Displaying 3 results from an estimated 3 matches for "logmodel".
2011 Jan 10
1
debug biglm response error on bigglm model
...------------------')
browser()
print(i)
print(table(iter.df$bin_varname1))
print(table(iter.df$bin_age))
print(table(iter.df$bin_util))
print(table(iter.df$bin_varname2))
#~ debug(predict.nov.2011[i] <-
#~ sum(predict(logModel.1, newdata=iter.df,
type='response')))
}
predict.nov.2011[i] <-
sum(predict(logModel.1, newdata=iter.df, type='response'))
print(predict.nov.2011[i])
}
Output
==========
[1] 36.56073
[1] 561.4516
[1] 4.83483
[1] 5.01398
[1] 7.984146
[1] "---------...
2006 Mar 10
1
How to compare fit of linear and nonlinear models
Dear statistics experts,
I'm looking for a way to compare the fit of the following three models:
LinModel <- lm(y ~ x)
LogModel <- nls(y ~ SSlogis(x, Asym, xmid, scal))
PotModel <- nls(y ~ a * x^n, start=list(a=1, n=1))
I am only interested in whether one of these models has substantial advances in
explaining the variance of y. So my original idea was simply to compare the
adjusted R squared values. This however seem...
2011 Jan 07
0
Error in x %*% coef(object) : non-conformable arguments
...------------------')
browser()
print(i)
print(table(iter.df$bin_varname1))
print(table(iter.df$bin_age))
print(table(iter.df$bin_util))
print(table(iter.df$bin_varname2))
#~ debug(predict.nov.2011[i] <-
#~ sum(predict(logModel.1, newdata=iter.df,
type='response')))
}
predict.nov.2011[i] <-
sum(predict(logModel.1, newdata=iter.df, type='response'))
print(predict.nov.2011[i])
}
Output
==========
[1] 36.56073
[1] 561.4516
[1] 4.83483
[1] 5.01398
[1] 7.984146
[1] "--------...