thanks Jorge. I appreciate your multiple improvements.
This still involves hard coding the co-efficients. I wonder if this is what
glm and lm are doing.
for e.g.
m<-lm(K~a+b,data=data)
m$coefficients would have 0 for all variables except a and b and then R must be
multiplying the weights the same way as your function.
I will try to use your code with the coefficients matrix from the model and see
if that works and report back what I find tomorrow.
Then if I can add code to return the names of the columns with the resulting
highest 3 values of the numbers then I should be done.
thanks a lot Jorge.
regards,
Dhruv
--- On Mon 07/07, Jorge Ivan Velez < [EMAIL PROTECTED] > wrote:
From: Jorge Ivan Velez [mailto: [EMAIL PROTECTED]
To: [EMAIL PROTECTED]
Date: Mon, 7 Jul 2008 21:42:54 -0400
Subject: Re: [R] question on lm or glm matrix of coeficients X test data terms
That's R: you come out with solutions every time. I hope don't bother you with
this. Try also:# data set (10 rows, 10
columns)set.seed(123)X=matrix(rpois(100,10),ncol=10)# Function to estimate your
outcome
outcome=function(x,betas){if(length(x)!=length(betas)) stop("x and beta have
different lengths!")y=x*betassum(y)}# let's assume that you want to include x1,
x4, x7 and x9 only# by using beta1=0.5, beta4=0.6, beta7=-0.1, beta9=0.3
betas=c(0.5,0,0,0.6,0,0,-0.1,0,0.3,0)# Resultsapply(X,1,outcome,
betas=betas)HTH,JorgeOn Mon, Jul 7, 2008 at 9:31 PM, Jorge Ivan Velez <[EMAIL
PROTECTED]> wrote:
Sorry, I forgot to the the sum over the rows:# data set (10 rows, 10 columns)
set.seed(123)X=matrix(rpois(100,10),ncol=10)# Function to estimate your
outcomeoutcome=function(x,betas){if(length(x)!=length(betas)) stop("x and beta
have different lengths!")
y=x*betasy}# let's assume that you want to include x1, x4, x7 and x9 only# by
using beta1=0.5, beta4=0.6, beta7=-0.1,
beta9=0.3betas=c(0.5,0,0,0.6,0,0,-0.1,0,0.3,0)
# Resultsapply(t(apply(X,1,outcome, betas=betas)),1,sum)
HTH,JorgeOn Mon, Jul 7, 2008 at 9:23 PM, Jorge Ivan Velez <[EMAIL PROTECTED]>
wrote:
Dear Dhruv,It's me again. I've been thinking about a little bit. If you want to
include/exclude variables to estimate your outcome, you could try something
like this:# data set (10 rows, 10 columns)
set.seed(123)X=matrix(rpois(100,10),ncol=10)# Function to estimate your
outcomeoutcome=function(x,betas){if(length(x)!=length(betas)) stop("x and beta
have different lengths!")
y=x*betasy}# let's assume that you want to include x1, x4, x7 and x9 only# by
using beta1=0.5, beta4=0.6, beta7=-0.1,
beta9=0.3betas=c(0.5,0,0,0.6,0,0,-0.1,0,0.3,0)# Resultst(apply(X,1,outcome,
betas=betas))
HTH,JorgeOn Mon, Jul 7, 2008 at 9:11 PM, Jorge Ivan Velez <[EMAIL PROTECTED]>
wrote:
Dear Dhruv,The short answer is not, because the function I built doesn't work
for more variables than coefficients (see the "stop" I introduced). You should
do some modifications such as coefficients equals to 1 or 0. For example:
# data set (10 rows, 10 columns)set.seed(123)X=matrix(rpois(100,10),ncol=10)X#
Function to estimate your
outcomeoutcome=function(x,betas,val){k=length(x)nb=length(betas)
if(length(x)!=length(betas)) betas=c(betas, rep(val,k-nb))
y=x*betasy}# beta1=1, beta2=2, the rest is equal to
zerot(apply(X,1,outcome,betas=c(1,2),val=0))# beta1=0.5, beta2=0.6, the rest is
equal to 1
t(apply(X,1,outcome,betas=c(1,2),val=1))
HTH,JorgeOn Mon, Jul 7, 2008 at 8:57 PM, DS <[EMAIL PROTECTED]> wrote:
thanks Jorge. I appreciate your quick help.
Will this work if I have 20 columns of data but my regression only has 5
variables?
I am looking for something generic where I can give it my model and test data
and get back a vector of the multiplied coefficients (with no hard coding).
When predict is called with an input model and data, R must be multiplying all
co-efficients times variables and summing the number but is there a way to get
components of the regressiom terms stored in a matrix before they are added?
The idea is to build n models with various terms and after producing a
prediction list the top 3 variables that had the biggest impact in that
particular set of predictor values.
e.g. if I build a model to predict default of loans I would then need to list
the top factors in the model that can be used to explain why the loan is risky.
With 10-16 variables which can be present or not for each case there be a
different 2 or 3 variables that led to the said prediction.
Dhruv
--- On Mon 07/07, Jorge Ivan Velez < [EMAIL PROTECTED] > wrote:
From: Jorge Ivan Velez [mailto: [EMAIL PROTECTED]
To: [EMAIL PROTECTED]
Date: Mon, 7 Jul 2008 20:12:53 -0400
Subject: Re: [R] question on lm or glm matrix of coeficients X test data terms
Dear Dhruv,Try also:# data setset.seed(123)X=matrix(rpois(10,10),ncol=2)#
Function to estimate your
outcomeoutcome=function(x,betas){if(length(x)!=length(betas)) stop("x and betas
are of different length!")
y=x*betasy}# outcome for beta1=0.05 and
beta2=0.6t(apply(X,1,outcome,betas=c(0.05,0.6)))# outcome for beta1=5 and
beta2=6
t(apply(X,1,outcome,betas=c(5,6)))
HTH,JorgeOn Mon, Jul 7, 2008 at 7:56 PM, DS <[EMAIL PROTECTED]> wrote:
Hi,
is there an easy way to get the calculated weights in a regression equation?
for e.g.
if my model has 2 variables 1 and 2 with coefficient .05 and .6
how can I get the computed values for a test dataset for each coefficient?
data
var1,var2
10,100
so I want to get .5, 60 back in a vector. This is a one row example but I
would want to get a matrix of multiplied out coefficients and terms for use in
comparing contribution of variables to final score. As in a scorecard using
logistic regression.
Please advise.
thanks
Dhruv
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______________________________________________
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