Hello!
I have parameter estimates for a generalized linear model and would like
to produce fitted values i.e. fitted(). This can be easily done in R,
but my problem lies in fact that I have a vector of parameters from some
other software, that is is not constrained i.e. I have the following
estimates for model with one factor with 4 levels
beta = c(intercept group1 group2 group3 group4)
where group1:4 are estimated deviations from intercept i.e. sum to zero
contraint, but all parameter estimates are there! How can I create a
model matrix that will not contain any constraints since I would like to
compute
model.matrix("some_formula") %*% beta
I.e. I would like to have model.matrix of the form
1 1 0 0 0
1 0 1 0 0
1 0 0 1 0
1 0 0 0 1
and not of the following form with contr.treatment or any other contraints
1 0 0 0
1 1 0 0
1 0 1 0
1 0 0 1
I could remove group4 from beta and use sum to zero contraint for
contrast in fomula, but I would like to overcome this, as my model can
be richer in number or parameters. The following example, will show what
I would like to do:
## Setup
groupN <- 4
NPerGroup <- 5
min <- 1
max <- 5
g <- runif(n = groupN, min = min, max = max)
## Simulate
group <- factor(rep(paste("G", 1:groupN, sep = ""), each = NPerGroup))
y <- vector(mode = "numeric", length = groupN * NPerGroup)
j <- 1
for (i in 1:groupN) {
y[j:(i * NPerGroup)] <- rpois(n = NPerGroup, lambda = g[i])
j <- (i * NPerGroup) + 1
}
## GLM
contrasts(group) <- contr.sum(groupN)
fit <- glm(y ~ group, family = "poisson")
coef(fit)
## Now this goes nicely
model.matrix(y ~ group) %*% coef(fit)
## But pretend I have the following vector of estimated parameters
beta <- c(coef(fit), 0 - sum(coef(fit)[-1]))
names(beta) <- c(names(beta)[1:4], "group4")
## I can not apply this as model matrix does not conform to beta
model.matrix(y ~ group) %*% beta
## Is there any general way of constructing full design/model matrix
## without any
Thanks!

5/9/06, Gregor Gorjanc <gregor.gorjanc (AT) gmail (DOT) comwrote:
Hello!

I have parameter estimates for a generalized linear model and would like
to produce fitted values i.e. fitted(). This can be easily done in R,
but my problem lies in fact that I have a vector of parameters from some
other software, that is is not constrained i.e. I have the following
estimates for model with one factor with 4 levels

beta = c(intercept group1 group2 group3 group4)

where group1:4 are estimated deviations from intercept i.e. sum to zero
contraint, but all parameter estimates are there! How can I create a
model matrix that will not contain any constraints since I would like to
compute

model.matrix("some_formula") %*% beta

I.e. I would like to have model.matrix of the form

1 1 0 0 0
1 0 1 0 0
1 0 0 1 0
1 0 0 0 1

and not of the following form with contr.treatment or any other contraints

1 0 0 0
1 1 0 0
1 0 1 0
1 0 0 1

I could remove group4 from beta and use sum to zero contraint for
contrast in fomula, but I would like to overcome this, as my model can
be richer in number or parameters. The following example, will show what
I would like to do:

## Setup

groupN <- 4
NPerGroup <- 5
min <- 1
max <- 5
g <- runif(n = groupN, min = min, max = max)

## Simulate

group <- factor(rep(paste("G", 1:groupN, sep = ""), each = NPerGroup))
y <- vector(mode = "numeric", length = groupN * NPerGroup)
j <- 1
for (i in 1:groupN) {
y[j:(i * NPerGroup)] <- rpois(n = NPerGroup, lambda = g[i])
j <- (i * NPerGroup) + 1
}

## GLM

contrasts(group) <- contr.sum(groupN)
fit <- glm(y ~ group, family = "poisson")
coef(fit)

## Now this goes nicely
model.matrix(y ~ group) %*% coef(fit)

## But pretend I have the following vector of estimated parameters
beta <- c(coef(fit), 0 - sum(coef(fit)[-1]))
names(beta) <- c(names(beta)[1:4], "group4")

## I can not apply this as model matrix does not conform to beta
model.matrix(y ~ group) %*% beta

Try this:

model.matrix(y ~ group-1) %*% beta[-1] + beta[1]

## Is there any general way of constructing full design/model matrix
## without any

Thanks!

Hello!

Thank you very much for the response. Please read within the lines

Gabor Grothendieck wrote:
5/9/06, Gregor Gorjanc <gregor.gorjanc (AT) gmail (DOT) comwrote:

>Hello!
>>
>I have parameter estimates for a generalized linear model and would like
>to produce fitted values i.e. fitted(). This can be easily done in R,
>but my problem lies in fact that I have a vector of parameters from some
>other software, that is is not constrained i.e. I have the following
>estimates for model with one factor with 4 levels
>>
>beta = c(intercept group1 group2 group3 group4)
>>
>where group1:4 are estimated deviations from intercept i.e. sum to zero
>contraint, but all parameter estimates are there! How can I create a
>model matrix that will not contain any constraints since I would like to
>compute
>>
>model.matrix("some_formula") %*% beta
>>
>I.e. I would like to have model.matrix of the form
>>
>1 1 0 0 0
>1 0 1 0 0
>1 0 0 1 0
>1 0 0 0 1
>>
>and not of the following form with contr.treatment or any other
>contraints
>>
>1 0 0 0
>1 1 0 0
>1 0 1 0
>1 0 0 1
>>
>I could remove group4 from beta and use sum to zero contraint for
>contrast in fomula, but I would like to overcome this, as my model can
>be richer in number or parameters. The following example, will show what
>I would like to do:
>>
>## Setup
>>
>groupN <- 4
>NPerGroup <- 5
>min <- 1
>max <- 5
>g <- runif(n = groupN, min = min, max = max)
>>
>## Simulate
>>
>group <- factor(rep(paste("G", 1:groupN, sep = ""), each = NPerGroup))
>y <- vector(mode = "numeric", length = groupN * NPerGroup)
>j <- 1
>for (i in 1:groupN) {
>y[j:(i * NPerGroup)] <- rpois(n = NPerGroup, lambda = g[i])
>j <- (i * NPerGroup) + 1
>}
>>
>## GLM
>>
>contrasts(group) <- contr.sum(groupN)
>fit <- glm(y ~ group, family = "poisson")
>coef(fit)
>>
>## Now this goes nicely
>model.matrix(y ~ group) %*% coef(fit)
>>
>## But pretend I have the following vector of estimated parameters
>beta <- c(coef(fit), 0 - sum(coef(fit)[-1]))
>names(beta) <- c(names(beta)[1:4], "group4")
>>
>## I can not apply this as model matrix does not conform to beta
>model.matrix(y ~ group) %*% beta

Try this:

model.matrix(y ~ group-1) %*% beta[-1] + beta[1]

This is a nice hack here, but does not work in general. Say I have
another factor

sex <- factor(rep(paste("S", 1:2, sep = ""), times=10))

model.matrix(y ~ group + sex - 1)

produces a matrix of 5 columns and not 6 as I want to.

>>
>## Is there any general way of constructing full design/model matrix
>## without any
>>
>Thanks!
>>
>--
>Lep pozdrav / With regards,
>Gregor Gorjanc
>>
>
>University of Ljubljana PhD student
>Biotechnical Faculty
>Zootechnical Department URI:
>Groblje 3 mail: gregor.gorjanc <atbfro.uni-lj.si
>>
>SI-1230 Domzale tel: +386 (0)1 72 17 861
>Slovenia, Europe fax: +386 (0)1 72 17 888
>>
>
>" must learn by doing the thing; for though you think you know it,
>you have no certainty until you try." Sophocles ~ 450 B.C.
>>
>
>R-help (AT) stat (DOT) math.ethz.ch mailing list
>
>PLEASE do read the posting guide!
>
>>

5/9/06, Gregor Gorjanc <gregor.gorjanc (AT) gmail (DOT) comwrote:
Hello!

Thank you very much for the response. Please read within the lines

Gabor Grothendieck wrote:
5/9/06, Gregor Gorjanc <gregor.gorjanc (AT) gmail (DOT) comwrote:
>
>Hello!
>>
>I have parameter estimates for a generalized linear model and would like
>to produce fitted values i.e. fitted(). This can be easily done in R,
>but my problem lies in fact that I have a vector of parameters from some
>other software, that is is not constrained i.e. I have the following
>estimates for model with one factor with 4 levels
>>
>beta = c(intercept group1 group2 group3 group4)
>>
>where group1:4 are estimated deviations from intercept i.e. sum to zero
>contraint, but all parameter estimates are there! How can I create a
>model matrix that will not contain any constraints since I would like to
>compute
>>
>model.matrix("some_formula") %*% beta
>>
>I.e. I would like to have model.matrix of the form
>>
>1 1 0 0 0
>1 0 1 0 0
>1 0 0 1 0
>1 0 0 0 1
>>
>and not of the following form with contr.treatment or any other
>contraints
>>
>1 0 0 0
>1 1 0 0
>1 0 1 0
>1 0 0 1
>>
>I could remove group4 from beta and use sum to zero contraint for
>contrast in fomula, but I would like to overcome this, as my model can
>be richer in number or parameters. The following example, will show what
>I would like to do:
>>
>## Setup
>>
>groupN <- 4
>NPerGroup <- 5
>min <- 1
>max <- 5
>g <- runif(n = groupN, min = min, max = max)
>>
>## Simulate
>>
>group <- factor(rep(paste("G", 1:groupN, sep = ""), each = NPerGroup))
>y <- vector(mode = "numeric", length = groupN * NPerGroup)
>j <- 1
>for (i in 1:groupN) {
>y[j:(i * NPerGroup)] <- rpois(n = NPerGroup, lambda = g[i])
>j <- (i * NPerGroup) + 1
>}
>>
>## GLM
>>
>contrasts(group) <- contr.sum(groupN)
>fit <- glm(y ~ group, family = "poisson")
>coef(fit)
>>
>## Now this goes nicely
>model.matrix(y ~ group) %*% coef(fit)
>>
>## But pretend I have the following vector of estimated parameters
>beta <- c(coef(fit), 0 - sum(coef(fit)[-1]))
>names(beta) <- c(names(beta)[1:4], "group4")
>>
>## I can not apply this as model matrix does not conform to beta
>model.matrix(y ~ group) %*% beta
>
>
Try this:

model.matrix(y ~ group-1) %*% beta[-1] + beta[1]

This is a nice hack here, but does not work in general. Say I have
another factor

sex <- factor(rep(paste("S", 1:2, sep = ""), times=10))

model.matrix(y ~ group + sex - 1)

produces a matrix of 5 columns and not 6 as I want to.

Try this:

cbind(1, model.matrix(~group-1), model.matrix(~sex-1))


>
>>
>## Is there any general way of constructing full design/model matrix
>## without any
>>
>Thanks!
>>
>--
>Lep pozdrav / With regards,
>Gregor Gorjanc
>>
>
>University of Ljubljana PhD student
>Biotechnical Faculty
>Zootechnical Department URI:
>Groblje 3 mail: gregor.gorjanc <atbfro.uni-lj.si
>>
>SI-1230 Domzale tel: +386 (0)1 72 17 861
>Slovenia, Europe fax: +386 (0)1 72 17 888
>>
>
>" must learn by doing the thing; for though you think you know it,
>you have no certainty until you try." Sophocles ~ 450 B.C.
>>
>
>R-help (AT) stat (DOT) math.ethz.ch mailing list
>
>PLEASE do read the posting guide!
>
>>
>
>

Gabor thanks! Last suggestion works like charm!

Gabor Grothendieck wrote:
5/9/06, Gabor Grothendieck <ggrothendieck (AT) gmail (DOT) comwrote:

>5/9/06, Gregor Gorjanc <gregor.gorjanc (AT) gmail (DOT) comwrote:
>Hello!
>>
>Thank you very much for the response. Please read within the lines
>>
>Gabor Grothendieck wrote:
>5/9/06, Gregor Gorjanc <gregor.gorjanc (AT) gmail (DOT) comwrote:
>>
>>Hello!
>>>
>>I have parameter estimates for a generalized linear model and
>would like
>>to produce fitted values i.e. fitted(). This can be easily done
>in R,
>>but my problem lies in fact that I have a vector of parameters
>from some
>>other software, that is is not constrained i.e. I have the following
>>estimates for model with one factor with 4 levels
>>>
>>beta = c(intercept group1 group2 group3 group4)
>>>
>>where group1:4 are estimated deviations from intercept i.e. sum
>to zero
>>contraint, but all parameter estimates are there! How can I create a
>>model matrix that will not contain any constraints since I would
>like to
>>compute
>>>
>>model.matrix("some_formula") %*% beta
>>>
>>I.e. I would like to have model.matrix of the form
>>>
>>1 1 0 0 0
>>1 0 1 0 0
>>1 0 0 1 0
>>1 0 0 0 1
>>>
>>and not of the following form with contr.treatment or any other
>>contraints
>>>
>>1 0 0 0
>>1 1 0 0
>>1 0 1 0
>>1 0 0 1
>>>
>>I could remove group4 from beta and use sum to zero contraint for
>>contrast in fomula, but I would like to overcome this, as my
>model can
>>be richer in number or parameters. The following example, will
>show what
>>I would like to do:
>>>
>>## Setup
>>>
>>groupN <- 4
>>NPerGroup <- 5
>>min <- 1
>>max <- 5
>>g <- runif(n = groupN, min = min, max = max)
>>>
>>## Simulate
>>>
>>group <- factor(rep(paste("G", 1:groupN, sep = ""), each =
>NPerGroup))
>>y <- vector(mode = "numeric", length = groupN * NPerGroup)
>>j <- 1
>>for (i in 1:groupN) {
>>y[j:(i * NPerGroup)] <- rpois(n = NPerGroup, lambda = g[i])
>>j <- (i * NPerGroup) + 1
>>}
>>>
>>## GLM
>>>
>>contrasts(group) <- contr.sum(groupN)
>>fit <- glm(y ~ group, family = "poisson")
>>coef(fit)
>>>
>>## Now this goes nicely
>>model.matrix(y ~ group) %*% coef(fit)
>>>
>>## But pretend I have the following vector of estimated parameters
>>beta <- c(coef(fit), 0 - sum(coef(fit)[-1]))
>>names(beta) <- c(names(beta)[1:4], "group4")
>>>
>>## I can not apply this as model matrix does not conform to beta
>>model.matrix(y ~ group) %*% beta
>>
>>
>Try this:
>>
>model.matrix(y ~ group-1) %*% beta[-1] + beta[1]
>>
>This is a nice hack here, but does not work in general. Say I have
>another factor
>>
>sex <- factor(rep(paste("S", 1:2, sep = ""), times=10))
>>
>model.matrix(y ~ group + sex - 1)
>>
>produces a matrix of 5 columns and not 6 as I want to.
>>
>Try this:
>>
>cbind(1, model.matrix(~group-1), model.matrix(~sex-1))
>>

In thinking about this a bit more, try this:

attr(group, "contrasts") <- diag(nlevels(group))
attr(sex, "contrasts") <- diag(nlevels(sex))
model.matrix(~ group + sex)