Hi everyone,
I have a question about mediation when there is an interaction between 2
categorical variables, one with two levels and one with three levels.
Using the âmediationâ package by Tingley and colleagues in R, Iâm trying
to
conduct a bootstrapped mediation analysis for an
I'd like to create several tables in which my data are stored.
I did the follow:
x-mydata
splitted-split(x[,1],x[,2])
#Create my function for the tables
fxdata-function(){
y.row=c(0,1,2,3,4,5)
yy=c(rep(0,6))
w=data.frame(A3=yy,A2=yy,A1=yy,A0=yy)
NOELIA LEGAL noelialegal at hotmail.com writes:
Please I need your help.I'm intrested to know if there is any R-package
for fit a Pearson III extreme value
distribution to data.Thanks a lot
Noelia
[[alternative HTML version deleted]]
On Fri, May 9, 2014 at 10:42 PM, Hadley Wickham h.wick...@gmail.com wrote:
Beware of the is.* functions:
* is.object() does not test the usual definition of objects
* is.vector() does not test the usual definition of vectors
* is.numeric() does not work the same way as is.character() or
On 10/05/2014, 6:46 AM, Barry Rowlingson wrote:
On Fri, May 9, 2014 at 10:42 PM, Hadley Wickham h.wick...@gmail.com wrote:
Beware of the is.* functions:
* is.object() does not test the usual definition of objects
* is.vector() does not test the usual definition of vectors
* is.numeric() does
HI,
If you want to try other ways:
fun1 - function(mat, rowN) {
dm - dim(mat)[1]
rowN1 - rowN - 1
indx - rep(1:rowN, dm - rowN1) + rep(seq(0, dm - rowN), each = rowN)
indx1 - (seq_along(indx)-1)%/%rowN+1
as.vector(tapply(indx, list(indx1), FUN = function(i) sum(mat[i, ])))
}
Thankyou very much arun. Its always nice to hear from you.
Eliza
Date: Sat, 10 May 2014 03:55:29 -0700
From: smartpink...@yahoo.com
Subject: Re: [R] adding rows
To: r-help@r-project.org
CC: ruipbarra...@sapo.pt; eliza_bo...@hotmail.com
HI,
If you want to try other ways:
fun1 -
On 10 May 2014, at 12:54 , Duncan Murdoch murdoch.dun...@gmail.com wrote:
On 10/05/2014, 6:46 AM, Barry Rowlingson wrote:
On Fri, May 9, 2014 at 10:42 PM, Hadley Wickham h.wick...@gmail.com wrote:
Beware of the is.* functions:
* is.object() does not test the usual definition of objects
*
Dear Group,
I have data like the following
id contacts_list number of contacts
---
1 3 4 2
2 1 3 43
34 2 1 3
411
On 10/05/2014, 7:46 AM, Ragia Ibrahim wrote:
Dear Group,
I have data like the following
id contacts_list number of contacts
---
1 3 4 2
2 1 3 43
34 2 1
Thanks, modifying the predict function to allow for new levels was what was
required
Brian
--
View this message in context:
http://r.789695.n4.nabble.com/Extracting-the-names-of-coefficients-of-random-effects-tp4689109p4690298.html
Sent from the R help mailing list archive at Nabble.com.
Hi,
Try:
dat - data.frame(id=1:4, contacts_list=I(list(3:4,c(1,3,4), c(4,2,1), 1)),
`number of contacts`=c(2,3,3,1),check.names=FALSE)
attr(dat$contacts_list,class) - NULL #if needed
dat
A.K.
Dear Group,
I have data like the following
id contacts_list number of contacts
One answer to your question might be ?get. A way better answer is to never use
assign to create xx1 etc in the first place but to store those results in a
list that you can simply index later. If you were to create a reproducible
example [1] by defining mydata you would be more likely to get an
d - data.frame(id=1:4, no.contacts=c(2,3,3,1))
d$contacts_list - list(3:4, c(1,3,4), c(4,2,1), 1 )
If you store that information in a longer format, with each row being
an edge to the relationship graph, it can make further processing
easier:
d2 - with(d,
data.frame(id=rep(id,
On May 10, 2014, at 3:54 AM, Duncan Murdoch wrote:
On 10/05/2014, 6:46 AM, Barry Rowlingson wrote:
On Fri, May 9, 2014 at 10:42 PM, Hadley Wickham h.wick...@gmail.com wrote:
Beware of the is.* functions:
* is.object() does not test the usual definition of objects
* is.vector() does not
Hello,
I have tried to apply this approach -- which worked for this example
-- to a larger dataset. Although I have not received error messages,
the variable I have set (my.data) is empty.
Do you have any tip about this?
Many thanks
Luigi
# code::
my.data-structure(list(
column_1 = 1:120,
Hi,
---
2767.493803,4796.33016,12292.93705,3864.657567,9380.673835,14886.44683,8457.88646,26050.47191))),#
.Names = c(row, stimulation, positivity, group, copy),
row.names = c(NA, -120L), class = data.frame)
str(my.data)
'data.frame': 120
You made the same error as the first time - a misplaced parenthesis.
Make the last 2 lines
.Names = c(row, stimulation, positivity, group, copy)),
row.names=c(NA, 120L), class=data.frame)
instead of
.Names = c(row, stimulation, positivity, group, copy),
row.names = c(NA, -120L)), class
And I misplaced the closing parenthesis in my reponse - it goes before
the .Names attribute as well.
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Sat, May 10, 2014 at 11:36 AM, William Dunlap wdun...@tibco.com wrote:
You made the same error as the first time - a misplaced parenthesis.
Make
Hello,
I'm a novice R user. I'd like to estimate the linear trends (b) and their
statistical significance (p-value) of quite a few univariate time series.
Because several time series show autocorrelation and heteroskedasticity,
the ordinary least squares method lm(), as I understand it, isn't the
Hello
I am using xyplot (lattice) to plot a xts variable. There are 20 variables
within the xts variable (index by a datetime vector sampled every 1 minute).
The results are very nice and help me to understand what is going on with the
data. However since the names (labels of each variable)
On May 10, 2014, at 2:26 PM, Eduardo M. A. M. Mendes wrote:
Hello
I am using xyplot (lattice) to plot a xts variable. There are 20 variables
within the xts variable (index by a datetime vector sampled every 1 minute).
The results are very nice and help me to understand what is going on
Hello
Many thanks.
par.main.text is the main title (on top of all plots). I need to change the
each of the titles in the, say, subplots.
Ed
On May 10, 2014, at 7:05 PM, David Winsemius dwinsem...@comcast.net wrote:
On May 10, 2014, at 2:26 PM, Eduardo M. A. M. Mendes wrote:
Hello
On May 10, 2014, at 3:30 PM, Eduardo M. A. M. Mendes wrote:
Hello
Many thanks.
par.main.text is the main title (on top of all plots). I need to change the
each of the titles in the, say, subplots.
One gets the response the fills ones needs when those needs are clearly
expressed and
Yes, you are right. I meant panels, xyplot generates 22 panels, which one
with y-axis, x-axis and a title. I need to change the font size of title in
the panels.
Ed
On May 10, 2014, at 8:04 PM, David Winsemius dwinsem...@comcast.net wrote:
On May 10, 2014, at 3:30 PM, Eduardo M. A.
On May 10, 2014, at 5:47 PM, Eduardo M. A. M. Mendes wrote:
Yes, you are right. I meant panels, xyplot generates 22 panels, which one
with y-axis, x-axis and a title. I need to change the font size of title in
the panels.
It appears to me that the answer is probably in :
library(zoo);
library(lattice);
a=matrix(runif(100),25,4);
b=zoo(a,seq(1,25));
names(b)=c(How to change the font size?,2,3,4);
xyplot(b)
Ed
On May 10, 2014, at 9:52 PM, David Winsemius dwinsem...@comcast.net wrote:
On May 10, 2014, at 5:47 PM, Eduardo M. A. M. Mendes wrote:
Yes, you are
On May 10, 2014, at 6:06 PM, Eduardo M. A. M. Mendes wrote:
library(zoo);
library(lattice);
a=matrix(runif(100),25,4);
b=zoo(a,seq(1,25));
names(b)=c(How to change the font size?,2,3,4);
xyplot(b)
xyplot(b, strip=strip.custom( par.strip.text=list(cex=.5)))
--
David.
Ed
On May
cex=0.85 did the job for me.
many thanks
Ed
On May 10, 2014, at 10:13 PM, David Winsemius dwinsem...@comcast.net wrote:
On May 10, 2014, at 6:06 PM, Eduardo M. A. M. Mendes wrote:
library(zoo);
library(lattice);
a=matrix(runif(100),25,4);
b=zoo(a,seq(1,25));
names(b)=c(How to
Hi
Just an addition for the future :
If you wanted to have different main title plots to the same device then
par.settings can be used for each plot
xyplot(1 ~1,
par.settings = list(par.main.text = 0.85,
par.sub.text = 0.85)
Eduardo M. A. M. Mendes emammendes at gmail.com writes:
Hello
I am using xyplot (lattice) to plot a xts variable. There are 20 variables
within the xts variable (index by a
datetime vector sampled every 1 minute). The results are very nice and help
me to understand what is going
on
Hola a todos y todas,
Gracias por vuestro apoyo en cantidad de preguntas anteriores, de nuevo os
escribo para compartir una duda:
Estoy trabajando con un modelo bien sencillo, es una regresión simple, pero
me gustarÃa comprobar la significación estadÃstica de cada uno de los
coeficientes de
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