Your function 'll' only returns a single value when passed a vector:
> x <- seq(0,2,.1)
> ll(x)
[1] -7.571559
'plot' expects to pass a vector to the function and have it return a
vector of the same length; e.g.,
> sin(x)
[1] 0.00000000 0.09983342 0.19866933 0.29552021 0.38941834 0.47942554
0.56464247 0.64421769 0.71735609
[10] 0.78332691 0.84147098 0.89120736 0.93203909 0.96355819 0.98544973
0.99749499 0.99957360 0.99166481
[19] 0.97384763 0.94630009 0.90929743>
So you either have to rewrite your function, or have a loop that will
evaluate the function at each individual point and then plot it.
On Feb 5, 2008 7:06 PM, John Smith <zmring at gmail.com>
wrote:> Dear R-users,
>
> Suppose I have defined a likelihood function as ll(tau), how can I plot
this
> likelihood function by calling it by plot?
>
> I want to do it like this:
>
> ll <- function(tau)
> {
> w <- 1 / (s^2 + tau^2)
> mu <- sum(theta * w) / sum(w)
> -1/2*sum((theta-mu)^2*w -log(w))
> }
> plot(ll, 0, 2)
>
>
>
> But have the following error:
> Error in xy.coords(x, y, xlabel, ylabel, log) :
> 'x' and 'y' lengths differ
> In addition: Warning messages:
> 1: In s^2 + tau^2 :
> longer object length is not a multiple of shorter object length
> 2: In theta * w :
> longer object length is not a multiple of shorter object length
> 3: In (theta - mu)^2 * w :
> longer object length is not a multiple of shorter object length
>
>
> Thanks
>
> [[alternative HTML version deleted]]
>
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>
--
Jim Holtman
Cincinnati, OH
+1 513 646 9390
What is the problem you are trying to solve?