Hi
here is another approach.
> cbind(mydata, idx=(rowSums(mydata==2, na.rm=T)>0)*1)
X125 X255 X558 X2366 X177 X255.1 idx
aa010NA0 0 0
bb110NA0 1 0
cs212 10 0 1
de010NA0 0 0
gh20
Hi
Not sure about elegancy/efficiency.
library(reshape2)
dat3 <- melt(dat2)
dat3
mid variablevalue
1 m10 -19.5482
2 m20 -0.5120
3 m30 -0.4920
4 m11 0.0070
5 m21 3.2410
6 m31 -2.2560
7 m12 1.2230
8 m22 -4.490
Josh,
I have found other demos on demo().. any way thanks for the help and for
supporting this project :)
boredstog
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Hey thanks josh,
I was looking for such a demo programme can you point to some more such demo
programmes if available
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[I might have sent the following request to a wrong email address -
'r-help-requ...@r-project.org']
Hi,
I have two data frame dat1 and dat2.
dat1 <- data.frame(pid = paste('C', 1:5, sep = ''),
m1 = c(2, 2, 1, -1, 0),
###
# Clear memory and set the working directory and the seed
###
rm(list = ls())
setwd("/Dropbox/LMER/R ")
set.seed(7010)
###
??
?uniroot tells you exactly what uniroot() expects as arguments. Unless
I misunderstand, your query seems totally unrelated to the way
uniroot() works.
Also, stop posting to Nabble and post to this (plain text) list. We
have no idea of what the Nabble contexts are, and many of us won't
even bot
On 11/08/15 11:26, Dr. Thomas W. MacFarland wrote:
Everyone:
I am currently trying to reproduce the nonparametric Walsh Test, as
presented in Siegel (1956, pp. 83-87), and unfortunately I cannot
find a R function for this test.
I would certainly appreciate any pointers for the Walsh Test using
Everyone:
I am currently trying to reproduce the nonparametric Walsh Test, as presented
in Siegel (1956, pp. 83-87), and unfortunately I cannot find a R function for
this test.
I would certainly appreciate any pointers for the Walsh Test using R, or
perhaps a reasonable substitute if there is
Dear Lida,
Here is a solution. Please don't post in HTML. And provide an easy to use
example of the data. E.g. the output of dput(mydata)
set.seed(1234)
mydata <- matrix(
sample(
c(0, 1, 2, NA),
size = 30,
replace = TRUE,
prob = c(2, 1, 1, 1)
),
ncol = 6
)
idx <- apply(myda
Easy enough (note that your column names are problematic, though)
> mydata <- structure(list(X125 = c(0L, 1L, 2L, 0L, 2L), X255 = c(1L, 1L,
+ 1L, 1L, 0L), X558 = c(0L, 0L, 2L, 0L, 0L), X2366 = c(NA, NA,
+ 1L, NA, 0L), X177 = c(0L, 0L, 0L, 0L, 0L), X255.1 = c(0L, 1L,
+ 0L, 0L, 0L)), .Names = c("X12
Hi there,
I have a matrix contain 0,1,2, NA elements.
I want to add a column to this matrix with name of "idx" . then for each
row, I should put 1 in this column (idx) if there is at least one 2 in that
row otherwise I should put 0 in this column!
for example mydata:
125 255 558 23
Hi, I have a 3D finite element mesh where each element (cell) is defined by 8
vertices. Each element is a regular polyhedron. The overall domain is a block
in shape, but its horizontal principal axes are not coincident with x and y
(i.e. the domain is rotated about the z-axis).
I want to pl
Avril,
The more direct way to access these columns from the matrix is:
ci[, "lwr"]
ci[, "upr"]
Jean
On Mon, Aug 10, 2015 at 3:20 AM, alc wrote:
>
>
> Dear all,
>
> I'm wondering how can I access the confidence interval values ('upr' and
> 'lwr' values) produced by the 'predict' function. For
On Mon, Aug 10, 2015 at 11:02 AM, boredstoog via R-help
wrote:
> Thanks Joshua for the quick reply to the mail and once more sorry for
> bothering with another doubt. So i have modified your code :) for
> backtesting and this is the code
>
> *
> library(quantmod)
> library(tseries)
> require(quant
Thanks Joshua for the quick reply to the mail and once more sorry for
bothering with another doubt. So i have modified your code :) for
backtesting and this is the code
*
library(quantmod)
library(tseries)
require(quantstrat)
library(PerformanceAnalytics)
sym <- get(getSymbols('SPY'))["2013::"]
sy
The || operator will always return a result of type 'logical' and length 1.
You gave it two operands of length 0, so it returned the logical value NA,
meaning it had no idea what the result should be. If you give it operands
of length > 1, it will use the only the first elements of them. (S and S
Mayukh,
I apologize for taking so long to get back to your problem. I expect you may
have found the solution. If so I would be interested. I have developed a hack
to solve the problem, but I expect if someone knew how to handle JSON objects
or even text parsing better they could develop a more
Since 2008, Revolution Analytics (and now Microsoft) staff and guests have
written about R every weekday at the Revolutions blog:
http://blog.revolutionanalytics.com
and every month I post a summary of articles from the previous month of
particular interest to readers of r-help.
In case you mi
Dear R-Experts,
I am trying to get the bootstrapped confidence intervals of R-squared
(Nagelkerke) for an ordinal logistic regression. Something is going wrong at
the end of my script. Many thanks for your help.
Here is my reproducible example.
install.packages("rms")
library(rms)
x=c(1,2,3,2
Dear Avril
I think you will find that predict.lm returns a matrix not a data frame.
I find str() useful when R does things I did not expect or quite understand.
Michael
On 10/08/2015 09:20, alc wrote:
Dear all,
I'm wondering how can I access the confidence interval values ('upr' and
'lwr'
Thanks Krishna!
You are correct, but even when I manually add m<-17 just after x I still get
the same error?
Uniroot falls over when it tries to take in the argument MLEobj=fit, thus I
think the error is occurring before LamOpt() gets called by uniroot().
Which is why the error you mentioned di
Hi there,
I'd like to be able to pass an entire object to uniroot() as one of the
arguments. Unfortunately to recreate my precise error would be rather
involved. So I present a simplified version below:
To create a structured object let the output from the lm() function use this
simple code:
Dear all,
I'm wondering how can I access the confidence interval values ('upr' and
'lwr' values) produced by the 'predict' function. For example, I fitted
a linear regression line using:
fit <- lm(y ~ x)
I then wanted to calculate a 95% confidence interval for the line, and
did this using
I don't see any conversion of your time data from character to a time type, so
it is probably converting to factor within the ggplot function. Something like
Sys.setenv(TZ="Etc/GMT+5") # you need to study time types, including ?strptime
Sandy$Deal1 <- as.POSIXct( Sandy$Deal1, format="%m/%d/%Y %H
> My preference is to start in different working directories depending on which
> project I am working on. R_USER is not a project directory. One way to do that
> is to double-click on an RData file located where you want to start.
Saving an empty 'empty.RData' image in my project directories when
> >I would like the startup working directory to be that pointed to by
> >R_USER.
In Windows, right click on your R desktop shortcut (or create a new one
pointing to Rgui.exe) and on the properties tab, change "Start in:" to
%R_USER%
R will then start 'in' the R_USER directory if R_USER exists (
Dear list members,
I am building a model such as:
Y1 = Y2*X1 + X2
Y2 = Y1*X1 + X2
X2 is the exogenous variable
Z1 is the instrument of Y1
Z2 is the instrument of Y2
This is a simultaneous equation model. I know how to build a simultaneous
equation model without interaction terms:
library(syste
Hi
Your question is without reproducible example and I find it a bit cryptic. You
do not uncover what is i. If it is a number I wonder why your atempts fail.
Find answer in line below.
> -Original Message-
> From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of Ragia
> Ibrahim
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