:
I got two questions on factors in regression:
Q1.
In a table, there a few categorical/factor variables, a few
numerical
variables and the response variable is numeric. Some factors are
important
but others not.
How to determine which categorical variables are significant to the
response variable?
Seems that you should engage the services of a consulting
statistician
for that sort of question. Or post in a venue where statistical
consulting is supposed to occur, such as one of the sci.stat.*
newsgroups.
I googled sci.stat.* and got sci.stat.math and sci.stat.consult.
Are they good?
The quality of responses varies. You may get what you pay for. On the
other hand sometimes you get high-quality advice for free.
I have no idea to do this. So any clue will be appreciated.
http://groups.google.com/?hl=en
Q2.
As we knew, lm can deal with categorical variables.
I thought, when there is a categorical predictor, we may use lm
directly
without quantifying these factors and assigning different values to
factors
would not change the fittings as shown:
The "numbers" that you are attempting to assign are really just
labels
for the factor levels. The regression functions in R will not use
them
for any calculations. They should not be thought of as having
"values". Even if the factor is an ordered factor, the labels may not
be interpretable as having the same numerical order as the string
values might suggest.
x <- 1:20 ## numeric predictor
yes.no <- c("yes","no")
factors <- gl(2,10,20,yes.no) ##factor predictor
factors.quant <- rep(c(18.8,29.9),c(10,10)) ##quantificatio of
factors
Not sure what that is supposed to mean. It is not a factor object
even
though you may be misleading yourself in to believing it should be.
It's a numeric vector.
Yes, levels are not numeric but just labels. But
after the levels factors being assigned to numeric values as
factors.quant
and factors.quant.1,
lm(response ~ x + factors.quant) and lm(response ~ x + factors.quant1)
produced the same fitted curve as lm(response ~ x + factors). This
is what
I could not understand.
In for the factor variable case and the numeric variable case there
was no variation in the predictor variable within a level. So the
predictions will all be the same within levels in each case. There
will be differences in the coefficients arrived at to achieve that
result, however.
str(factors.quant)
num [1:20] 18.8 18.8 18.8 18.8 18.8 18.8 18.8 18.8 18.8 18.8 ...
factors.quant.1 <- rep(c(16.9,38.9),c(10,10))
##second quantificatio of factors
response <- 0.8*x + 18 + factors.quant + rnorm(20) ##response
lm.quant <- lm(response ~ x + factors.quant) ##lm with
quantifications
lm.fact <- lm(response ~ x + factors) ##lm with factors
lm.quant
Call:
lm(formula = response ~ x + factors.quant)
Coefficients:
(Intercept) x factors.quant
14.9098 0.5385 1.2350
lm.fact
Call:
lm(formula = response ~ x + factors)
Coefficients:
(Intercept) x factorsno
38.1286 0.5385 13.7090
lm.quant.1 <- lm(response ~ x + factors.quant.1) ##lm with
quantifications
lm.quant.1
Call:
lm(formula = response ~ x + factors.quant.1)
Coefficients:
(Intercept) x factors.quant.1
27.5976 0.5385 0.6231
lm.fact.1 <- lm(response ~ x + factors) ##lm with factors
par(mfrow=c(2,2)) ## comparisons of two fittings
plot(x, response)
lines(x,fitted(lm.quant),col="blue")
grid()
plot(x,response)
lines(x,fitted(lm.fact),col = "red")
grid()
plot(x, response)
lines(x,fitted(lm.quant.1),lty =2,col="blue")
grid()
plot(x,response)
lines(x,fitted(lm.fact.1),lty =2,col = "red")
grid()
par(mfrow = c(1,1))
So, is it right that we can assign any numeric values to factors,
for example, c(yes, no) = c(18.8,29.9) or (16.9,38.9) in the above,
before doing lm, glm, aov, even nls?
You can give factor levels any name you like, including any sequence
of digit characters. Unlike "ordinary R where unquoted numbers cannot
start variable names, factor functions will coerce numeric vectors to
character vectors when assigning level names. But you seem to be
conflating factors with numeric vectors that have many ties. Those
two
entities would have different handling by R's regression functions.
--
David Winsemius, MD
Heritage Laboratories
West Hartford, CT
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