Displaying 2 results from an estimated 2 matches for "press_x".
2010 Mar 29
1
getting CI's for certain y of nls fitted curve
...235.7, 120.4, 69.4,
318.3, 271.6, 97.5, 59.3, 476.5, 204.8, 49.5, 41.2)), .Names = c("run",
"press", "tr_rel", "tr"), row.names = c(NA, -12L), class = "data.frame")))
summary(mod1<-nls(tr ~ SSlogis( log(press), Asym, xmid, scal),data=por))
press_x <- seq(10, 40, length = 100)
predict(mod1, data.frame(press = press_x))
with(por, plot(press,tr,xlim=c(10,35),ylim=c(0,500)))
lines(press_x, predict(mod1, data.frame(press = press_x)))
###http://finzi.psych.upenn.edu/R/Rhelp02/archive/42932.html
###calc. of conf. intervalls, by linear approx...
2010 Mar 25
1
nls, predict() problem
...69.4,
318.3, 271.6, 97.5, 59.3, 476.5, 204.8, 49.5, 41.2)), .Names = c("run",
"press", "tr_rel", "tr"), row.names = c(NA, -12L), class = "data.frame")))
attach(por)
###nls:
summary(mod1<-nls(tr ~ SSlogis( log(press), Asym, xmid, scal)))
press_x <- seq(15, 30, length = 100)
####predict() somehow doesn't recognize the list and returns values
####as if list argument was absent:
predict(mod1,list(Pressure=press_x))
####consequently plotting of fitted line doesn't work either:
plot(press,tr)
lines(press_x,predict(mod1,list(pres...