Displaying 4 results from an estimated 4 matches for "fmodel".
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2010 Feb 11
2
Question about rank() function
...am doing wrong. Please help.
I ran the following commands:
data = read.table("test1.csv", head=T, as.is=T, na.string=".", row.nam=NULL)
X1 = as.factor(data[[3]])
X2 = as.factor(data[[4]])
X3 = as.factor(data[[5]])
Y = data[[2]]
model = lm(Y ~ X1*X2*X3, na.action = na.exclude)
fmodel = fitted(model)
fmodel
(First line is shown below.....)
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9 10 11 12 13 14 15 16
17
180.3763 180.3763 180.3763 180.3763 180.4546 180.3763 177.9245 177.9245
181.3859 180.3763 NA...
2009 Aug 21
1
Possible bug with lrm.fit in Design Library
...thought I'd try lrm.fit to see if the results where better.
Since the model WAS created by a function of the Design library, I can't
see how it would then be rejected.
Below is the session transcript:
--------------------------------------------------------------------------------
> fmodel <- lrm.fit(cbind(trainlogdata$v1, trainlogdata$v2),
trainlogdata$label)
> fmodel
Logistic Regression Model
lrm.fit(x = cbind(trainlogdata$v1, trainlogdata$v2), y = trainlogdata$label)
Frequencies of Responses
0 1
20988 6684
Obs Max Deriv Model L.R. d.f....
2009 Aug 21
1
Repost - Possible bug with lrm.fit in Design Library
...thought I'd try lrm.fit to see if the results where better.
Since the model WAS created by a function of the Design library, I can't
see how it would then be rejected.
Below is the session transcript:
--------------------------------------------------------------------------------
> fmodel <- lrm.fit(cbind(trainlogdata$v1, trainlogdata$v2),
trainlogdata$label)
> fmodel
Logistic Regression Model
lrm.fit(x = cbind(trainlogdata$v1, trainlogdata$v2), y = trainlogdata$label)
Frequencies of Responses
0 1
20988 6684
Obs Max Deriv Model L.R. d.f....
2010 May 19
0
Piecewise nls w/ boundary as a fitting parameter
...x <- 0:8
y <- c(0.5, 0.5, 0.5, 0.5, 1.914214, 2.5, 2.949490, 3.328427, 3.662278)
plot(x, y)
# Neither of these seem to work
f <- function(x, x0, y0, k) { ifelse(x < x0, y0, y0 + sqrt(k*(x-x0))) }
# f <- function(x, x0, y0, k) { (x < x0)*y0 + (x >= x0)*(y0 + sqrt(k*(x-x0))) }
fmodel <- nls(y ~ f(x, x0, y0, k), start = list(x0 = 1, y0 = 1, k = 1))
What's the proper way to deal with such a case? Am I supposed to
somehow obtain the break point separately, and then fit only y0 and k?
What is the preferred method for obtaining break point (I am not even
sure if I am using t...