I don't know if you've gotten any follow up, but here are some quick reactions:
1) You make reference to the columns of y but your dput(y) does not
provide columns,
2) It's still not clear to me what all this data actually means? Do
you have multiple observations of the dependent variable corresp
When I tried dput function, the result was this:
> dput(x)
c(20, 200, 2000, 2)
> dput(y)
c(0.45, 0.05, 0.5, 0.4, 0, 0.5, 0.4, 0.05, 0.4, 0.25, 0.35, 0.5,
0.05, 0.4, 0.5, 0.5, 0.5, 0.25, 0.85, 0.5, 0.5, 0.5, 0.25, 0.4,
0.25, 0.25, 0.4, 0.25, 0.5, 0.15, 0.25, 0.1, 0.25, 0.25, 0.015,
0.4, 0.5
Could you dput() the structure of x and y: I'm having trouble
visualizing how your data is set up.
Michael
On Mon, Oct 24, 2011 at 12:07 PM, Julie wrote:
> The variable y is made of four columns, each paired to 20, 200, 2000 or 20
> 000.
>> y <- c(rdiktator20, rDiktator200, rDikt2000, rDikt2
The variable y is made of four columns, each paired to 20, 200, 2000 or 20
000.
> y <- c(rdiktator20, rDiktator200, rDikt2000, rDikt2)
So I guess the problem is in the fact that I did not specify it correctly,
is it so? How can I tell R properly that one part of y matches to one part
of x?
Th
X and y must have the same number of elements, and NA values must be removed
(?na.omit)
---
Jeff Newmiller The . . Go Live...
DCN: Basics: ##.#. ##.#. Live Go...
Live: OO#.. Dead: OO#.. Playing
Research Engineer (Solar
You are trying to regress ~372 observations of the dependent against
~4 observations of the independent variable. Ask yourself again if
this makes sense.
A further hint might be given by this
y = rnorm(5); x = y[1:4]
lm(y~x)
Michael
On Mon, Oct 24, 2011 at 11:13 AM, Julie wrote:
> Hello,
> I a
Hello,
I am trying to get a linear model of y ~ log(x).
*> lm (y~log(x))*
However, I always get an error report:
/Error in model.frame.default(formula = y ~ log(x), drop.unused.levels =
TRUE) :
variable lengths differ (found for 'log(x)')/
*Here was my y:*
> y
[1]0.4500.050
7 matches
Mail list logo
