On Aug 29, 2013, at 23:56 , John Fox wrote:
> Dear Gerard,
>
> Without your data, it's not possible to reproduce your problem exactly, but
> it's clear that it isn't specific to the scatterplot() function in the car
> package. For example, try
>
> plot(1:10)
> title(main=bquote(paste("Hypothesi
This is my a part of my data set
> D[1:15,c(1,5:10)]
X. media IE.2005 IE.2006 IE.2007 IE.2008 IE.2009 IE.2010
1 1108 22.060.0 39 4.0 8.016.0 5.0
2 1479 110.0 NA NA53.0 1166.0 344.8 110.0
3 1591 86.6 247.0 87
Thanks
On Fri, Aug 30, 2013 at 1:49 AM, Jim Lemon wrote:
> On 08/30/2013 01:28 AM, Shane Carey wrote:
>
>> Hello all,
>>
>> I have decided to go ahead with gap.boxplot. I am trying to suppress the
>> axis labels, both x and y labels. I tried using axis.labels=NULL but it
>> would not work.
>>
>
Hello together i have a little problem with the combine of two data.frames.
I have 2 data.frames, which look like this one:
first dataframe:
ID Name
1 Jack
2 John
3 Jill
second dataframe
ID Days Type
13 Training
21 Management
34 Training
At the end i want to
Hello,
Suposing that your data frames are named dat1 and dat2, the following
works, but it's a bit complicated, maybe there are simpler solutions.
dat1 <- read.table(text = "
ID Name
1 Jack
2 John
3 Jill
", header = TRUE, stringsAsFactors = FALSE)
dat2 <- read.table(text = "
ID Days Ty
On 29.08.2013 16:33, Luvalle, Michael J (Michael) wrote:
I upgraded R from 2.12.1 to 3.0.0 (on windows XP(, and as soon as I saved the 3.0.0 workspace, was
unable to access .Rdata from 2.12.1. The message in the R console is "Error in
loadNamesSpace(name): there is no package called parallel
Thanks first.
that doesn't look bad. But if have a equal ID in dat2, the days are no
longer correct.
the correct data.frame has to look like this one:
ID Name Management Training
1 1 Jack 13
2 2 John 10
3 3 Jill 04
not this one:
ID Name
Hi Ma Teresa,
Sorry, but I can't understand what you're trying to achieve.
On a statistical note, I'd tend to think more in terms of medians and would
think hard before replacing any outliers, but that's another matter.
Here I created the dataframe dd with the means column of D in its first colu
Hi,
xlab="",
ylab"",
Would not work.
Thanks
On Fri, Aug 30, 2013 at 9:37 AM, Shane Carey wrote:
> Thanks
>
>
> On Fri, Aug 30, 2013 at 1:49 AM, Jim Lemon wrote:
>
>> On 08/30/2013 01:28 AM, Shane Carey wrote:
>>
>>> Hello all,
>>>
>>> I have decided to go ahead with gap.boxplot. I am trying
On 30-08-2013, at 11:49, Shane Carey wrote:
> Hi,
>
> xlab="",
> ylab"",
>
You were told to use xlab="", ylab=""
You seem to have omitted the = after ylab
Berend
> Would not work.
>
> Thanks
>
>
> On Fri, Aug 30, 2013 at 9:37 AM, Shane Carey wrote:
>
>> Thanks
>>
>>
>> On Fri, Aug
This is what I put in:
gap.boxplot(DATA$Conductivity~factor(DATA$UnitName_1),ylim=c(LOWER_Y_Conductivity,UPPER_Y_Conductivity_int),gap=gap_Conductivity,
col=colours,outwex=one,whisklty = "solid",whisklwd=lwth,outcol=
"black", outpch=dtsym, outcex=dtsize,
range=1.5,xlab="",y
Hello again,
I have a string which I need to put in some legitimate date format.
My string is: "MAY-14"
And output format would be "05/01/2014", this should be of Date class,
so that I can make some sensible calculation with it.
I have tried this without any success:
> as.Date("MAY-14", format
On 08/30/2013 07:57 PM, Shane Carey wrote:
This is what I put in:
gap.boxplot(DATA$Conductivity~factor(DATA$UnitName_1),ylim=c(LOWER_Y_Conductivity,UPPER_Y_Conductivity_int),gap=gap_Conductivity,
col=colours,outwex=one,whisklty =
"solid",whisklwd=lwth,outcol= "black", outpch=dtsym,
Hello,
Ok, try instead
library(reshape2)
tmp <- dcast(data = dat2, ID ~ Type, value.var = "Type")
tmp[-1] <- lapply(tmp[-1], function(x){
y <- integer(length(x))
s <- as.character(x[!is.na(x)])[1]
idx <- which(as.character(dat2[["Type"]]) == s)
y[!is.na(x)] <-
On 30.08.2013 11:59, Christofer Bogaso wrote:
Hello again,
I have a string which I need to put in some legitimate date format.
My string is: "MAY-14"
And output format would be "05/01/2014", this should be of Date class,
so that I can make some sensible calculation with it.
I have tried thi
Oooff, Right, I will give it a go and see how I get on.
Thanks
On Fri, Aug 30, 2013 at 11:17 AM, Jim Lemon wrote:
> On 08/30/2013 07:57 PM, Shane Carey wrote:
>
>> This is what I put in:
>> gap.boxplot(DATA$Conductivity~**factor(DATA$UnitName_1),ylim=**
>> c(LOWER_Y_Conductivity,UPPER_**Y_Cond
It worked perfectly, your a star!!!
Thanks
On Fri, Aug 30, 2013 at 11:40 AM, Shane Carey wrote:
> Oooff, Right, I will give it a go and see how I get on.
>
> Thanks
>
>
> On Fri, Aug 30, 2013 at 11:17 AM, Jim Lemon wrote:
>
>> On 08/30/2013 07:57 PM, Shane Carey wrote:
>>
>>> This is what I p
Thanks a lot!
It works like I want to!
Am 29.08.2013 15:44 schrieb "arun kirshna [via R]" <
ml-node+s789695n4674877...@n4.nabble.com>:
> HI,
> May be this helps:
>
> ts1<- ts(1:20)
> ts2<- ts(1:25)
> ts1[-(1:3)]<- ts1[-(1:3)]+ts2[1:17]
>
> as.numeric(ts1)
> # [1] 1 2 3 5 7 9 11 13 15 17
Hi
Do you have 2 versions of .RData? If yes you probably need to install packages
which were installed and used before saving old .RData file to be able to
access this file with new R version.
If this does not help, you could return to previous R version, save necessary
objects to separate .r
Hi
It would be good if you provided some more specific inforamation. What function
do you use for transfering data to R?.
E.g. read.table has option to specify NA values during data transfer.
If you want to set some values to NA in already read objects you can do it
object[object=="some value"
Hi,
I have a problem when I try to generate the Documentation pdf (from .rda
files)in Spanish during the package creation. Could you tell me the way I can
do it?.
Thanks in advance.
Regards.
Eva
[[alternative HTML version deleted]]
__
R-help
Hi
as.Date(paste("MAY-14","-01", sep=""), format = "%b-%y-%d")
Shall be OK. Change output format by ?format.
Regards
Petr
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
> project.org] On Behalf Of Christofer Bogaso
> Sent: Friday, August 30, 2013 12
I have three datasets that I want to compute the errors between them using
linear regression.for this, I want to iterate to reach certain criteria for
the calibration. if changes become smaller than eps the iteration is
successful, hence stop and write parameters into cal:eps=0.1 if number
of i
Hi
see ?merge
something like
merge(df1, df2)
Petr
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
> project.org] On Behalf Of Mat
> Sent: Friday, August 30, 2013 10:37 AM
> To: r-help@r-project.org
> Subject: [R] create new column to combine 2 data.
Hi
Sorry, I did not read your question to the end
library(reshape)
merge(dat1,cast(melt(dat2, c("ID", "Type")), ID~Type))
ID Name Management Training
1 1 Jack 13
2 2 John 1 NA
3 3 Jill NA4
Is close enough, you can easily change NA to 0 if you
On 30-08-2013, at 09:44, Jonsson wrote:
> I have three datasets that I want to compute the errors between them using
> linear regression.for this, I want to iterate to reach certain criteria for
> the calibration. if changes become smaller than eps the iteration is
> successful, hence stop and w
thanks. Works perfectly
you made my day :-)
--
View this message in context:
http://r.789695.n4.nabble.com/create-new-column-to-combine-2-data-frames-tp4674963p4674994.html
Sent from the R help mailing list archive at Nabble.com.
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R-help@r-projec
Hi,
I haven't tried the code yet. Is there a way to parse this data
using R and create bar plots so that each program's 'RAM used' figures are
grouped together.
So 'uuidd' bars will be together. The data will have about 50 sets. So if
there are 100 processes each will have about 50 bar
Hi,
You can use 'fill=0' in ?cast()
merge(dat1,cast(melt(dat2, c("ID", "Type")), ID~Type,fill=0))
# ID Name Management Training
#1 1 Jack 1 3
#2 2 John 1 0
#3 3 Jill 0 4
A.K.
- Original Message -
From: PIKAL Petr
To: Mat ; "r-help@r-p
HI,
Also,
dd1<-matrix(cbind(D[,1],(D[-c(1:2)]/D[,2]>4)*1),dimnames=NULL,ncol=7)
identical(dd,dd1)
#[1] TRUE
A.K.
- Original Message -
From: Jose Iparraguirre
To: Mª Teresa Martinez Soriano ;
"r-help@r-project.org"
Cc:
Sent: Friday, August 30, 2013 5:39 AM
Subject: Re: [R] Outlie
Hi,
Better would be to show a reproducible example using ?dput() and the codes you
used. Assuming that you tried something like this:
lst1<- list(1:10,c(5,4,3),4:15)
mean(lst1)
#[1] NA
#Warning message:
#In mean.default(lst1) : argument is not numeric or logical: returning NA
sapply(lst1,mean)
Hello,
This memory usage should be graphed with time. Are there
examples of scatterplots that can clearly show usage vs time ? This is
for memory leak detection.
Thanks,
Mohan
From: PIKAL Petr
To: "mohan.radhakrish...@polarisft.com"
, "r-help@r-project.org"
Date:
forgot to send it back to the list.
Jim Holtman
Data Munger Guru
What is the problem that you are trying to solve?
Tell me what you want to do, not how you want to do it.
-- Forwarded message --
From: jim holtman
Date: Fri, Aug 30, 2013 at 8:10 AM
Subject: Re: ddply for compa
Hi Ramón,
May be this helps:
tags_totals<-matrix(c(15,11,23,7,5),ncol=1,dimnames=list(c("Wikis","Glosarios","Grupos","Bases
de datos","Taller"),NULL))
tags_totals[order(tags_totals[,1],decreasing=TRUE),,drop=FALSE]
# [,1]
#Grupos 23
#Wikis 15
#Glosarios 1
Here is how to parse the data and put it into groups. Not sure what
the 'timing' of each group is since not time information was given.
Also not sure is there is an 'MiB' qualifier on the data, but you have
the matrix of data which is easy to do with as you want.
> input <- readLines(textConnect
Hi Jim et al,
I want to remove the upper bounding box,
I did this by
#box()
in the gap.plot function. It still leaves me with two horizontal lines. I
would like to remove them also, where are the created within the function?
Thanks, this is great, exactly what I need.
Cheers
On Fri, Aug 30, 20
You can also look at the XLConnect package.
Jim Holtman
Data Munger Guru
What is the problem that you are trying to solve?
Tell me what you want to do, not how you want to do it.
On Thu, Aug 29, 2013 at 9:40 AM, Zsurzsa Laszlo wrote:
> I understand you response but it does not solve the problem
HI,
You could also parse the data by:
input1<- input
library(stringr)
input2<-str_trim(gsub("[=+]","",input1))
dat1<-read.table(text=word(input2[!grepl("---",input2)& input2!="" &
!grepl("RAM|MiB",input2)],8,15),sep="",header=FALSE,stringsAsFactors=FALSE)
lst1<-split(dat1,cumsum(dat1$V3=="uuidd")
On 30.08.2013 15:27, Joana Costa wrote:
good afternoon, I´m writting because I´m having some problems with R
installation. I would like to install R essentials with spss Version 20 but I
can´t find wher in Cran site I can do the dowload speciffically of Essentials.
Can you help me? I do not u
Hi R-friends,
Can anyone explain the following strange behavior to me?
> as.Date( "4/25/71","%m/%d/%y")
[1] "1971-04-25"
> as.Date( "4/25/62","%m/%d/%y")
[1] "2062-04-25"
so 71 is converted to 1971, while 62 is converted to 2062? Does anyone know
why? And is there a simple way to specify the date?
It would be easier to diagnose the problem if you included an example
illustrating exactly what you did. I'll guess:
> a <- list(3,4,5)
> mean(a)
[1] NA
Warning message:
In mean.default(a) : argument is not numeric or logical: returning NA
> mean(as.numeric(a))
[1] 4
But that's just a guess, as
On Aug 30, 2013, at 8:27 AM, Joana Costa wrote:
> good afternoon, I´m writting because I´m having some problems with R
> installation. I would like to install R essentials with spss Version 20 but I
> can´t find wher in Cran site I can do the dowload speciffically of Essentials.
> Can you help
Take a look at the documentation as to what "%y" means:
‘%y’ Year without century (00-99). On input, values 00 to 68 are
prefixed by 20 and 69 to 99 by 19 - that is the behaviour
specified by the 2004 and 2008 POSIX standards, but they do
also say ‘it is expected
From: gabriel.wajnb...@hotmail.com
To: r-help@r-project.org
Subject: Fisher test for a more than two group of genesþ
Date: Wed, 28 Aug 2013 15:53:59 -0300
Good Afternoon,
My name is Gabriel, I'm doing an analysis if there is increase or decrease in
dependence on the mutated genes, using
Use is.numeric(list) on the list to see if it is a list of numbers or if it is
list of characters. If it is a list of characters (which could be the case if
you read it in using read.csv or the like) and it makes sense to convert to
numeric, then do something like:
list<-as.numeric(list)
-R
Maybe you want to check out the prevalence package
(http://cran.r-project.org/web/packages/prevalence/index.html), and its
development version on GitHub (https://github.com/brechtdv/prevalence/).
When you have two tests, and neither can be considered to be a 'gold
standard' (ie, SE & SP = 100%)
Hi Joana,
As far as I can see, R Essentials has nothing to do with R in terms of being an
official R package so you will not find it on CRAN.
It seems to be an IBM add-in for SPSS that let's you run R code within SPSS.
It should be available from http://sourceforge.net/projects/ibmspssstat/
G
On Aug 30, 2013, at 10:26 AM, Gerard Smits wrote:
> Hi All,
>
> This is a variant of a problem I posted yesterday (see below) where I found I
> had a large gap between my N= and he number I had evaluated using .(x). I
> seem to have trouble with newlines in a main title. I find now that all
Hi,
Using the same datasets:
dat1<- read.table(text="
V1 V2 V3
2 6 8
4 3 4
1 9 8
",sep="",header=TRUE)
dat2<- read.table(text="
V1 V2 V3
6 8 4
2 0 7
8 1 3
",sep="",header=TRUE)
sapply(seq_len(ncol(dat1)),function(i) cor(dat1[,i],dat2[,i],method="spearman"))
#[1] -1.000 0.5
When I try to apply mean to a list, I get the answer :
argument is not numeric or logical: returning NA
Could you help me?
(I am a very beginner)
--
View this message in context: http://r.789695.n4.nabble.com/mean-tp4674999.html
Sent from the R help mailing list archive at Nabble.com.
__
good afternoon, I´m writting because I´m having some problems with R
installation. I would like to install R essentials with spss Version 20 but I
can´t find wher in Cran site I can do the dowload speciffically of Essentials.
Can you help me? I do not understand musch of R software and I just nee
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R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-p
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R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-p
Hi All,
This is a variant of a problem I posted yesterday (see below) where I found I
had a large gap between my N= and he number I had evaluated using .(x). I seem
to have trouble with newlines in a main title. I find now that all works as
expected (no unsightly gap between my N= and the val
On 8/30/2013 3:36 AM, Uwe Ligges wrote:
>
>
> On 30.08.2013 11:59, Christofer Bogaso wrote:
>> Hello again,
>>
>> I have a string which I need to put in some legitimate date format.
>>
>> My string is: "MAY-14"
>>
>> And output format would be "05/01/2014", this should be of Date class,
>> so that
Maybe there is some built-in assumption about dating and when to change to the
new century. It seems to kick-in at 1968
as.Date("25/04/69", "%d/%m/%y")
as.Date("25/04/68", "%d/%m/%y")
as.Date("25/04/60", "%d/%m/%y")
John Kane
Kingston ON Canada
> -Original Message-
> From: frans.marc
Hi
From: mohan.radhakrish...@polarisft.com
[mailto:mohan.radhakrish...@polarisft.com]
Sent: Friday, August 30, 2013 3:16 PM
To: PIKAL Petr
Cc: r-help@r-project.org
Subject: RE: [R] Memory usage bar plot
Hello,
This memory usage should be graphed with time. Are there
examples of
Hi,
You may try:
x1<- c("4/25/71","4/20/64")
fun1<- function(x, year){
m1<-as.numeric(format(as.Date(x1,"%m/%d/%y"),"%y"))
m1<- ifelse(m1>year%%100,1900+m1,2000+m1)
m2<- paste0(gsub("(.*\\/).*$","\\1",x),m1)
as.Date(m2,"%m/%d/%Y")
}
fun1(x1,1950)
#[1] "1971-04-25" "1964-04-20"
str(fun1(x1,1950)
Hi
For reading data into R you shall look to read.table and similar.
For plotting ggplot could handle it. However I wonder if 100 times 50 bars is
the way how to present your data. You shall think over what do you want to show
to yourself or your audience. Maybe boxplots or scatterplots could b
This might do it:
> lhs=c('a','a','a','b')
> rhs=c('a','b','b','b')
>
>
> # function to determine differences
> f_diff <- function(l, r){
+ t_l <- table(l)
+ t_r <- table(r)
+ # compare 'l' to 'r'
+ sapply(names(t_l), function(x){
+ if (is.na(t_r[x])) return(t_l[x])
+
That explains it. Thanks for the info. Gerard
On Aug 30, 2013, at 8:42 AM, Marc Schwartz wrote:
>
> On Aug 30, 2013, at 10:26 AM, Gerard Smits wrote:
>
>> Hi All,
>>
>> This is a variant of a problem I posted yesterday (see below) where I found
>> I had a large gap between my N= and he n
Thank you a lot A.K.!
One more question:
I'd like to compute the Spearman's rank correlation coefficients for V1
(from dat1) and V1 (from dat2) and so on... Do you know how to do that?
I managed to write the Pearson's correlation product moment coefficient with
your sapply-approach, but I have no
## Here is a plot. The input was parsed with Jim Holtman's code.
## The panel.dumbell is something I devised to show differences.
## Rich
input <- readLines(textConnection("
Private + Shared = RAM used Program
96.0 KiB + 11.5 KiB = 107.5 KiB uuidd
108.0 KiB + 12.5 KiB = 12
Thanks Berend.
Yes that is right. I should get 5 values(results) of x_e because I have
five values of X. I wonder how can I fix it?
On 30/08/2013 13:13, Berend Hasselman wrote:
On 30-08-2013, at 09:44, Jonsson wrote:
I have three datasets that I want to compute the errors between them using
Hi,
In case, some of the list elements are vectors, this procedure would not work.
a1<- list(c(3,5),4,5:6)
as.numeric(a1)
#Error: (list) object cannot be coerced to type 'double'
The OP didn't provide any info as to how the data looks like. So, these are
just assumptions.
mean(unlist(a1)) #if
I am posting here the brilliant solutions, gently provided by Prof JC Nash
, to me; so that people struggling in the future with
the same issue can find a way through.
FYI, compared to the original Matlab implementation: 1) "it does not handle
the case with more than one input, and
2) (m > n)
It sounds like that column of data is not of type "date" at all. You cannot
have one element of a column different from the rest of the column. In a
data.frame you can have different types of data in different columns but not in
the same column.
Where mydata is your data.frame do :
str(mydata
Hi,
May be this helps:
rle(try)$values
#[1] 1 2 3 1 2 4
#or
aggregate(try,list(cumsum(c(1,abs(diff(try),unique)[,2]
#[1] 1 2 3 1 2 4
#or
res<-tapply(try,cumsum(c(1,abs(diff(try,head,1)
attr(res,"dimnames")<-NULL
res
#[1] 1 2 3 1 2 4
A.K.
I am trying to delete repeated values in an
Hi there,
I am attempting to access an Allegro Graph site, using SPARQL function in
RStudio and am getting this error message "Error: XML content does not seem to
be XML: ''. However, the same query that I am using within the SPARQL function
can get back the results when directly from the browse
Hi,
I'm just starting to learn how to use R and I'm trying to create a
histogram with 7 breaks. This is my code so far:
dat=read.table("titanic.csv",header=TRUE,sep=",",na.string=".")
age=dat$Age
breaks=seq(min(age),max(age),length=7)
hist(age,breaks,freq=FALSE)
I don't know why, but on the four
I appreciate your time, thank you in advance :)
I'd like to start 2 rstudio-server in one Linux machine, one daemon listen port
8787 running R-2.15, on daemon listen port 8788 running R-3.01.
Can I have named rstudio-server configurations so I can start rstudio-server
like this :
:rst
so I have to create a for loop of the geometric sequence
h(x,n)=1+x+x^2+x^3^4...x^n. I know that it would be easier to simply
vectorize the sequence to x^(0:n), but I am required to make the loop, and I
can't wrap my brain around how to loop it because the equation is so
simple.
--
View this m
Hi Casey,
Yes, if there are missing values than you don't strictly speaking know
what the minimum value is. You need to tell min() and max() to exclude
missing, i.e.,
breaks=seq(min(age, na.rm=TRUE),max(age, na.rm=TRUE),length=7)
Best,
Ista
On Fri, Aug 30, 2013 at 9:41 PM, Casey Zhang wrote:
>
The Rstudio support forum is at http://support.rstudio.com
Best,
Ista
On Fri, Aug 30, 2013 at 12:41 PM, qiulin wrote:
> I appreciate your time, thank you in advance :)
>
> I'd like to start 2 rstudio-server in one Linux machine, one daemon listen
> port 8787 running R-2.15, on daemon listen p
Hello,
Check the result of
> min(c(1,2,3,4,NA,6))
And read
> ?min
Hope this helps,
Pascal
2013/8/31 Casey Zhang
> Hi,
>
> I'm just starting to learn how to use R and I'm trying to create a
> histogram with 7 breaks. This is my code so far:
>
> dat=read.table("titanic.csv",header=TRUE,sep
Hello,
Here is the R-help mailing list. For Rstudio-sever support, please see
http://support.rstudio.org/help/discussions
Regards,
Pascal
2013/8/31 qiulin
> I appreciate your time, thank you in advance :)
>
> I'd like to start 2 rstudio-server in one Linux machine, one daemon listen
> port 8
On 08/30/2013 05:13 PM, Eva Prieto Castro wrote:
Hi,
I have a problem when I try to generate the Documentation pdf (from .rda
files)in Spanish during the package creation. Could you tell me the way I can
do it?.
Thanks in advance.
Hi Eva,
Without knowing what the error is, it is very difficu
>> I have a multi-panel lattice figure. It has an even number of
>> equal-width columns. I would like to center text across the columns.
>>
>> The xlab argument often handles this job nicely. But I want a more
>> flexible solution. (I have many strings and want to position them at
>> different heig
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