similar to: Verify the linear regression model used in R ( fundamental theory)

Displaying 20 results from an estimated 4000 matches similar to: "Verify the linear regression model used in R ( fundamental theory)"

2010 Jun 21
2
How to predict the mean and variance of the dependent variable after regression
Hi, folks, As seen in the following codes: x1=rlnorm(10) x2=rlnorm(10,mean=2) y=rlnorm(10,mean=10)### Fake dataset linmod=lm(log(y)~log(x1)+log(x2)) After the regression, I would like to know the mean of y. Since log(y) is normal and y is lognormal, I need to know the mean and variance of log(y) first. I tried mean (y) and mean(linmod), but either one is what I want. Any tips? Thanks in
2010 Jul 21
2
Variance of the prediction in the linear regression model (Theory and programming)
Hi, folks, Here are the codes: ############## y=1:10 x=c(1:9,1) lin=lm(log(y)~x) ### log(y) is following Normal distribution x=5:14 prediction=predict(lin,newdata=x) ##prediction=predict(lin) ############### 1. The codes do not work, and give the error message: Error in eval(predvars, data, env) : numeric 'envir' arg not of length one. But if I use the code after the pound sign, it
2010 Jun 24
2
count data with a specific range
I would like to prepare the data for barplot. But I only have the data frame now. x1=rnorm(10,mean=2) x2=rnorm(20,mean=-1) x3=rnorm(15,mean=3) data=data.frame(x1,x2,x3) If there a way to put data within a specific range? The expected result is as follows: range x1 x2 x3 -10-0 2 5 1 (# points in this
2012 Aug 29
3
Help on calculating spearman rank correlation for a data frame with conditions
Dear all, Suppose my data frame is as follows: id price distance 1 2 4 1 3 5 ... 2 4 8 2 5 9 ... n 3 7 n 8 9 I would like to calculate the rank-order correlation between price and distance for each id. cor(price,distance,method = "spearman") calculate a correlation for all. Then I tried to use apply(data,list='id',cor(price , distance , method =
2010 Jun 23
1
How to 'understand' R functions besides reading R codes
Apologize for not being clearer earlier. I would like to ask again. Thank Joris and Markleeds for response. Two examples: 1. Function 'var'. In R, it is the sum of square divided by (n-1) but not by n. (I know this in R class) 2. Function 'lm'. In R, it is the residual sum of square divied by (n-2) not by n, the same as in the least squares estimate. But the assumption following
2010 Jul 02
2
how to save summary(lm) and anova (lm) in format?
Hi, folks, I would like to copy the output of summary(lm) and anova (lm) in R to my word file. But the output will be a mess if I just copy after I call summary and anova. ##################### x=rnorm(10) y=rnorm(10,mean=3) lm=lm(y~x) summary(lm) Call: lm(formula = y ~ x) Residuals: Min 1Q Median 3Q Max -1.278567 -0.312017 0.001938 0.297578 1.310113
2010 Jun 18
1
How to calculate the robust standard error of the dependent variable
Hi, folks linmod=y~x+z summary(linmod) The summary of linmod shows the standard error of the coefficients. How can we get the sd of y and the robust standard errors in R? Thanks! [[alternative HTML version deleted]]
2010 Jun 26
1
All a column to a data frame with a specific condition
Hi, folks, Please first look at the codes: plan_a=c('apple','orange','apple','apple','pear','bread') plan_b=c('bread','bread','orange','bread','bread','yogurt') value=1:6 data=data.frame(plan_a,plan_b,value) library(plyr) library(reshape) mm=melt(data, id=c('plan_a','plan_b'))
2010 Sep 02
1
How to generate integers from uniform distribution with fixed mean
Hi, folks, runif (n,min,max) is the typical code for generate R.V from uniform dist. But what if we need to fix the mean as 20, and we want the values to be integers only? Thanks [[alternative HTML version deleted]]
2010 Jun 25
2
Delete rows in the data frame by limiting values in two columns
Hi, folks, Finally Friday~~ Here comes the question: x=c('germany','poor italy','usa','england','poor italy','japan') y=c('Spain','germany','usa','brazil','england','chile') s=1:6 z=3:8 test=data.frame(x,y,s,z) #Now I only concern the countries ('germany','england','brazil').
2010 Jun 29
3
How to delete the replicate rows by summing up the numeric columns
Hi, folks, I am sorry that I did not state the problem correctly yesterday. Please let me address the problem by the following codes: first=c('u','b','e','k','j','c','u','f','c','e')
2004 Mar 03
1
Bug in plot.lm (PR#6640)
Dear all, I noticed the following behaviour of plot.lm: > fm1 <- lm(time~dist, data=hills, weights=c(0,0,rep(1,33))) > par(mfrow=c(2,2)) > plot(fm1) Warning messages: 1: longer object length is not a multiple of shorter object length in: res/(sd * (1 - hat)) 2: longer object length is not a multiple of shorter object length in: (res/(sd * (1 - hat)))^2 * hat which seems to be
2003 Oct 02
4
using a string as the formula in rlm
Hi, I am trying to build a series of rlm models. I have my data frame and the models will be built using various coulmns of the data frame. Thus a series of models would be m1 <- rlm(V1 ~ V2 + V3 + V4, data) m2 <- rlm(V1 ~ V2 + V5 + V7, data) m3 <- rlm(V1 ~ V2 + V8 + V9, data) I would like to automate this. Is it possible to use a string in place of the formula? I tried doing: fmla
2004 Apr 07
4
Problems with rlm
Dear all, When calling rlm with the following data, I get an error. (R v.1.8.1, WinXP Pro 2002 with service pack 1.) > d <- na.omit(data.frame(CPRATIO, HEIGHTZ, FAMILYID)) > c <- tapply(d$CPRATIO, d$FAMILYID, mean) > h <- tapply(d$HEIGHTZ, d$FAMILYID, mean) > c 1 2 3 6 7 9 10 11 6.000000 2.500000 3.250000
2004 Oct 11
3
split and rlm
Hello, I'm trying to do a little rlm of some data that looks like this: UNIT COHORT perdo adjodds 1010 96 0.39890 1.06894 1010 97 0.48113 1.57500 1010 98 0.36328 1.21498 1010 99 0.44391 1.38608 It works fine like this: rlm(perdo ~ COHORT, psi=psisquare) But the problem is that I have about 100 UNITs, and I want to do a
2007 Nov 29
1
relative importance of predictors
Hei Group, I want to compare the relative importance of predictors in a multiple linear regression y~a+bx1+cx2... However, bptest indicates heteroskedasticity of my model. I therefore perform a robust regression (rlm), in combination with bootstrapping (as outlined in J. Fox, Bootstrapping Regression Models). Now I want to compare the relative importance of my predictors. Can I rely on the
2009 Dec 03
2
Avoiding singular fits in rlm
I keep coming back to this problem of singular fits in rlm (MASS library), but cannot figure out a good solution. I am fitting a linear model with a factor variable, like lm( Y ~ factorVar) and this works fine. lm knows to construct the contrast matrix the way I would expect, which puts the first factor as the baseline level. But when I try rlm( Y ~ factorVar) I get the message "'x'
2011 Mar 14
1
discrepancy between lm and MASS:rlm
Dear R-devel, There seems to be a discrepancy in the order in which lm and rlm evaluate their arguments. This causes rlm to sometimes produce an error where lm is just fine. Here is a little script that illustrate the issue: > library(MASS) > ## create data > n <- 100 > dat <- data.frame(x=rep(c(-1,0,1), n), y=rnorm(3*n)) > > ## call lm, works fine > summary(lm(y ~
2005 Mar 24
1
Robust multivariate regression with rlm
Dear Group, I am having trouble with using rlm on multivariate data sets. When I call rlm I get Error in lm.wfit(x, y, w, method = "qr") : incompatible dimensions lm on the same data sets seem to work well (see code example). Am I doing something wrong? I have already browsed through the forums and google but could not find any related discussions. I use Windows XP and R
2010 Nov 08
1
Add values of rlm coefficients to xyplot
Hello, I have a simple xyplot with rlm lines. I would like to add the a and b coefficients (y=ax+b) of the rlm calculation in each panel. I know I can do it 'outside' the xyplot command but I would like to do all at the same time. I found some posts with the same question, but no answer. Is it impossible ? Thanks in advance for your help. Ptit Bleu. x11(15,12) xyplot(df1$col2 ~