Thank you for the clarification, this makes sense now! dose.p from MASS also
does the job perfectly, in returning x values for specified proportions. I'm
curious, if I use the other parameter in predict.glm type="link" (instead of
type="response"), in the case of a logistic binomial, what does this
predict?



On Wed, Jan 26, 2011 at 11:41 PM, Gavin Simpson <gavin.simp...@ucl.ac.uk>wrote:

> On Wed, 2011-01-26 at 19:25 -1000, Ahnate Lim wrote:
> > Even when I try to predict y values from x, let's say I want to predict y
> at
> > x=0. Looking at the graph from the provided syntax, I would expect y to
> be
> > about 0.85. Is this right:
> >
> > predict(mylogit,newdata=as.data.frame(0),type="response")
>
> Your original problem was the use of `newdata = as.data.frame(0.5)`.
> There are two problems here: i) if you don't name the input (x = 0.5,
> say) then you get a data frame with the name(s) "0.5":
>
> > as.data.frame(0.5)
>   0.5
> 1 0.5
>
> and ii) if you do name it, you still get a data frame with name(s) "0.5"
>
> > as.data.frame(x = 0.5)
>  0.5
> 1 0.5
>
> In both cases, predict wants to find a variable with the name `x` in the
> object supplied to `newdata`. It finds `x` but your `x` in the global
> workspace, but warns because it knows that `newdata` was a data frame
> with a single row - so there was a mismatch and you likely made a
> mistake.
>
> In these cases, `data.frame()` is preferred to `as.data.frame()`:
>
> predict(mylogit, newdata = data.frame(x = 0), type = "response")
>
> or we can use a list, to save a few characters:
>
> predict(mylogit, newdata = list(x = 0), type = "response")
>
> which give:
>
> > predict(mylogit, newdata = list(x = 0), type = "response")
>       1
> 0.813069
> > predict(mylogit, newdata = data.frame(x = 0), type = "response")
>       1
> 0.813069
>
> In summary, use `data.frame()` or `list()` to create the object passed
> as `newdata` and make sure you give the component containing the new
> values a *name* that matches the predictor in the model formula.
>
> HTH
>
> G
>
> >
> > # I get:
> >
> > Warning message:
> > 'newdata' had 1 rows but variable(s) found have 34 rows
> >
> > # Why do I need 34 rows? So I try:
> >
> > v=vector()
> > v[1:34]=0
> > predict(mylogit,newdata=as.data.frame(v),type="response")
> >
> > # And I get a matrix of 34 values that appear to fit a logistic function,
> > instead of 0.85..
> >
> >
> >
> >
> > On Wed, Jan 26, 2011 at 6:59 PM, David Winsemius <dwinsem...@comcast.net
> >wrote:
> >
> > >
> > > On Jan 26, 2011, at 10:52 PM, Ahnate Lim wrote:
> > >
> > >  Dear R-help,
> > >>
> > >> I have fitted a glm logistic function to dichotomous forced choices
> > >> responses varying according to time interval between two stimulus. x
> > >> values
> > >> are time separation in miliseconds, and the y values are proportion
> > >> responses for one of the stimulus. Now I am trying to extrapolate x
> values
> > >> for the y value (proportion) at .25, .5, and .75. I have tried several
> > >> predict parameters, and they don't appear to be working. Is this
> correct
> > >> use
> > >> and understanding of the predict function? It would be nice to know
> the
> > >> parameters for the glm best fit, but all I really need are the
> > >> extrapolated
> > >> x values for those proportions. Thank you for your help. Here is the
> code:
> > >>
> > >> x <-
> > >> c(-283.9, -267.2, -250.5, -233.8, -217.1, -200.4, -183.7, -167,
> > >> -150.3, -133.6, -116.9, -100.2, -83.5, -66.8, -50.1, -33.4, -16.7,
> > >> 16.7, 33.4, 50.1, 66.8, 83.5, 100.2, 116.9, 133.6, 150.3, 167,
> > >> 183.7, 200.4, 217.1, 233.8, 250.5, 267.2, 283.9)
> > >>
> > >> y <-
> > >> c(0, 0.333333333333333, 0, 0, 0, 0, 0, 0, 0, 0.333333333333333,
> > >> 0, 0.133333333333333, 0.238095238095238, 0.527777777777778,
> > >> 0.566666666666667,
> > >> 0.845238095238095, 0.55, 1, 0.888888888888889, 1, 1, 1, 1, 1,
> > >> 1, 1, 1, 1, 1, 1, 1, 1, 1, 0.5)
> > >>
> > >> weight <-
> > >> c(1, 3, 2, 5, 4, 4, 3, 5, 5, 4, 5, 11, 22, 11, 15, 16, 11, 7,
> > >> 14, 10, 16, 19, 11, 5, 4, 5, 6, 9, 4, 2, 5, 5, 2, 2)
> > >>
> > >> mylogit <- glm(y~x,weights=weight, family = binomial)
> > >>
> > >> # now I try plotting the predicted value, and it looks like a good
> fit,
> > >> hopefully I can access what the glm is doing
> > >>
> > >> ypred <- predict(mylogit,newdata=as.data.frame(x),type="response")
> > >> plot(x, ypred,type="l")
> > >> points(x,y)
> > >>
> > >> # so I try to predict the x value when y (proportion) is at .5, but
> > >> something is wrong..
> > >>
> > >
> > > Right. Predict goes in the other direction ... x predicts y.
> > >
> > > Perhaps if you created a function of y w.r.t. x and then inverted it.
> > >
> > > ?approxfun  # and see the posting earlier this week "Inverse Prediction
> > > with splines" where it was demonstrated how to reverse the roles of x
> and y.
> > >
> > >>
> > >> predict(mylogit,newdata=as.data.frame(0.5))
> > >>
> > >>        [[alternative HTML version deleted]]
> > >>
> > >> ______________________________________________
> > >> R-help@r-project.org mailing list
> > >> https://stat.ethz.ch/mailman/listinfo/r-help
> > >> PLEASE do read the posting guide
> > >> http://www.R-project.org/posting-guide.html
> > >> and provide commented, minimal, self-contained, reproducible code.
> > >>
> > >
> > > David Winsemius, MD
> > > West Hartford, CT
> > >
> > >
> >
> >       [[alternative HTML version deleted]]
> >
> > ______________________________________________
> > R-help@r-project.org mailing list
> > https://stat.ethz.ch/mailman/listinfo/r-help
> > PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> > and provide commented, minimal, self-contained, reproducible code.
>
> --
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>  Dr. Gavin Simpson             [t] +44 (0)20 7679 0522
>  ECRC, UCL Geography,          [f] +44 (0)20 7679 0565
>  Pearson Building,             [e] gavin.simpsonATNOSPAMucl.ac.uk
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>
>

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