Re: [R] Extrapolating values from a glm fit
Hi: Are you trying to find something analogous to the median dose? If so, look at function dose.p() in the MASS package and see if it does what you're looking for. It can take a vector of quantiles as input. HTH, Dennis On Wed, Jan 26, 2011 at 7:52 PM, Ahnate Lim ahn...@gmail.com 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, 0.333, 0, 0, 0, 0, 0, 0, 0, 0.333, 0, 0.133, 0.238095238095238, 0.528, 0.567, 0.845238095238095, 0.55, 1, 0.889, 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.. 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. [[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.
Re: [R] Extrapolating values from a glm fit
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.netwrote: 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, 0.333, 0, 0, 0, 0, 0, 0, 0, 0.333, 0, 0.133, 0.238095238095238, 0.528, 0.567, 0.845238095238095, 0.55, 1, 0.889, 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. -- %~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~% Dr. Gavin Simpson [t] +44 (0)20 7679 0522 ECRC, UCL Geography, [f] +44 (0)20 7679 0565 Pearson Building, [e]
Re: [R] Extrapolating values from a glm fit
On Thu, 2011-01-27 at 00:10 -1000, Ahnate Lim wrote: 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? This is the predicted value(s) on the scale of the linear predictor, or the link function, depending on terminology, hence link. Recall that in the GLM the response and the linear predictor are related through a link function: g(y) = eta so for your model logit(y) = eta where eta is the linear predictor: beta_0 + beta_1 * x (in your example). beta_0 + beta_1 * x gives us the fitted value but on the untransformed scale. This is the value given by predict() when type = link is used. To get the predicted value on the response scale, we apply the inverse of the link function g(): y = g(eta)^-1 The inverse of the logit function is the inverse-logit: logit(eta)^-1 = exp(eta) / (1 + exp(eta)) So in R, we can see the relationship through a few simple commands. First, the prediction for x = 0.5 on the response scale: predict(mylogit, newdata = list(x = 0.5), type = response) 1 0.8149848 Then we compute the same prediction but on the scale of the link function: (p1 - predict(mylogit, newdata = list(x = 0.5), type = link)) 1 1.482732 to which we apply the inverse-logit function giving us the same value we got earlier for type = response: exp(p1) / (1 + exp(p1)) 1 0.8149848 There is a function to do this for us, however, called plogis() plogis(p1) 1 0.8149848 One reason why you might want predictions on the scale of the link function is for computation of confidence intervals using normal theory (e.g. 2-sigma ~95% confidence intervals). On the response scale these should be asymmetric and respect the scale of the response (bounding etc), so you compute them on the link scale and apply the inverse of the link function to get them on to the response scale. HTH G 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.netwrote: On Jan 26, 2011, at 10:52 PM,
Re: [R] Extrapolating values from a glm fit
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.ukwrote: 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, 0.333, 0, 0, 0, 0, 0, 0, 0, 0.333, 0, 0.133, 0.238095238095238, 0.528, 0.567, 0.845238095238095, 0.55, 1, 0.889, 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
[R] Extrapolating values from a glm fit
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, 0.333, 0, 0, 0, 0, 0, 0, 0, 0.333, 0, 0.133, 0.238095238095238, 0.528, 0.567, 0.845238095238095, 0.55, 1, 0.889, 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.. 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.
Re: [R] Extrapolating values from a glm fit
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, 0.333, 0, 0, 0, 0, 0, 0, 0, 0.333, 0, 0.133, 0.238095238095238, 0.528, 0.567, 0.845238095238095, 0.55, 1, 0.889, 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 __ 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.
Re: [R] Extrapolating values from a glm fit
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) # 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.netwrote: 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, 0.333, 0, 0, 0, 0, 0, 0, 0, 0.333, 0, 0.133, 0.238095238095238, 0.528, 0.567, 0.845238095238095, 0.55, 1, 0.889, 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.
