Driss,
David is right - you should include data/code.
Nevertheless, the error you get suggests that what you
call a "matrix" is in fact a data.frame; that's usually
a crucial distinction. There are two things you might try:
Dennis' way:
lm(as.matrix(a) ~ b)
David's way:
for(i in seq_along(a)) print(r<- lm(a[ ,i] ~ b) )
#Note the comma!
-Peter Ehlers
On 2010-03-31 7:42, David Winsemius wrote:
On Mar 31, 2010, at 9:13 AM, Driss Agramelal wrote:
Hello and thank you both for your answers!
Dennis, I tried to simply run
lm(a ~ b)
after re-importing "a" as a matrix, but I get the following error
message:
Error in model.frame.default(formula = a ~ b, drop.unused.levels =
TRUE) :
invalid type (list) for variable 'a'
so maybe I have to specify something in the arguments? What do you
think?
David,
I tried your syntax as well, and received quasi-the same error
statement:
for(i in seq_along(a)) print(r<- lm(a[i] ~ b) )
Error in model.frame.default(formula = a[i] ~ b, drop.unused.levels
= TRUE) :
invalid type (list) for variable 'a[i]'
I am not too familiar with the use of accolades, square brackets and
parentheses, the order in which they have to come
in the function and the role they play, but I think they might be
important...
I also tried to use "lapply"; it works wonderfully for a basic
function like:
lapply(a, mean)
I get a list of results with names and values..perfect! But with the
lm() function... I just
don't know how to write the arguments... tried several options
without success...
Any idea that could help me solve this only seemingly easy task
would be most welcome!!
Do you think we can figure this out when you have provided no sample
data and have not provided even the results of str on the data object
you are working with? Generally one gains insight by parring the
problem down to smaller test cases and working with them. Perhaps the
first 5 elements of "a" rather than all 150?
--
Peter Ehlers
University of Calgary
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