Displaying 6 results from an estimated 6 matches for "intercept1".
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intercept
2011 Jul 07
3
coefficients lm of data.frame
...655
9 0.08065333 0.22168589
10 0.25196536 0.84619914
11 -0.59536986 -0.08243074
12 1.09115054 0.49822977
I need to add two columns as result of the fitting of linear model
based on a preset numbers of row.
For example if I need to compute a lm each 4 rows, I get the
data.frame below, where intercept1 and coeff1 is obtained from V1 and
V2 of first 4 rows lm(V2 ~ V1), and so on...
V1 V2 "intercept" "coeff"
1 0.6931694 0.05797771 intercept1 coeff1
2 -1.4069786 0.23983307 intercept1 coeff1
3 -1.4901708 0.45079601 intercept1 coeff1
4 0.2215696...
2003 Jan 22
1
something wrong when using pspline in clogit?
...4*xvar)/(1+exp(0.5+0.4*xvar))))
library(Nnet)
mult.fit<-multinom(data.mult~xvar)
coef(mult.fit)
library(survival)
choice<-c(ifelse(data.mult==0,1,0),ifelse(data.mult==1,1,0),ifelse(data.mult==2,1,0),ifelse(data.mult==3,1,0))
temp<-list(time=2-choice,
status=choice,
intercept1=c(rep(0,length(data.mult)),rep(1,length(data.mult)),rep(0,length(data.mult)),rep(0,length(data.mult))),
intercept2=c(rep(0,length(data.mult)),rep(0,length(data.mult)),rep(1,length(data.mult)),rep(0,length(data.mult))),
intercept3=c(rep(0,length(data.mult)),rep(0,length(dat...
2007 Jun 09
1
How to plot vertical line
Hi,I have a result from polr which I fit a univariate variable (of ordinal data) with probit function. What I would like to do is to overlay the plot of my fitted values with the different intercept for each level in my ordinal data. I can do something like:lines(rep(intercept1, 1000), seq(from=0,to=max(fit),by=max(fit)/1000))where my intercept1 is, for example, the intercept that breaks between y=1 and y=2 labels and the max(fit) is the maximum of overall fitted values or maximum of all ordinal y labels. I'm wondering if there is better way to do this? If you could l...
2005 Jan 20
1
Windows Front end-crash error
...u<-c(100,150,200,250)
Sigma<-matrix(c(400,80,80,80,80,400,80,80,80,80,400,80,80,80,80,400),4,4
)
mu2<-c(0,0,0)
LE<-8^2 #Linking Error
Sigma2<-diag(LE,3)
sample.size<-5000
N<-100 #Number of datasets
#Take a single draw from VL distribution
vl.error<-mvrnorm(n=N, mu2, Sigma2)
intercept1 <- 0
slope1 <- 0
intercept2 <- 0
slope2 <- 0
for(i in 1:N){
temp <- data.frame(ID=seq(1:sample.size),mvrnorm(n=sample.size,
mu,Sigma))
temp$X5 <- temp$X1
temp$X6 <- temp$X2 + vl.error[i,1]
temp$X7 <- temp$X3 + vl.error[i,2]
temp$X8 <- temp$X4 + vl.error[i,3...
2017 Dec 20
1
Nonlinear regression
...>##The data
>ce <- c(15.17, 42.15, 69.12, 237.7, 419.77)
>qe <- c(17.65, 30.07, 65.36, 81.7, 90.2)
>
>##The linearized data
>celin <- 1/ce
>qelin <- 1/qe
>
>plot(ce, qe, xlim = xlim, ylim = ylim)
>
>##The linear model
>fit1 <- lm(qelin ~ celin)
>intercept1 <- fit1$coefficients[1]
>slope1 <- fit1$coeffecients[2]
>summary(fit1)
>
>Qmax <- 1/intercept1
>Kl <- .735011*Qmax
>
>xlim <- range(ce, celin)
>ylim <- range(qe, qelin)
>
>abline(lm(qelin ~ celin))
>
>c <- seq(min(ce), max(ce))
>q <- (Q...
2017 Dec 20
0
Nonlinear regression
...7, 419.77)
> >qe <- c(17.65, 30.07, 65.36, 81.7, 90.2)
> >
> >##The linearized data
> >celin <- 1/ce
> >qelin <- 1/qe
> >
> >plot(ce, qe, xlim = xlim, ylim = ylim)
> >
> >##The linear model
> >fit1 <- lm(qelin ~ celin)
> >intercept1 <- fit1$coefficients[1]
> >slope1 <- fit1$coeffecients[2]
> >summary(fit1)
> >
> >Qmax <- 1/intercept1
> >Kl <- .735011*Qmax
> >
> >xlim <- range(ce, celin)
> >ylim <- range(qe, qelin)
> >
> >abline(lm(qelin ~ celin))
>...