similar to: regression with ordered arguments

Displaying 20 results from an estimated 20000 matches similar to: "regression with ordered arguments"

2007 Dec 02
3
Help with a Loop
I am having trouble getting a loop to work for the following problem. Any help would be much appreciated. Thanks. I need to find the slope and intercept from the linear regression of Drug Level on Day by Participant. There are a total of 37 Participants. I need to store the Participant, Label, Slope, and Intercept in a new data frame. This data is ordered by Participant number 37 total
2011 Aug 25
4
{R} How to extract correctly from vector?
Dear list, I have problem that I cannot solve and would like to ask your opinion. I tried to ask a few days ago already but got no answer and all my attempts to solve it by myself since then failed. Sorry for repeated posting! Here the problem broken down a bit. My problem basically is, that I want to use the elements of a character vector as names for objects and by recalling only the
2011 Aug 15
1
update() ignores object
Hi all, I'm extracting the name of the term in a regression model that dropterm specifies as the least significant one, and I'm assigning this name to an object. However, when I use update(), it ignores this object. Is there a way I can make it not ignore it? A reproducible example is below: > lm(x1~1+y1*y2+y3+y4,data=anscombe)->my.lm >
2005 Jan 21
2
Selecting a subplot of pairs
Hello, I'm trying to plot a set of 3 dependant variables (y) against 4 predictors (x) in a matrix-like plot, sharing x- an y-axis for all the plot on the same column/line : y1/x1 y1/x2 y1/x3 y1/x4 y2/x1 y2/x2 y2/x3 y2/x4 y3/x1 y3/x2 y3/x3 y3/x4 In fact, this plot is a rectangular selection of the result of pairs(), limited to the relations between x's and y's
2011 Jan 19
2
Reshape
Hi - I'm up against a complicated reshape problem. I have data of the form X1,Y1,hr1,hr2,hr3 X1,Y2,hr1,hr2,hr3 X1,Y3,hr1,hr2,hr3 X2,Y1,hr1,hr2,hr3 X2,Y2,hr1,hr2,hr3 X2,Y3,hr1,hr2,hr3 where X and Y are factors and the hr(1,2,3) are values. I need it as ,X1, X2 Y1,hr1,hr1 Y1,hr2,hr2 Y1,hr3,hr3 Y2,hr1,hr1 Y2,hr2,hr2 Y2,hr3,hr3 .., Any hints? I've been at it for hours. p -- View
2010 Oct 21
1
All other variables in upper scope arg for stepAIC
Hi - I am trying to substitute for "the_other_y" in the code below. I want y2 and y3 to be there when i is 1, y1 and y3 to be there when i is 2 and y1 and y2 to be there when i is 3. I'm sure it's to do with what format the data should be in and I've tried alldata[,-i], but it fits all the columns of alldata except i rather than each column one at a time. I've tried
2005 Apr 04
1
custom loss function + nonlinear models
Hi all; I'm trying to fit a reparameterization of the assymptotic regression model as that shown in Ratkowsky (1990) page 96. Y~y1+(((y2-y1)*(1-((y2-y3)/(y3-y1))^(2*(X-x1)/(x2-x1))))/(1-((y2-y3)/(y3-y1))^2)) where y1,y2,y3 are expected-values for X=x1, X=x2, and X=average(x1,x2), respectively. I tried first with Statistica v7 by LS and Gauss-Newton algorithm without success (no
2008 Dec 29
4
Merge or combine data frames with missing columns
Hi R-experts, suppose I have a list with containing data frame elements: [[1]] (Intercept) y1 y2 y3 y4 -6.64 0.761 0.383 0.775 0.163 [[2]] (Intercept) y2 y3 -3.858 0.854 0.834 Now I want to put them into ONE dataframe like this: (Intercept) y1
2010 Feb 28
4
Reducing a matrix
I wish to rearrange the matrix, df, such that all there are not repeated x values. Particularly, for each value of x that is reated, the corresponded y value should fall under the appropriate column. For example, the x value 3 appears 4 times under the different columns of y, i.e. y1,y2,y3,y4. The output should be such that for the lone value of 3 selected for x, the corresponding row entries
2008 Aug 13
1
summary.manova rank deficiency error + data
Dear R-users; Previously I posted a question about the problem of rank deficiency in summary.manova. As somebody suggested, I'm attaching a small part of the data set. #*************************************************** "test" <- structure(.Data = list(structure(.Data = c(rep(1,3),rep(2,18),rep(3,10)), levels = c("1", "2", "3"), class =
2007 Nov 14
1
Changing the text in the strips of lattice plots and y axis
Dear R-helpers, I am sorry for asking something I know has been asked before, I have tried different combinations in the strip function without success... I am using version 2.5.1 and work on a PC. I have barcharts generated from the following formula: barchart(y1+y2+y3~x | g) I need to change the names of the variables y1,y2 or y3 that currently appear in the strips, I have two
2007 Dec 06
2
How can I plot this graph
Hi, I am having trouble plotting the graph I need given the follow kind of data > xxx <- data.frame( "x"=c(1,2,3,4,5), "y1"=c(2,4,3,5,6), "y2"=c(3,4,6,3,1), "y3"=c(1,3,5,7,3), "z1"=c(1,NA,3,5,NA), "z2"=c(2,NA,4,6,NA) ) > xxx x y1 y2 y3
2005 Feb 14
1
corrupt data frame: columns will be truncated or padded with NAs in: format.data.frame(x, digits = digits)
Hello! I posted on saturday mail with the same subject on r-help seeking for help in my work, but now I realized that this list is more appropriate for this. I think I found I bug. Bellow are comments and reproducible examples: # Create a data frame (tmp <- data.frame(y1=1:4, f1=factor(c("A", "B", "C", "D")))) y1 f1 1 1 A 2 2 B 3 3 C 4 4 D #
2005 Feb 14
1
corrupt data frame: columns will be truncated or padded with NAs in: format.data.frame(x, digits = digits)
Hello! I posted on saturday mail with the same subject on r-help seeking for help in my work, but now I realized that this list is more appropriate for this. I think I found I bug. Bellow are comments and reproducible examples: # Create a data frame (tmp <- data.frame(y1=1:4, f1=factor(c("A", "B", "C", "D")))) y1 f1 1 1 A 2 2 B 3 3 C 4 4 D #
2010 Nov 19
3
assigment
Hello R Users, I have vectors x <- c("a2","b7","c8") y1 <- c(1,2,3,2) y2 <- c(4,2,7,5,4,3,8) y3 <- c(1:10) and I want to assign values form vector y1 to a new variable which name comes from the 1st value of the vector x etc. How to do it using only vector x. As a result I should have: a2 <- y1 b7 <- y2 c8 <- y3 More general problem: I have 2 txt
2010 Jan 05
1
Multivariate Poisson GLM??
Dear R Users, I'm working on a problem where I have a multivariate response vector of counts and a continuous predictor. I've thought about doing this the same way you would do a Multvariate regression model with normally distributed data, but since these data are counts, they are probably better modeled with a Poisson distribution. For example y1<-rpois(100,3.5) y2<-rpois(100,1.5)
2006 Jul 07
2
Multistage Sampling
Dear WizaRds, dear Thomas, First of all, I want to tell you how grateful I am for all your support. I wish I will be able to help others along one day the same way you do. Thank you so much. I am struggling with a multistage sampling design: library(survey) multi3 <- data.frame(cluster=c(1,1,1,1 ,2,2,2, 3,3), id=c(1,2,3,4, 1,2,3, 1,2), nl=c(4,4,4,4, 3,3,3, 2,2), Nl=c(100,100,100,100,
2002 Nov 26
1
Reshape by multiple variables
Dear list I'm using the reshape command and want to reshape a wide data set to a long one e.g. I have the variables y1,y2,y3,age1,age2,age3,sex,ethnic I want my new long data set to consist of the variables y (which has been created from y1,y2,y3), age (which has been created from age1,age2,age3), sex and ethnic I have tried to use the command:
2005 Dec 08
2
how to change a dataframe with characters to a numeric matrix?
hi netters, i have a dataframe TEST like this: Y1 Y2 Y3 X1 4 7 8 X2 6 2 Z X3 8 0 1 i would like to change it to a numeric matrix, replacing "Z" with NA Y1 Y2 Y3 X1 4 7 8 X2 6 2 NA X3 8 0 1 i've tried the function data.matrix but it didn't work. is there any easy way to do this? thanks a lot!
2010 Feb 25
2
Rearranging entries in a matrix
I have a matrix, called data. I used the code below to rearrange the data such that the first column remains the same, but the y value falls under either columns 2, 3 or 4, depending on the value of z. If z=1 for example, then the value of y will fall under column 2, if z=2, the value of y falls under column 3, and so on. data x y z [1,] 50 13 1 [2,] 14 8 2 [3,] 3 7 3 [4,] 4 16 1 [5,] 6