tag:blogger.com,1999:blog-1699790186532417278.post5861820982748112323..comments2023-06-17T07:40:37.867+02:00Comments on theBioBucket*: Fit Sigmoid Curve with Confidence IntervalsUnknownnoreply@blogger.comBlogger8125tag:blogger.com,1999:blog-1699790186532417278.post-57704249000728472842015-01-14T17:36:53.162+01:002015-01-14T17:36:53.162+01:00Division by zero!! See http://stackoverflow.com/qu...Division by zero!! See http://stackoverflow.com/questions/19544942/how-to-get-rid-of-error-with-multiple-regression-in-rKayhttps://www.blogger.com/profile/09320307979146470105noreply@blogger.comtag:blogger.com,1999:blog-1699790186532417278.post-19128544405569532152015-01-14T16:08:15.809+01:002015-01-14T16:08:15.809+01:00Hi
I hope this tread is still active. I get the i...Hi<br /><br />I hope this tread is still active. I get the issue: "Error in lm.fit(x, y, offset = offset, singular.ok = singular.ok, ...) : <br /> NA/NaN/Inf in 'x'"<br /><br />if I only have 0's associated with a particular x value, anyway to get round this?<br />Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-1699790186532417278.post-52925594956895205102014-01-17T17:39:06.193+01:002014-01-17T17:39:06.193+01:00OK - many thanks Kay - i'll try on R-help
chee...OK - many thanks Kay - i'll try on R-help<br />cheersAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-1699790186532417278.post-2031613608646183602014-01-15T17:35:49.794+01:002014-01-15T17:35:49.794+01:00This is old code - there might be some issues with...This is old code - there might be some issues with package updates! Please post to R-Help or suchlike with your question and example code!Kayhttps://www.blogger.com/profile/09320307979146470105noreply@blogger.comtag:blogger.com,1999:blog-1699790186532417278.post-70621355605600770712014-01-15T17:01:11.476+01:002014-01-15T17:01:11.476+01:00Hi
i found very useful this script - which runs ve...Hi<br />i found very useful this script - which runs very well<br /><br />i would like to apply it with an analysis involving SSfpl<br /><br />here is the first part of my code<br /><br />##<br />summary(mod1<-nls(DTest ~ SSfpl(ltc, a,b,c,d),data=dataMRr))<br />ltc_x <- seq(71, 224, length = 100)<br />predict(mod1, data.frame(ltc = ltc_x))<br />with(dataMRr, plot(ltc,DTest,xlim=c(70,230),ylim=c(0,40)), cex=0.2)<br />lines(ltc_x, predict(mod1, data.frame(ltc = ltc_x)), lwd=2)<br />se.fit <- sqrt(apply(attr(predict(mod1, data.frame(ltc = ltc_x)),"gradient"),1,<br />function(x) sum(vcov(mod1)*outer(x,x))))<br />##<br /><br />which works well without any error message<br /><br />but when i run<br /><br />##<br />matplot(ltc_x, data.frame(ltc = ltc_x))+outer(se.fit,qnorm(c(.5, .025,.975)), type="l", lty=c(1,3,3))<br />##<br /><br />i have the following message:<br />##<br />Error in outer(se.fit, qnorm(c(0.5, 0.025, 0.975)), type = "l", lty = c(1, : <br /> using ... with FUN = "*" is an error<br />##<br /><br />Any help would be greatly appreciated<br />Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-1699790186532417278.post-40931478636508431412012-04-09T09:47:23.111+02:002012-04-09T09:47:23.111+02:00Its all there, see line 54 and 60!Its all there, see line 54 and 60!Kayhttps://www.blogger.com/profile/09320307979146470105noreply@blogger.comtag:blogger.com,1999:blog-1699790186532417278.post-59111132326551690172012-04-09T08:45:12.821+02:002012-04-09T08:45:12.821+02:00Thank you very much. But I have two question:
1) ...Thank you very much. But I have two question:<br /><br />1) I just put my data and after "summary" appear this error "Error in nls(y ~ 1/(1 + exp((xmid - x)/scal)), data = xy, start = list(xmid = aux[1L], : <br /> increasing factor 0.000488281 below 'minFactor' 0.000976562".....what is the meaning? I can't apply my data? Also with other kind of equation that are present in ?nls, it doesn't work. But if I exchange the data of x with y the equation works. What is the meaning?<br /><br />2) If I want to find x from y in this script there is the equation "x_tenth=exp((xmid-log(Asym/y_tenth-1)*scal)). I don't understand the meaning of y_tenth. I mean if I put one value of waterpot (25 for example) I can find x with this equation? x=exp((xmid-log(Asym/25-1)*scal))<br /><br />I'm sorry for all this questions, but I lost a lot of time without results<br /><br />Thank you in advanceAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-1699790186532417278.post-11605703440907950472012-04-02T21:22:23.497+02:002012-04-02T21:22:23.497+02:00Hi
good job.. Can you show me the sigmoidal equat...Hi<br /><br />good job.. Can you show me the sigmoidal equation. I'm going to apply this model to my data<br /><br /><br />thank you in advancedAnonymousnoreply@blogger.com