Re: [R] User defined function within a formula
Thanks Bill for your quick reply.
I tried your solution and it did work for the simple user defined function
xploly. But when I try with other function, it gave me error again:
OPoly-function(x,degree=1,weight=1){
weight=round(weight,0)# weight need to be integer
if(length(weight)!=length(x))weight=rep(1,length(x))
p=poly(4*(rep(x,weight)-mean(range(x)))/diff(range(x)),degree)
Z-(t(t(p[cumsum(weight),])*sqrt(attr(p,coefs)$norm2[-seq(2)]))[,degree])
class(Z)-OPoly;Z
}
##this OPoly is an FORTRAN orthogonal polynomial routine, it first maps the
x to range[-2,2] then do QR, then return the results with sqrt(norm2).
Comparing with poly, this transformation will make the model coefficients
within a similar range as other variables, the R poly routine will usually
give you a very large coefficients. I did not find such routine in R, so I
have to define this as user defined function.
###
I also have following function as you suggested:
makepredictcall.OPoly-function(var,call)
{
if (is.call(call)) {
if (identical(call[[1]], quote(OPoly))) {
if (!is.null(tmp - attr(var, coefs))) {
call$coefs - tmp
}
}
}
call
}
But I still got error for following:
g3=glm(lot1 ~ log(u) + OPoly(u,1), data = clotting, family = Gamma)
predict(g3,dc)Error in poly(4 * (rep(x, weight) -
mean(range(x)))/diff(range(x)), degree) :
missing values are not allowed in 'poly'
I thought it might be due to the /diff(range(x) in the function. But even
I remove that part, it will still give me error. Any idea?
Many thanks in advance.
Alex
On Thu, Jul 16, 2015 at 2:09 PM, William Dunlap wdun...@tibco.com wrote:
Read about the 'makepredictcall' generic function. There is a method,
makepredictcall.poly(), for poly() that attaches the polynomial
coefficients
used during the fitting procedure to the call to poly() that predict()
makes.
You ought to supply a similar method for your xpoly(), and xpoly() needs
to return an object of a a new class that will cause that method to be used.
E.g.,
xpoly - function(x,degree=1,...){ ret - poly(x,degree=degree,...);
class(ret) - xpoly ; ret }
makepredictcall.xpoly - function (var, call)
{
if (is.call(call)) {
if (identical(call[[1]], quote(xpoly))) {
if (!is.null(tmp - attr(var, coefs))) {
call$coefs - tmp
}
}
}
call
}
g2 - glm(lot1 ~ log(u) + xpoly(u,1), data = clotting, family = Gamma)
predict(g2,dc)
# 1 2 3 4
5
#-0.01398928608 -0.01398928608 -0.01398928608 -0.01398928608
#-0.01398928608
# 6 7 8 9
#-0.01398928608 -0.01398928608 -0.01398928608 -0.01398928608
You can see the effects of makepredictcall() in the 'terms' component
of glm's output. The 'variables' attribute of it gives the original
function
calls and the 'predvars' attribute gives the calls to be used for
prediction:
attr(g2$terms, variables)
list(lot1, log(u), xpoly(u, 1))
attr(g2$terms, predvars)
list(lot1, log(u), xpoly(u, 1, coefs = list(alpha = 40, norm2 = c(1,
9, 8850
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Thu, Jul 16, 2015 at 12:35 PM, Kunshan Yin yinkuns...@gmail.com
wrote:
Hello, I have a question about the formula and the user defined function:
I can do following:
###Case 1:
clotting - data.frame(
+ u = c(5,10,15,20,30,40,60,80,100),
+ lot1 = c(118,58,42,35,27,25,21,19,18),
+ lot2 = c(69,35,26,21,18,16,13,12,12))
g1=glm(lot1 ~ log(u) + poly(u,1), data = clotting, family = Gamma)
dc=clotting
dc$u=1
predict(g1,dc)
1 2 3 4 5
6 7 8 9
-0.01398929 -0.01398929 -0.01398929 -0.01398929 -0.01398929 -0.01398929
-0.01398929 -0.01398929 -0.01398929
However, if I just simply wrap the poly as a user defined function ( in
reality I would have my own more complex function) then I will get error:
###Case 2:
xpoly-function(x,degree=1){poly(x,degree)}
g2=glm(lot1 ~ log(u) + xpoly(u,1), data = clotting, family = Gamma)
predict(g2,dc)
Error in poly(x, degree) :
'degree' must be less than number of unique points
It seems that the predict always treat the user defined function in the
formula with I(). My question is how can I get the results for Case2
same
as case1?
Anyone can have any idea about this?
Thank you very much.
Alex
[[alternative HTML version deleted]]
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R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
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PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[[alternative HTML version deleted]]
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Re: [R] User defined function within a formula
Thank you very much. It worked. I think I need to digest this further
later. Thanks again for the help.
On Thu, Jul 16, 2015 at 4:51 PM, William Dunlap wdun...@tibco.com wrote:
This might do what you want:
OPoly - function(x, degree=1, weight=1, coefs=NULL, rangeX=NULL){
weight - round(weight,0)# weight need to be integer
if(length(weight)!=length(x)) {
weight - rep(1,length(x))
}
if (is.null(rangeX)) {
rangeX - range(x)
}
p - poly(4*(rep(x,weight)-mean(rangeX))/diff(rangeX), degree=degree,
coefs=coefs)
# why t(t(...))? That strips the attributes.
Z - t( t(p[cumsum(weight),]) * sqrt(attr(p,coefs)$norm2[-seq(2)]) )[,
degree, drop=FALSE]
class(Z) - OPoly
attr(Z, coefs) - attr(p, coefs)
attr(Z, rangeX) - rangeX
Z
}
makepredictcall.OPoly-function(var,call)
{
if (is.call(call)) {
if (identical(call[[1]], quote(OPoly))) {
if (!is.null(tmp - attr(var, coefs))) {
call$coefs - tmp
}
if (!is.null(tmp - attr(var, rangeX))) {
call$rangeX - tmp
}
call$weight - 1 # weight not relevant in predictions
}
}
call
}
d - data.frame(Y=1:8, X=log(1:8), Weight=1:8)
fit - lm(data=d, Y ~ OPoly(X, degree=2, weight=Weight))
predict(fit)[c(3,8)]
predict(fit, newdata=data.frame(X=d$X[c(3,8)])) # same result
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Thu, Jul 16, 2015 at 4:39 PM, William Dunlap wdun...@tibco.com wrote:
OPoly-function(x,degree=1,weight=1){
weight=round(weight,0)# weight need to be integer
if(length(weight)!=length(x))weight=rep(1,length(x))
p=poly(4*(rep(x,weight)-mean(range(x)))/diff(range(x)),degree)
Z-(t(t(p[cumsum(weight),])*sqrt(attr(p,coefs)$norm2[-
seq(2)]))[,degree])
class(Z)-OPoly;Z
}
You need to make OPoly to have optional argument(s) that give
the original-regressor-dependent information to OPoly and then
have it return, as attributes, the value of those arguments.
makepredictcall
will take the attributes and attach them to the call in predvars so
predict uses values derived from the original regressors, not value
derived
from the data to be predicted from.
Take a look at a pair like makepredictcall.scale() and scale() for an
example:
scale has optional arguments 'center' and 'scale' that it returns as
attributes
and makepredictcall.scale adds those to the call to scale that it is
given.
Thus when you predict, the scale and center arguments come from the
original data, not from the data you are predicting from.
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Thu, Jul 16, 2015 at 3:43 PM, Kunshan Yin yinkuns...@gmail.com
wrote:
Thanks Bill for your quick reply.
I tried your solution and it did work for the simple user defined
function xploly. But when I try with other function, it gave me error again:
OPoly-function(x,degree=1,weight=1){
weight=round(weight,0)# weight need to be integer
if(length(weight)!=length(x))weight=rep(1,length(x))
p=poly(4*(rep(x,weight)-mean(range(x)))/diff(range(x)),degree)
Z-(t(t(p[cumsum(weight),])*sqrt(attr(p,coefs)$norm2[-seq(2)]))[,degree])
class(Z)-OPoly;Z
}
##this OPoly is an FORTRAN orthogonal polynomial routine, it first maps
the x to range[-2,2] then do QR, then return the results with sqrt(norm2).
Comparing with poly, this transformation will make the model coefficients
within a similar range as other variables, the R poly routine will usually
give you a very large coefficients. I did not find such routine in R, so I
have to define this as user defined function.
###
I also have following function as you suggested:
makepredictcall.OPoly-function(var,call)
{
if (is.call(call)) {
if (identical(call[[1]], quote(OPoly))) {
if (!is.null(tmp - attr(var, coefs))) {
call$coefs - tmp
}
}
}
call
}
But I still got error for following:
g3=glm(lot1 ~ log(u) + OPoly(u,1), data = clotting, family = Gamma)
predict(g3,dc)Error in poly(4 * (rep(x, weight) -
mean(range(x)))/diff(range(x)), degree) :
missing values are not allowed in 'poly'
I thought it might be due to the /diff(range(x) in the function. But
even I remove that part, it will still give me error. Any idea?
Many thanks in advance.
Alex
On Thu, Jul 16, 2015 at 2:09 PM, William Dunlap wdun...@tibco.com
wrote:
Read about the 'makepredictcall' generic function. There is a method,
makepredictcall.poly(), for poly() that attaches the polynomial
coefficients
used during the fitting procedure to the call to poly() that predict()
makes.
You ought to supply a similar method for your xpoly(), and xpoly()
needs to return an object of a a new class that will cause that method to
be used.
E.g.,
xpoly - function(x,degree=1,...){ ret - poly(x,degree=degree,...);
class(ret) - xpoly ; ret }
makepredictcall.xpoly - function (var, call)
{
if (is.call(call)) {
if (identical(call[[1]],
Re: [R] User defined function within a formula
This might do what you want:
OPoly - function(x, degree=1, weight=1, coefs=NULL, rangeX=NULL){
weight - round(weight,0)# weight need to be integer
if(length(weight)!=length(x)) {
weight - rep(1,length(x))
}
if (is.null(rangeX)) {
rangeX - range(x)
}
p - poly(4*(rep(x,weight)-mean(rangeX))/diff(rangeX), degree=degree,
coefs=coefs)
# why t(t(...))? That strips the attributes.
Z - t( t(p[cumsum(weight),]) * sqrt(attr(p,coefs)$norm2[-seq(2)]) )[,
degree, drop=FALSE]
class(Z) - OPoly
attr(Z, coefs) - attr(p, coefs)
attr(Z, rangeX) - rangeX
Z
}
makepredictcall.OPoly-function(var,call)
{
if (is.call(call)) {
if (identical(call[[1]], quote(OPoly))) {
if (!is.null(tmp - attr(var, coefs))) {
call$coefs - tmp
}
if (!is.null(tmp - attr(var, rangeX))) {
call$rangeX - tmp
}
call$weight - 1 # weight not relevant in predictions
}
}
call
}
d - data.frame(Y=1:8, X=log(1:8), Weight=1:8)
fit - lm(data=d, Y ~ OPoly(X, degree=2, weight=Weight))
predict(fit)[c(3,8)]
predict(fit, newdata=data.frame(X=d$X[c(3,8)])) # same result
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Thu, Jul 16, 2015 at 4:39 PM, William Dunlap wdun...@tibco.com wrote:
OPoly-function(x,degree=1,weight=1){
weight=round(weight,0)# weight need to be integer
if(length(weight)!=length(x))weight=rep(1,length(x))
p=poly(4*(rep(x,weight)-mean(range(x)))/diff(range(x)),degree)
Z-(t(t(p[cumsum(weight),])*sqrt(attr(p,coefs)$norm2[-
seq(2)]))[,degree])
class(Z)-OPoly;Z
}
You need to make OPoly to have optional argument(s) that give
the original-regressor-dependent information to OPoly and then
have it return, as attributes, the value of those arguments.
makepredictcall
will take the attributes and attach them to the call in predvars so
predict uses values derived from the original regressors, not value derived
from the data to be predicted from.
Take a look at a pair like makepredictcall.scale() and scale() for an
example:
scale has optional arguments 'center' and 'scale' that it returns as
attributes
and makepredictcall.scale adds those to the call to scale that it is given.
Thus when you predict, the scale and center arguments come from the
original data, not from the data you are predicting from.
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Thu, Jul 16, 2015 at 3:43 PM, Kunshan Yin yinkuns...@gmail.com wrote:
Thanks Bill for your quick reply.
I tried your solution and it did work for the simple user defined
function xploly. But when I try with other function, it gave me error again:
OPoly-function(x,degree=1,weight=1){
weight=round(weight,0)# weight need to be integer
if(length(weight)!=length(x))weight=rep(1,length(x))
p=poly(4*(rep(x,weight)-mean(range(x)))/diff(range(x)),degree)
Z-(t(t(p[cumsum(weight),])*sqrt(attr(p,coefs)$norm2[-seq(2)]))[,degree])
class(Z)-OPoly;Z
}
##this OPoly is an FORTRAN orthogonal polynomial routine, it first maps
the x to range[-2,2] then do QR, then return the results with sqrt(norm2).
Comparing with poly, this transformation will make the model coefficients
within a similar range as other variables, the R poly routine will usually
give you a very large coefficients. I did not find such routine in R, so I
have to define this as user defined function.
###
I also have following function as you suggested:
makepredictcall.OPoly-function(var,call)
{
if (is.call(call)) {
if (identical(call[[1]], quote(OPoly))) {
if (!is.null(tmp - attr(var, coefs))) {
call$coefs - tmp
}
}
}
call
}
But I still got error for following:
g3=glm(lot1 ~ log(u) + OPoly(u,1), data = clotting, family = Gamma)
predict(g3,dc)Error in poly(4 * (rep(x, weight) -
mean(range(x)))/diff(range(x)), degree) :
missing values are not allowed in 'poly'
I thought it might be due to the /diff(range(x) in the function. But
even I remove that part, it will still give me error. Any idea?
Many thanks in advance.
Alex
On Thu, Jul 16, 2015 at 2:09 PM, William Dunlap wdun...@tibco.com
wrote:
Read about the 'makepredictcall' generic function. There is a method,
makepredictcall.poly(), for poly() that attaches the polynomial
coefficients
used during the fitting procedure to the call to poly() that predict()
makes.
You ought to supply a similar method for your xpoly(), and xpoly() needs
to return an object of a a new class that will cause that method to be used.
E.g.,
xpoly - function(x,degree=1,...){ ret - poly(x,degree=degree,...);
class(ret) - xpoly ; ret }
makepredictcall.xpoly - function (var, call)
{
if (is.call(call)) {
if (identical(call[[1]], quote(xpoly))) {
if (!is.null(tmp - attr(var, coefs))) {
call$coefs - tmp
}
}
}
call
}
g2 - glm(lot1 ~ log(u) + xpoly(u,1), data = clotting, family = Gamma)
Re: [R] User defined function within a formula
Read about the 'makepredictcall' generic function. There is a method,
makepredictcall.poly(), for poly() that attaches the polynomial coefficients
used during the fitting procedure to the call to poly() that predict()
makes.
You ought to supply a similar method for your xpoly(), and xpoly() needs to
return an object of a a new class that will cause that method to be used.
E.g.,
xpoly - function(x,degree=1,...){ ret - poly(x,degree=degree,...);
class(ret) - xpoly ; ret }
makepredictcall.xpoly - function (var, call)
{
if (is.call(call)) {
if (identical(call[[1]], quote(xpoly))) {
if (!is.null(tmp - attr(var, coefs))) {
call$coefs - tmp
}
}
}
call
}
g2 - glm(lot1 ~ log(u) + xpoly(u,1), data = clotting, family = Gamma)
predict(g2,dc)
# 1 2 3 4 5
#-0.01398928608 -0.01398928608 -0.01398928608 -0.01398928608
#-0.01398928608
# 6 7 8 9
#-0.01398928608 -0.01398928608 -0.01398928608 -0.01398928608
You can see the effects of makepredictcall() in the 'terms' component
of glm's output. The 'variables' attribute of it gives the original
function
calls and the 'predvars' attribute gives the calls to be used for
prediction:
attr(g2$terms, variables)
list(lot1, log(u), xpoly(u, 1))
attr(g2$terms, predvars)
list(lot1, log(u), xpoly(u, 1, coefs = list(alpha = 40, norm2 = c(1,
9, 8850
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Thu, Jul 16, 2015 at 12:35 PM, Kunshan Yin yinkuns...@gmail.com wrote:
Hello, I have a question about the formula and the user defined function:
I can do following:
###Case 1:
clotting - data.frame(
+ u = c(5,10,15,20,30,40,60,80,100),
+ lot1 = c(118,58,42,35,27,25,21,19,18),
+ lot2 = c(69,35,26,21,18,16,13,12,12))
g1=glm(lot1 ~ log(u) + poly(u,1), data = clotting, family = Gamma)
dc=clotting
dc$u=1
predict(g1,dc)
1 2 3 4 5
6 7 8 9
-0.01398929 -0.01398929 -0.01398929 -0.01398929 -0.01398929 -0.01398929
-0.01398929 -0.01398929 -0.01398929
However, if I just simply wrap the poly as a user defined function ( in
reality I would have my own more complex function) then I will get error:
###Case 2:
xpoly-function(x,degree=1){poly(x,degree)}
g2=glm(lot1 ~ log(u) + xpoly(u,1), data = clotting, family = Gamma)
predict(g2,dc)
Error in poly(x, degree) :
'degree' must be less than number of unique points
It seems that the predict always treat the user defined function in the
formula with I(). My question is how can I get the results for Case2 same
as case1?
Anyone can have any idea about this?
Thank you very much.
Alex
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] User defined function within a formula
Hello, I have a question about the formula and the user defined function:
I can do following:
###Case 1:
clotting - data.frame(
+ u = c(5,10,15,20,30,40,60,80,100),
+ lot1 = c(118,58,42,35,27,25,21,19,18),
+ lot2 = c(69,35,26,21,18,16,13,12,12))
g1=glm(lot1 ~ log(u) + poly(u,1), data = clotting, family = Gamma)
dc=clotting
dc$u=1
predict(g1,dc)
1 2 3 4 5
6 7 8 9
-0.01398929 -0.01398929 -0.01398929 -0.01398929 -0.01398929 -0.01398929
-0.01398929 -0.01398929 -0.01398929
However, if I just simply wrap the poly as a user defined function ( in
reality I would have my own more complex function) then I will get error:
###Case 2:
xpoly-function(x,degree=1){poly(x,degree)}
g2=glm(lot1 ~ log(u) + xpoly(u,1), data = clotting, family = Gamma)
predict(g2,dc)
Error in poly(x, degree) :
'degree' must be less than number of unique points
It seems that the predict always treat the user defined function in the
formula with I(). My question is how can I get the results for Case2 same
as case1?
Anyone can have any idea about this?
Thank you very much.
Alex
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] User defined function within a formula
OPoly-function(x,degree=1,weight=1){
weight=round(weight,0)# weight need to be integer
if(length(weight)!=length(x))weight=rep(1,length(x))
p=poly(4*(rep(x,weight)-mean(range(x)))/diff(range(x)),degree)
Z-(t(t(p[cumsum(weight),])*sqrt(attr(p,coefs)$norm2[-
seq(2)]))[,degree])
class(Z)-OPoly;Z
}
You need to make OPoly to have optional argument(s) that give
the original-regressor-dependent information to OPoly and then
have it return, as attributes, the value of those arguments.
makepredictcall
will take the attributes and attach them to the call in predvars so
predict uses values derived from the original regressors, not value derived
from the data to be predicted from.
Take a look at a pair like makepredictcall.scale() and scale() for an
example:
scale has optional arguments 'center' and 'scale' that it returns as
attributes
and makepredictcall.scale adds those to the call to scale that it is given.
Thus when you predict, the scale and center arguments come from the
original data, not from the data you are predicting from.
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Thu, Jul 16, 2015 at 3:43 PM, Kunshan Yin yinkuns...@gmail.com wrote:
Thanks Bill for your quick reply.
I tried your solution and it did work for the simple user defined function
xploly. But when I try with other function, it gave me error again:
OPoly-function(x,degree=1,weight=1){
weight=round(weight,0)# weight need to be integer
if(length(weight)!=length(x))weight=rep(1,length(x))
p=poly(4*(rep(x,weight)-mean(range(x)))/diff(range(x)),degree)
Z-(t(t(p[cumsum(weight),])*sqrt(attr(p,coefs)$norm2[-seq(2)]))[,degree])
class(Z)-OPoly;Z
}
##this OPoly is an FORTRAN orthogonal polynomial routine, it first maps
the x to range[-2,2] then do QR, then return the results with sqrt(norm2).
Comparing with poly, this transformation will make the model coefficients
within a similar range as other variables, the R poly routine will usually
give you a very large coefficients. I did not find such routine in R, so I
have to define this as user defined function.
###
I also have following function as you suggested:
makepredictcall.OPoly-function(var,call)
{
if (is.call(call)) {
if (identical(call[[1]], quote(OPoly))) {
if (!is.null(tmp - attr(var, coefs))) {
call$coefs - tmp
}
}
}
call
}
But I still got error for following:
g3=glm(lot1 ~ log(u) + OPoly(u,1), data = clotting, family = Gamma)
predict(g3,dc)Error in poly(4 * (rep(x, weight) -
mean(range(x)))/diff(range(x)), degree) :
missing values are not allowed in 'poly'
I thought it might be due to the /diff(range(x) in the function. But even
I remove that part, it will still give me error. Any idea?
Many thanks in advance.
Alex
On Thu, Jul 16, 2015 at 2:09 PM, William Dunlap wdun...@tibco.com wrote:
Read about the 'makepredictcall' generic function. There is a method,
makepredictcall.poly(), for poly() that attaches the polynomial
coefficients
used during the fitting procedure to the call to poly() that predict()
makes.
You ought to supply a similar method for your xpoly(), and xpoly() needs
to return an object of a a new class that will cause that method to be used.
E.g.,
xpoly - function(x,degree=1,...){ ret - poly(x,degree=degree,...);
class(ret) - xpoly ; ret }
makepredictcall.xpoly - function (var, call)
{
if (is.call(call)) {
if (identical(call[[1]], quote(xpoly))) {
if (!is.null(tmp - attr(var, coefs))) {
call$coefs - tmp
}
}
}
call
}
g2 - glm(lot1 ~ log(u) + xpoly(u,1), data = clotting, family = Gamma)
predict(g2,dc)
# 1 2 3 4
5
#-0.01398928608 -0.01398928608 -0.01398928608 -0.01398928608
#-0.01398928608
# 6 7 8 9
#-0.01398928608 -0.01398928608 -0.01398928608 -0.01398928608
You can see the effects of makepredictcall() in the 'terms' component
of glm's output. The 'variables' attribute of it gives the original
function
calls and the 'predvars' attribute gives the calls to be used for
prediction:
attr(g2$terms, variables)
list(lot1, log(u), xpoly(u, 1))
attr(g2$terms, predvars)
list(lot1, log(u), xpoly(u, 1, coefs = list(alpha = 40, norm2 = c(1,
9, 8850
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Thu, Jul 16, 2015 at 12:35 PM, Kunshan Yin yinkuns...@gmail.com
wrote:
Hello, I have a question about the formula and the user defined function:
I can do following:
###Case 1:
clotting - data.frame(
+ u = c(5,10,15,20,30,40,60,80,100),
+ lot1 = c(118,58,42,35,27,25,21,19,18),
+ lot2 = c(69,35,26,21,18,16,13,12,12))
g1=glm(lot1 ~ log(u) + poly(u,1), data = clotting, family = Gamma)
dc=clotting
dc$u=1
predict(g1,dc)
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