R help - Jun 2012 - Problem with predict?

Hello,

I am trying to fit a model to some "death over time" data that does
not fit the criteria for the usual LD50 type models (the counts are too large).
I am using a simple linear model in an attempt to plot a nice line on a scatter
plot and calculate some LD values to use in designing an experiment. Here is the
basic idea of what I'm doing:


head(mort)

Time    Density
0        2233333333
0        2100000000
0        1933333333
5        1900000000
5        1433333333
5         900000000


plot(Density~Time)

This plots something that looks a lot like a decay rate

mod<-lm(log(Density)~Time)

xv<-seq(0,60,0.1)
yv<-exp(predict(mod,list(time=xv)))
lines(xv,yv)

Everything seems to work fine until I try to plot the lines, but then I get the
error message:

Error in xy.coords(x, y) : 'x' and 'y' lengths differ

Checking the lengths of x and y confirms that somehow the object yv is not using
xv to predict the data, and is only predicting as many data points as there are
rows in the data frame.

Any ideas on why this might be would be very much appreciated.

Thank you,

Julie





	[[alternative HTML version deleted]]

Hi
> 
> Hello,
> 
> I am trying to fit a model to some "death over time" data that
does not
> fit the criteria for the usual LD50 type models (the counts are too 
> large). I am using a simple linear model in an attempt to plot a nice 
line
> on a scatter plot and calculate some LD values to use in designing an 
> experiment. Here is the basic idea of what I'm doing:
> 
> 
> head(mort)
> 
> Time    Density
> 0        2233333333
> 0        2100000000
> 0        1933333333
> 5        1900000000
> 5        1433333333
> 5         900000000
> 
> 
> plot(Density~Time)
> 
> This plots something that looks a lot like a decay rate
> 
> mod<-lm(log(Density)~Time)
> 
> xv<-seq(0,60,0.1)
> yv<-exp(predict(mod,list(time=xv)))
>From help page
Usage
## S3 method for class 'lm'
predict(object, newdata, se.fit = FALSE, scale = NULL, df = Inf,
        interval = c("none", "confidence",
"prediction"),
        level = 0.95, type = c("response", "terms"),
        terms = NULL, na.action = na.pass,
        pred.var = res.var/weights, weights = 1, ...)

Arguments
object Object of class inheriting from "lm"
 
newdata An optional data frame in which to look for variables with which 
to predict. If omitted, the fitted values are used.
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 

yv<-exp(predict(mod, data.frame(time=xv)))

shall work.

Regards
Petr
> lines(xv,yv)
> 
> Everything seems to work fine until I try to plot the lines, but then I 
> get the error message:
> 
> Error in xy.coords(x, y) : 'x' and 'y' lengths differ
> 
> Checking the lengths of x and y confirms that somehow the object yv is 
not
> using xv to predict the data, and is only predicting as many data points 
> as there are rows in the data frame.
> 
> Any ideas on why this might be would be very much appreciated.
> 
> Thank you,
> 
> Julie
> 
> 
> 
> 
> 
>    [[alternative HTML version deleted]]
> 
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