Hi All,
I have a quick question on random forests. Simply, I am not sure how to read
the values of independent variables related to the highest value of a response
variable from all trees generated from random forests. It is easy to do this in
a single regression tree. But I am not clear how to
Hi!
An alternative with 'assign':
for ( i in 71:75) {
setwd(paste("C:/Awork/geneAssociation/removed8samples/neuhausen", i,
sep=""))
temp.df<-read.csv("seg.pr3.csv", head=T)
temp.df$id<-paste0("sn",i,sep="")
assign(paste0("seg",i,sep=""),temp.df)
}
rm(temp.df,i) # Clean up
HTH,
Kimmo
Hi!
Not just an gmail issue. After my last reply I have gotten tons of
spams from "Samantha Smith". Keep hitting my "rank as spam"-button in
the hope that my MUA could learn :)
Best,
Kimmo
ti, 2018-04-17 kello 19:34 +, Ding, Yuan Chun kirjoitti:
> No, I do not use gmail, still got dirty spam
Hi!
Seems to be an encoding problem. This worked for me (have not full-
checked the output, though):
fromJSON(encodeString(zWebObj))
HTH,
Kimmo
2018-05-08 12:49 -0700, David Winsemius wrote:
> >
> > On May 8, 2018, at 10:08 AM, Evans, Richard K. (GRC-H000) > k.ev...@
Hi!
How about this:
--- snip --
for (i in 1:(length(split_str)-1)) {
assign(paste("DF",i,sep=""),DF[
c((which(DF$name==split_str[i])+1):(which(DF$name==split_str[i+1])-1)),
])
}
--- snip ---
'assign' creates for each subset a new data.frame DFn, where n ist a
count (1,2,...).
But note: i
Hi!
2017-09-18 07:13 -0500, Therneau, Terry M., Ph.D. wrote:
> This question likely has a 1 line answer, I'm just not seeing
> it. (2, 3, or 10 lines is
> fine too.)
>
> For a vector I can do group <- match(x, unqiue(x)) to get a vector
> that labels each
> element of x.
Actually, you get a v
Dear R-helpers!
I have a data frame storing data for word co-occurrences, average
distances and co-occurence frequency:
Group.1Group.2 x Freq
1 deutschland achtziger 2.001
2 deutschlandalt 1.254
3 deutschland anfang -2.001
4 deutschlandansehen 1.00
Hi!
Many thanks to Duncan and Jim for their quick replies.
27.05.2016, 01:08, Jim Lemon wrote:
Hi Kimmo,
par(mar=c(5,7,4,2))
dotchart(kedf$x)
mtext(kedf$Group.2,side=2,at=1:6,line=0.5,
las=2,cex=log(abs(kedf$Freq))+1)
Jim
This 'dotchart' solution worked fine and I got what I wanted :) Howev
Hi Juho!
01.06.2016, 14:40, Juho Kiuru wrote:
Hi all, I am new to R and TwitteR and would love to get some advice from
you.
I managed to get list of tweets containing word 'innovation' tweeted in
Helsinki with following script:
searchTwitter('innovation', n=1, geocode='60.1920,24.9458,30mi
Hi!
Some sample data could help us to help you...
But have you read '?xyf' in order to ensure that your 'Y' is what 'xyf'
expects it to be?
What kind of error messages do you get?
Regards,
Kimmo
16.06.2016, 15:13, ch.elahe via R-help wrote:
Is there any answer?
Hi all, I have a df and I
4 3 3 2 1 4 1 2 3 ..
and I want to cluster my data based on speed, to see the coming costumer's
protocols fall into which speed group and I think I need to bring this speed
column in Y element of xyf
On Thursday, June 16, 2016 2:29 PM, K. Elo wrote:
Hi!
Some sample data could help us t
Hi again!
21.06.2016, 21:33, chalabi.el...@yahoo.de wrote:
Hi Kimmo, Thanks for your reply. I think now my problem is that I
don't understand what does factor(df.classes[training]) do?
Sorry, my mistake, should habe been 'df$speed'. Please try the following:
--- snip ---
set.seed(7) training
Hi!
22.06.2016, 22:00, chalabi.el...@yahoo.de wrote:
Dear Kimmo,
I already used df$speed[training] in df.xyf but I get this error:
Error in xyf(Xtraining,factor(df$speed[training]),grid=somgrid(5, :
NA/NaN/Inf in foreign function call (arg 1)
Please check for zeros (0) and NAs in df$spee
clustering purpose.
*data=read.table("my_distance_matrix", header=FALSE)[-1]*
*attach(data)*
*head(data)*
*d=as.dist(data);*
*hc.complete=hclust(d,method="complete")*
*cutree(hc.complete, k=6)*
*groups<- cutree(hc.complete, k=6)*
*x<-cbind(groups)*
*x*
*x1<- subset
I am completely new to R and am trying to utilize its capabilities as an
alternative to Minitab. I don't have any development ability at all, but
the R Commander GUI is able to give me the functionality I need with the
exception of control charts. I have installed the qcc package but when I
load
Hi!
10.06.2015, 13:20, khatri wrote:
My date column is in following format : "%m/%d H:M:S"
There is not mention of year in the data.So how can I read this using
strptime function.
I have tried strptime(dates,"%m/%d H:M:S") but this is returning NA.
Thanks
How about:
strptime(dates,"%m/%d %H:%
ot;,"perc","poisson"))
Error in boot.ci(db.fix.boot, index = 1, type = c("bca", "perc", "poisson")) :
object 'db.fix.boot' not found
--
-Original Message-
From: Aravi
Dear Sir or Madam,
i want to use the var.test() (f.test()) for n samples.
But in R the var.test() can only used for variances of two samples. In the
intruductions stands: Performs an F test to compare the variances of two
samples from normal populations.
I need a variance test for n samples. It
language.
talk soon
Thanking you
Regards
Namratha K
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Is there any method or built-in function for implementing a/b testing using
R language
Are there any function developed to implement a/b testing in R language?
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Hi!
Maybe not the most elegant solution, but a workaround is to have a
function:
> save2<-function(y, ...) { save(y,...)}
> save2(x1,x2,file="test.RData")
The point is to include the variables to be "renamed" as parameters (in
my example: y). The function will use the parameter variable names wh
Hi!
Have you tried to use 'fromJSON' with the parameter 'simplifyDataFrame'
set to TRUE?
See:
https://cran.r-project.org/web/packages/jsonlite/vignettes/json-aaquickstart.html
-> Section "Data Frames" explains how this affects the data frame
structure. IMHO this should solve your problem...
Be
Hi!
Maybe this would do the trick:
--- snip ---
library(reshape2) # Use 'reshape2'
library(dplyr)# Use 'dplyr'
datatransfer<-data %>% mutate(letter2=letter) %>%
dcast(id+letter~letter2, value.var="weight")
--- snip ---
Or did I misunderstood something?
Best,
Kimmo
2019-01-06, 13:16
Hi!
Not having a data chunk prevents me from testing abit, but maybe you
should take a look on:
?table
?xtabs
to start with.
But as already suggested by other users, a small data set would be of
great help :)
HTH,
Kimmo
su, 2019-01-06 kello 13:49 -0500, Rachel Thompson kirjoitti:
> Hi Rich,
>
Hi!
2019-02-27 22:51 -0500, Aimin Yan wrote:
> I have a question about assigning color based on the value of a
> matrix
>
> The following is my matrix.
>
> d
> lateRT earlyRT NAD ciLAD
> lateRT 1.0 0.00 0.006224017 0.001260241
> earlyRT 0.
Hi!
2019-03-25 kello 09:30 +0800, Steven Yen wrote:
> The second command is ugly. How can I print the 25 numbers into 2
> rows
> of ten plus a helf row of 5? Thanks.
Something like this?
x<-1:25; for (i in seq(1,length(x),10))
print(x[i:ifelse((i+9)>length(x),length(x),i+9)])
HTH,
Kimmo
_
Dear Team,
I am getting this error while running the boot-strapping functions.
==
mod.db.hub<-glm(TOTAL~1+IPD,family="poisson",data=db)
fit<-fitted(mod.db.hub)
e<-residuals(mod.db.hub)
X<-model.matrix(mod.db.hub)
boot.huber.fixed<-function(data,indi
Dear Team,
I am getting this error while running the boot-strapping functions.
==
mod.db.hub<-glm(TOTAL~1+IPD,family="poisson",data=db)
fit<-fitted(mod.db.hub)
e<-residuals(mod.db.hub)
X<-model.matrix(mod.db.hub)
boot.huber.fixed<-function(data,indi
Team,
Can some one help me in computing the R-squared value in glm.
Thanks
Aravindhan
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Team,
Has anyone looked at this question from me ? it will help me immensely if
someone can provide an answer to this.
Thanks
Aravindhan
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of K Aravindhan
Sent: Friday, August 08, 2014
Hi David,
I am using generalized linear models (glm command with family="poisson").
Thanks
Aravindhan
-Original Message-
From: David Winsemius [mailto:dwinsem...@comcast.net]
Sent: Saturday, November 15, 2014 1:16 PM
To: Aravindhan, K
Cc: R-help@r-project.org
Subject: Re:
Hi!
25.08.2015, 18:17, Sam Albers wrote:
Hi all,
This is a process question. How do folks efficiently identify column
numbers in a dataframe without manually counting them. For example, if I
want to choose columns from the iris dataframe I know of two options. I can
do this:
str(iris)'data.
Hi!
I have collected 500.000+ tweets with a Python script using 'tweepy',
which stored the data in JSON format. I would like to use R for data
analysis, but have encountered problems when trying to import the data
file with 'jsonlite'.
Here what I have tried:
> data.df<-fromJSON("example.js
Hi!
You can download the example file with this link:
https://www.dropbox.com/s/tlf1gkym6d83log/example.json?dl=0
BTW, I have used a JSON validator and the problem seems to related to
wrong/missing EOF.
--- snip ---
Error: Parse error on line 1:
...:"1436705823768"} {"created_at":"Sun J
-
Hi,
thanks to Duncan and Jeroen to quick replies. I was actually my thinking
error :) I suppoed 'fromJSON' to cope with a multi-line file or a list,
but this seems not to be the case. So I first read the file with
'readLines' into a list and processed all items with 'fromJSON' within a
for-lo
Dear all,
I am currently working a research project on social media interaction.
As a part of this project, mostly for teaching purposes, I should
develop a R-based approach for real-time visualisation of streamed data
(from Twitter).
My idea is simple (and working :) ): A Python-script stre
Hi!
18.10.2016, 14:38, Abhinaba Roy wrote:
Hi R helpers,
I have json inputs from an app which I want to convert to dataframes. Below
are the two inputs. Can someone help me in converting these to dataframes
[...]
IMHO, the best way is to use the package 'jsonlite', see:
* https://cran.r-pro
Hi!
Maybe this helps:
http://r.789695.n4.nabble.com/Error-in-normalizePath-path-with-McAfee-td2532324.html
Best,
Kimmo
15.12.2016, 08:18, Amelia Marsh via R-help wrote:
Hi
I had installed R studio Desktop 1.0.44. However whenever I wanted to write any
command, before I could complete, I was
Respected Sirs/Madam,
Good Morning
As part of a project I'm using elasticnet package in R for sparse pca. I
would be great help for me if you can advice me on how to select Optimum
number of principal components and number of observation in elasticnet
package (k, para ) in R.
--
*T
Hello,
I am a newcomer to R and therefore apologize for posting such a
basic question. I am
trying to multiply 2 matrices t(X1)%*%X1, where t(X1) is:
1 2 3 4 5 8 12 13 20 24 26 27 31 33 34 36 37 40 41 42 45 46
47 48 49
ones 1 1 1 1 1 1 1 1 1 11 1 1 1
Hello,
I am interested in plotting my regression analysis
results(regression coefficients and standard errors obtained through OLS and
Tobit models)
in the form of graphs.What is the best way to accomplish this? Thanks.
Murali Kuchibhotla
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Dear R-helpers,
I am desperately looking for a solution for how to print out the console
output to a standard printer. For example, I would like to print out the
summary.lm() output, the output of different ftable-functions etc. I use
R on a linux machine.
The only ways so far have been to c
] 0.2730090246 0.4462614490 0.2382041477 0.9826505063 0.1556554718
0.3746872961
[7] 0.6108254879 0.6617410595 0.6694177436 0.4650380281 0.0414420397
0.2307212995
[13] 0.5338913775 0.9186298891 0.0006410333 0.8046684864 0.6205502201
0.5352788521
[19] 0.4255279053 0.7711444888
On Wed, Aug 27, 2008 at 7:44
Hi,
as mentioned in my previous posting, I run R on a linux machine. So a
possible function for printing (in linux) could look like this:
copy2lpr<-function(..., PRINTER="lpr") {
LPR<-pipe(PRINTER,"w")
capture.output(..., file=LPR)
close(LPR)
}
This seems to work... An allows the user to c
We get an error while using the following function.Please advise.
newton.method(FUN = function(b) 10/b+sum(a)-10*sum((x^b)*a)/sum(x^b), init
= 5,tol = 0.001)
Error in deriv.default(as.expression(body(FUN)), nms, function.arg = TRUE)
:
Function 'sum' is not in the derivatives table
-
Dear Anna,
19.02.2010 08:17, Anna Carter wrote:
> (1) If the dataset contains some variables having all the entries = 0
> and while analysing I want to delete those pericular columns, how do
> acheive this. i.e.
Let's suppose 'df' is your data frame, then:
subset(df, select=which(colSums(df)!=0)
Hi!
Right, my solution did not take into accound paired negative values
summing up to zero.
This should work in all cases:
df[, which(colSums(df!=0)!=0)]
Kind regards,
Kimmo
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Hi!
22.02.2010 17:45, xin wei wrote:
>
> thank you for reply. I just type: hist(x) from SSH terminal, expecting a
> histogram to pop up like what i got under windows.instead I got the
> following error msg:
>
> Error in X11(d$display, d$width, d$height, d$pointsize, d$gamma,
> d$colortype,
Hi!
22.02.2010 19:53, xin wei wrote:
>
> hi, Kevin and K.Elo:
> thank you for the suggestion. Can you be more specific on these? (like how
> exactly get into x-switch or man ssh). I am totally ignorant about linux and
> SSH:( Memory limitation forces me to switch from windows to Linux
> c
Hi Jens!
23.03.2010 17:18, koj wrote:
The problem is: I want to group the data. I want to have ten groups. The
first two bars should be [1,1] and [2,1] together in one bar and in the
second bar of the first obervation should be [1,2] and [2,2] (stacked with
beside =TRUE). Therefore the first obs
Hi Jens!
24.03.2010 14:48, koj wrote
>
> Hi Kimmo, thank you for your answer, but this is not the thing I am searching
> for. Unfortunately, I have not described the problem very good. But just in
> this moment I have a good idea: I use add=TRUE and paint two plots. And so I
> am sure that I can
Hello,
I was wondering if someone could tell me how I can make text dependent on a
variable in a R function I have created.
The function will create plots, thus I would like each plot to have a unique
title based on the inputted variable as well as a unique file name when saved
using the sa
Hi all,
Is there function on R for calculating Modula generators? For example for
primes above 100, e.g 157, i want to know which number generates the group
under multiplication mod 157. i.e i want to find an element whose order is
156. The problem I occur is that modular arithmetic becomes in
Hi!
Wacek Kusnierczyk wrote:
> m = matrix(1:4, 2)
>
> apply(m, 1, cat, '\n')
> # 1 2
> # 3 4
> # NULL
>
> why the null?
Could it be the return value of 'cat'. See ?cat, where:
---snip ---
Value
None (invisible NULL).
---snip ---
Kind regrads,
Kimmo
__
Dear all,
I am using the RWeka package to append several arff files. Although it
works the resulting arff files always have "@relation R_data_frame",
and I have to change this manually to my desired relation name. Can
the package accomplish this for me instead?
Thank you,
Wil Koetsier
_
Hi,
> https://stat.ethz.ch/mailman/listinfo/r-help
and there You'll find the section:
"To unsubscribe from R-help, get a password reminder, or change your
subscription options enter your subscription email address:"
Hope this helps,
Kimmo
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Hi,
minben wrote:
> I am a new R-language user. I have set up a data frame "mydata",one of
> the colume of which is "skill". Now I want to select the observations
> whose "skill" value is equal to 1,by what command can I get it?
Try this:
mydata1<-mydatasubset(mydata, skill==1)
Maybe You should
Hi,
Crosby, Jacy R wrote:
>
> i.e. I'd like to have aov(Phen1~L1) use only Pat1-Pat4,and Pat 10.
> Similarly, aov(Phen1~L2) should use Pat1, 6, and 10.
> Etc.
>
> Is this something I can do in the aov function, or do I need to modify my
> dataset before running aov? In either case, I
hi,
I am trying to run a mixed effect gam using gamm (mgcv)
and I get the following error:
"Error in lme.formula(y ~ X - 1, random = rand, data = strip.offset(mf), :
nlminb problem, convergence error code = 1
message = iteration limit reached without convergence (9)"
I've read all about 'notExp
Hi!
mathallan wrote:
> How can I from the summary function, decide which glm (fit1, fit2 or fit3)
> fits to data best? I don't know what to look after, so I would please
> explain the important output.
Start with the AIC value (Akaike Information Criterion). The model
having the lowest AIC is the
Hi!
Jean-Baptiste Combes wrote:
> Hello,
>
> I use R 2.10, and I am new in R (I used to use SAS and lately Stata), I am
> using XP.
>
> I have a data which has a data.frame format called x.df (read from a csv
> file). I want to take from this data observations for which the variable
> "Code" sta
Hi!
Let's suppose the values for the x axis are stored in 'values'.
barplot(values, col=c(rep("Red",3),rep(1,length(values)-8),rep("Blue",5)))
HTH,
Kimmo
vikrant kirjoitti:
> Suppose I need to draw a Grouped bar plot with 100 values on the X axis. Now
> my question is If I need to highlight sup
it is very long since I have many variables. Basically I
> want to select all numerical variables (10 to 24), and all categorical
> variables except MEASUREM, SEL_FACET and SEL_MEAS without having to
> write each of them. I would also like to avoid writing the names, the
> indexes wo
On Mon, Jan 18, 2010 at 10:17 AM, Ivan Calandra <
ivan.calan...@uni-hamburg.de> wrote:
> Thanks for your answer, but it doesn't work...
>
> Here is what I get:
> > ssfamean <- aggregate(ssfa[[10:24]],ssfa[c("SPECSHOR", "BONE", "TO_POS",
> "FACETTE", "SHEARFAC", "ENA_BA")],mean)
> Error in .subset2
On Mon, Jan 18, 2010 at 10:33 AM, Ivan Calandra <
ivan.calan...@uni-hamburg.de> wrote:
> I didn't understand from the help what really does the function rowMeans
> but it looks like it doesn't take into account the categorical variables (I
> want to calculate the means when the values of all categ
Hello,
I know there must be a simple soluton to this problem but it eludes me
currently.
My data is partitioned into two subsets, each subset has a common column
factor but with varying levels:
levels(fdf_ghc$AgeDemo)
[1] "26TO35" "36TO45" "46TO55" "56TO65" "66TO75" "76TO85"
levels(fdf_ghcnull$A
Hi!
29.01.2010 12:49, soeren.vo...@eawag.ch wrote:
> Hello,
>
> I read the help as well as the examples, but I can not figure out why
> the following code does not produce the *given* row names, "x" and "y":
>
> x <- 1:20
> y <- 21:40
> rbind(
> x=cbind(N=length(x), M=mean(x), SD=sd(x)),
> y
number quality for both
the native built-in PRNG and any alternatives including the random package.
Thanks,
Ben K.
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PLEASE do
Hi Drake,
2019-05-04, 17:34 -0700, Drake Gossi wrote:
> Hello everyone,
>
> I'm trying to learn how to put together a citation network, and, in
> doing so, I'm playing around with a data set of my own making. I'm
> going back and forth between two .csv files. One has two columns and
> is simply l
Hi Rajesh,
2019-05-05 10:23 +0530, Rajesh Ahir_GJ wrote:
> Hello R users,
>
> I am getting an error while running following code.
>
> library(ggplot2)
> ggplot(hourly_data1,aes(hour, power))+
> geom_boxplot(aes(fill=monthname),outlier.shape=NA) +
> facet_wrap(~monthname)
> ggplot(hourly_data1,ae
Needing a < , > comparison for imaginary numbers
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and provide c
Hi,
cannot reproduce, either, on my Linuxmint 19.3 + R 3.6.2.
Here the outputs:
--- snip ---
> test(mean, 1:10)
[1] 5.5
> test(NULL, 1:10)
NULL
Error in FUN(args) : could not find function "FUN"
> test(mean, list(x=1:10, na.rm=TRUE))
[1] NA
Warning message:
In mean.default(args) : argument is n
Hi!
Let's assume your data is stored in a data frame called 'df'. So this
code should do the job:
df$Value[ (df$Value>=0.2 & df$Values<=0.4) | df$Value>=0.7 ]
Best,
Kimmo
pe, 2020-01-31 kello 09:21 -0500, pooja sinha kirjoitti:
> Hi All,
>
> I have a .csv file with four columns (Chrom, Start
Hi!
Oh, sorry, one "s" too much in my code. Here the correct one:
df$Value[ (df$Value>=0.2 & df$Value<=0.4) | df$Value>=0.7 ]
Best,
Kimmo
pe, 2020-01-31 kello 17:12 +0200, K. Elo kirjoitti:
> Hi!
>
> Let's assume your data is stored in a data frame call
nha kirjoitti:
> Thanks for providing the code but I also needed the output sheet in
> .csv format with all the four columns corresponding to the value
> (Chrom,
> Start_pos, End_pos & Value ranging from what I specified earlier).
>
> Puja
>
> On Fri, Jan 31, 2020 at 10:23
Hi!
Have you already read this:
https://cran.r-project.org/web/packages/rtweet/vignettes/auth.html
I think they explain rather well how to use Twitter tokens with
rtweet...
HTH,
Kimmo
ke, 2020-04-08 kello 17:19 +1200, Patrick Connolly kirjoitti:
> I'm using the rtweet package which makes use o
llo Kimmo,
>
> Yes. I did that and it worked fine -- as far as it goes. But it
> didn't cover what to do when using the same twitter account on a
> computer with a different user name -- which is what my question was
> about.
>
>
> On Wed, 08-Apr-2020 at 08:55AM
Hi!
With 'dplyr':
dt_count %>% mutate(STATUS=ifelse(STATUS %in%
c("Resolved","Closed"),"Resolved/Closed",STATUS)) %>% group_by(STATUS)
%>% summarise(n=sum(N))
Output:
1 Assigned 135
2 Cancelled 20
3 In Progress56
4 Pending75
5 Resolved/Closed 1180
HTH,
Ki
Dear All,
I am using the example from one of the tutorial about "Metafor" package and
"escalc" function, to learn how this package can be applied to do
meta-analysi; the code and the data is directly from the tutorials but
"weights=freq" option in the escalc function is given me error message
This
ychiatry and
>
> Neuropsychology | Maastricht University | P.O. Box 616 (VIJV1) | 6200 MD
>
> Maastricht, The Netherlands | +31 (43) 388-4170 | http://www.wvbauer.com
>
>
> >-Original Message-
> >From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of
t; benefits.
>
> Best,
> Wolfgang
>
> --
> Wolfgang Viechtbauer, Ph.D., Statistician | Department of Psychiatry and
>
> Neuropsychology | Maastricht University | P.O. Box 616 (VIJV1) | 6200 MD
>
> Maastricht, The Netherlands | +31 (43) 388-4170 | http://www.wvbauer.com
&
te:
> Dear Kobby,
>
> Please post the output of sessionInfo() and class(result.md).
>
> Best,
> Wolfgang
>
> >-Original Message-
> >From: K Amoatwi [mailto:amoatwi...@gmail.com]
> >Sent: Monday, 22 June, 2020 22:30
> >To: Viechtbauer, Wolfgang
ave loaded the 'meta' package after 'metafor' and then forest() will
> try to use the corresponding function from the meta package and not
> metafor. With:
>
> metafor::forest(result.md)
>
> it should work.
>
> Best,
> Wolfgang
>
> >-Origin
e. If you had loaded them in the other order
> it would have masked the other one.
>
> In fact it masked seven functions in total in this case as the message
> told you.
>
> Michael
>
> On 23/06/2020 16:29, K Amoatwi wrote:
> > Dear Wolfgang,
> > Yes! The "m
Hey all,
Has anyone ever altered an R package for image analysis to do optical mark
recognition? I'm trying to find a way to semi-automate data entry of
several thousand paper health surveys that are predominantly composed of
check boxes. All the boxes are uniform size and shape, so it seems as
Hello,
I'm a newbie to R, I have a quesiton.
I'm a bit confused on global variable assignments.
I have the following situation. I have 2 global variables current_idx and
previous_idx. *These 2 global variables have to be set by a method in a
reference class*.
Essentially, using <<- assignment
Dear colleagues, I have 2 points: One opinion and one question.
1)
In one paper in a peer-reviewed journal, I read about the idea of using a
logit regression as a surrogate for the log-binomial, just adding the
numerator to the denominator ...
It’s tempting to immediately get the RR instead of OR
I am sure, that this is not a pure Poisson! Huge overdispersion!
You get inflated confidence intervals!
(although, the point estimates of the regression coefficients stay the same)
Try to look for the causes of overdispersion! It may be geteroscedastisity?
What is the nature of the response, is it
Hello,
Does the (core) designation in the CRAN Task View indicate packages
that provide basic/recommended functionality in that domain?
I don't see it documented anywhere on the site.
Thanks for your help.
__
R-help@r-project.org mailing list
https://
Esteemed R UseRs,
Regarding specifying arguments in functions and calling functions
within functions:
## beginning ##
## some data
ex <- rnorm(10)
ex[5] <- NA
## example function
Scale <- function(x, method=c("mean", "median")) {
scl <- method
scl(x)
}
## both return NA
Scale(ex, method=m
cale(ex, fivenum)
> [1] 0.134 0.5190959 0.7050648 0.8304476 0.9370754
>> Scale(ex, hist)
>
> --
> David L Carlson
> Associate Professor of Anthropology
> Texas A&M University
> College Station, TX 77843-4352
>
>> ---
Hi!
28.09.2012 08:41, Atte Tenkanen wrote:
Sorry. I should have mentioned that the order of the components is important.
So c(1,4,6) is accepted as a subvector of c(2,1,1,4,6,3), but not of
c(2,1,1,6,4,3).
How to test this?
How about this:
--- code ---
g1<- c(2,1,1,4,6,3)
g2<- c(2,1,1,6,4
Hi!
28.09.2012 09:13, Bhupendrasinh Thakre wrote:
Statement I tried :
b <- unclass(Sys.time())
b = 1348812597
c_b <- rnorm(1,2,1)
Do you mean this:
--- code ---
> df<-data.frame("x"=0,"y"=0)
> colnames(df)
[1] "x" "y"
> colnames(df)[2]<-paste("b",unclass(Sys.time()),sep="_")
> colnames(df)
Data attached - didn't realize I could do that last night. Here's the data
inport piece of my code, change the pathname to your computer.
asym<-read.csv('/Users/kirstensimmons/Desktop/Asym04.csv')
asym
#put the data into a data matrix
asym_matrix<-data.matrix(asym)
On Thu, Oct 4, 2012 at 6:35
Hi All,
I'm trying to create two side-by-side contour plots with one legend by
modifying the code found here:
http://wiki.cbr.washington.edu/qerm/sites/qerm/images/b/bb/Example_4-panel_v1a.R
I've been able to set up the other two functions called in the code, but I
can't find reference to print.l
Stats beginner here.
I have a dataset composed of observations taken from 16 separate
experimental panels, each nested into one of 4 conditions (Treatment A
Level 1, Treatment A Level 2, Treatment B Level 1, Treatment B Level 2; see
photo: http://imgur.com/ZbzFPNq). There are 100 observations of t
Hi!
How about trying this:
data[ data$col1!=data$col2 & !is.na(data$col3), ]
col1 col2 col3
2a1 ST001
3b2 ST002
HTH, Kimmo
28.05.2014 15:35, jeff6868 wrote:
> Hi everybody,
>
> I have a little problem in my R-code which seems be easy to solve, but I
> wasn't able to find t
Rook apps can be run using the httpd built into R (great feature!), and can
be deployed using rApache server.
For those who prefer to deploy the apps under FastRWeb, below is the code
to do so:
-- in your web.R/myrookapp.R file, we ran the Rook app instance (stored in
variable app):
run <- f
Hi!
Do you mean something like this (df is your original data frame):
--- cut here ---
df1<-df
df1[[1]]<-paste("R",df[[1]],sep="_")
colnames(df1)<-c("SERIES","YEAR","value")
df1$value[ df1$YEAR==2009 ]<-5
for (i in c(2009:2007)) { df1$value[ df1$YEAR==(i-1) ]<-( df1$value[
df1$YEAR==i ]-df$DELTA
Hi,
I am using "*Maanova* package" to do anova. I have created *datafile* with
probeID as the first column, which is a tab limited text file and also
created *designfile*. I have created *readma object* which is named as abf1.
>From that readma object, i have to create data object by using
*crea
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