, although I
have read that it doesn't work in every platform. I don't know about
ubuntu. I would advise you to try it anyhow. Good luck with that!
All the best,
Alfredo
El jue., 23 may. 2019 a las 21:38, Bogdan Tanasa ()
escribió:
> Dear all,
>
> if you could help me please with a s
just letters put together, therefore R can't understand their
meaning, and returns NA.
I believe that your best shot is to go with gsub, so that you translate
your "string letters" to "string numbers", and then you can do as.numeric
no problem
Try this
vec<-c("1"
Hi Michael et al,
I solved by myself simply running the code below.
Thanks anyway for the answers
Alfredo
t <- read.csv(file="C:\\Temp\\radio_survey.csv", header=TRUE, sep=",")
t1 <- table(t$Preference, t$Sex)
t2 <- table(t$Preference, t$Age)
t3
Hi, I am very new to r and need help from you to do a correspondence
analysis because I don't know how to structure the following data:
Thank you.
Alfredo
library(ca,lib.loc=folder)
table <- read.csv(file="C:\\Temp\\Survey_Data.csv", header=TRUE, sep="
Dear all,
Is it possible to convert a data frame with variables (columns) and the
number of persons for each variable at a given time at the day (rows) to
DNAbin object type, so that I can calculate Tajima's divergence? I've been
trying, but I just can't find the way.
Thank you very much,
V1 V2
Hola, buenos dias.
Escribo porque me gustaría saber si puedo convertir un data frame que
consiste en variables (indicadas por columnas) y en filas (el número de
personas que tienen cada variable en determinada hora del dia.) en un tipo
de objeto DNAbin para poder calcular la divergencia de Tajima.
Hello.
I'm trying to install bioconductor software on my ubuntu linux computer system.
I've installed already the latest R version but when trying to install any
semiconductor package, I can't. After write the command on R, it says that I
have to upload my R to the new version but I have already
ty my question.
Best,
Alfredo
-Messaggio originale-
Da: Therneau, Terry M., Ph.D. [mailto:thern...@mayo.edu]
You will need to give more detail of exactly what you mean by "prune using a
validation set". THe prune.rpart function will prune at any value you want,
what I
I'd like to use a different data ( validation) set for pruning my
classification tree. Unfortunately there aren't arguments to get this in
prune.rpart().
Any suggestions?
Thanks!
Alfredo
[[alternative HTML version deleted]]
_
Hi Dave,
Thanks for your reply. I can't reproduce the last line of code your code.
The ifelse argument throws back an error saying it was not used.
A
On Fri, Feb 1, 2013 at 10:56 PM, David Winsemius wrote:
>
> On Feb 1, 2013, at 9:30 AM, Alfredo Tello wrote:
>
> Hi Folk
is=0.8,cex=0.8)
This labels all the tickmarks in my barplot, which is the default behavior.
I would like to label every other tick mark in the barplot -->
1,3,5,7 and so on.
Thank you very much for your help!
All the best,
A
--
Alfredo Tello (http://alfredotello.com)
Sustainable Aquacult
rm(list=ls())
2012/2/19 sagarnikam123
> i have variables stored in previous history,next time i log in,& load
> previous history,i want to use same variable name for using other values,
> & i don't know exactly how many are they ? & their names also..
>
> How to delete all variable created in cu
Hi Joel,
to replace the colnames:
colnames(dataframe <- )gsub("X","",colnames(dataframe))
to order by colnames:
dataframe <- dataframe[,colnames(dataframe)]
Alfredo
2012/2/17 Joel Fürstenberg-Hägg
> Dear all,
>
> I have a data frame in which the co
Hi,
could you paste the results?
Alfredo
2012/2/12 Suranga Kasthurirathne :
> Hi everyone,
>
> I'm an R newbie working with the poLCA module. I achieved my target without
> having to bother anyone, but It seems that I've got stuck at the last
> minute.
>
> My pro
Hi Dennis,
That works brilliantly! Thanks a lot!
All the best,
A
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Hi Folks,
Ran into something I'd really like to do in R simply/elegantly, but my R -
coding skills seem surpassed. This is the thing. Imagine the following data:
labs<-c("abcdef","abcgg","tgthefdk","tgtijuel","tgtnjmoi","gbnt","dlift")
dat<-c(0.5,0.25,1,2,16,0.250,4)
dframe<-data.frame(labs,dat)
0.1 4
[6,] 0.546000 2-0.1 5
[7,] 0.546000 3-0.1 2
[8,] 0.846000 2-0.1 3
[9,] 0.246000 2-0.1 1
[10,] 0.846000 2-0.4 6
Thanks in advance,
Alfredo
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354 intercept3 coeff3
10 -1.1011662 0.02894731 intercept3 coeff3
11 -0.4098710 -0.01231322 intercept3 coeff3
12 1.1511811 -0.63923140 intercept3 coeff3
Thanks in advance,
Alfredo
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Hi All,
I'd like to randomly sample a vector N times, where each successive random
sample increases in size. I have realised that the function sample does not
take vectors for arguments. For example,
x<-rnorm(20,0,1)
sample(x,c(1,2,3)) ## will only return one random sample of size 1.
The trick
Problem solved..it turned out there was a "NA" in one of the rows which
was truncating "rep"..
thanks anyway
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Thanks a lot to Steve Lianoglou and Peter Savicky for their help!
Alfredo
-Messaggio originale-
Da: Steve Lianoglou [mailto:mailinglist.honey...@gmail.com]
> I'd to match-merge 2 tables in such a manner that I keep all the rows in
> table 1, but not the rows that are in b
I'd to match-merge 2 tables in such a manner that I keep all the rows in table
1, but not the rows that are in both table 1 and 2.
Thank you for your help,
Alfredo
> master <- data.frame(ID=2001:2011)
> train <- data.frame(ID=2004:2006)
> valid <- ???
in this example
Hi,
I've a list of list.
I want to extract an element by the rownames.
I can extract it by:
data[[1]][[1]][[4]][1]
But I want to exctract it by a command like this:
data[[1]][["B0"]][["smac"]][["cont"]][1]
It's possible?
Thanks,
Alfredo
> Exist any function to associated with paste make this result more
> automatically.
you can try with gsub.
Alfredo
2011/4/8 Ronaldo Reis Junior :
> Hi,
>
> I have a very simple doubt.
>
> Look:
>
>> teste <- c("A","B","C")
>
Hi,
> a <- 4
> a*0.2
[1] 0.8
ok!!
Is there a method to obtain this:
> a*0.2
[1] 0.80
I need to round the number also with the zero.
Thanks in advance,
Alfredo
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ut
their generosity and kindness to the other list members.
Alfredo
On Sun, 27 Feb 2011 08:27:00 -0800, Mark Knecht
wrote:
> On Sat, Feb 26, 2011 at 10:11 PM, Bert Gunter wrote:
>> Are you a fan of James Joyce? Is the Caps key on your keyboard broken?
>>
>> -- Bert
>
&
Hi Folks,
I'm hoping someone could help me a bit. I have plotted the following:
###
tiff("plot7.tiff",units="cm",width=15,height=36,res=700)
layout(matrix(c(1,2,3,4,5,6),3,1),widths=lcm(c(15)),heights=lcm(c(12,12,12)),respect=TRUE)
plot.t1<-plot(data.dist
Hi and thanks for your replies (I'm sorry to be a bit late..),
I think I might be being a bit thick on this, but I truly would not be
asking if I had not already gone through the manuals and examples in the
web. I just want to be sure I got Jorge's example.
the function is a generic functio
Hi There,
Just a question regarding the function that is specified to boot (I have
read the help, the manual and online examples.). The description of
boot says that the second argument of "statistic" (non parametric bootstrap)
must be a vector of indices, frequencies or weights which def
r2")
> maxadjr(adjr)
12312
0.713 0.56
ok...
but whit Cp method don't work
> adjr <- leaps(x,y,method="Cp")
> maxadjr(adjr)
Errore in order(l$a) : l'argomento 1 non è di tipo vector
Thanks,
Alfredo
--
View this message in context:
http://r.78
Thanks A LOT guys for your posts and replying despite the fact that it might
have been a bit hard to get my point (I apologise for the latter).
Josh's and Jim's examples worked perfectly. A loop was certainly not the
most elegant solution.
What I want to do (for those who asked) is re-generate
Hi Everyone,
I have a 2-dim data.matrix(e.g., table1) in which row1 specifies a range of
values. row2 - rown specify the number of times I want to replicate each
corresponding value in row1. I can do this with the following function:
rep(c(table1[1,]),c(table1[X,])) #where X would go from 2 -
Hi,
I've three values. What is the best method to choice the lowest values
with an if function?
example:
a = 3
b = 1
c = 5
if (lowest(a,b,c) is a) {}
if (lowest(a,b,c) is b) {}
if (lowest(a,b,c) is c) {}
Thanks,
Al
Hi,
I'm using gsub, but I've a problem.
> print(i)
[1] "piante_venere.csv"
> gsub("\\.csv$", "", i)
[1] "piante_venere"
> gsub("^piante_", "", i)
[1] "venere.csv"
Can I combine the two expressions?
0.214 0.2099
Thanks,
Alfredo
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
t;> m <- merge(data1, data2, by = 0, all = TRUE, sort = FALSE)
>> sapply(unique(union(names(data1), names(data2))), function(n)Reduce('+',
>> m[grep(n, names(m))]))
>
> On Wed, Jan 20, 2010 at 10:59 AM, Alfredo Alessandrini
> wrote:
>> Hi,
>>
>
0 0.000
Thanks,
Alfredo
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Oops forgot to include the radius
ang<-runif(200,0,2*pi)
x<-cos(ang)*.1 +0.4
y<-sin(ang)*.1 +0.8
plot(x,y)
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Wolfgang Amadeus
Sent: Wednesday, January 20, 2010 5:26 AM
To: r-help
Subjec
ang<-runif(20,0,2*pi)
x<-cos(ang) +0.4
y<-sin(ang) +0.8
plot(x,y,ylim=c(-3,3),xlim=c(-3,3))
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Wolfgang Amadeus
Sent: Wednesday, January 20, 2010 5:26 AM
To: r-help
Subject: [R] question
Hi,
I've a problem with unmatched quotes with a bash script within a R script:
> system("awk 'NR>2 {FS="
> ";print$1","$2","$3","$4","$5","$6","$7","$8","$9","
What's the better method for obtain monthly value?
Thanks,
Alfredo
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and provide commente
.1370867 60.01 6 0.1122784
70.6358819 -12.1333488 82.41 7 0.4493264
8 -7.9629934 -15.6282544 60.01 8 0.1932665
9 -7.9962383 -7.1998453 62.01 9 0.1982783
10 -17.3282497 5.6302896 67.01 10 NA
11 -3.2569414 5.0152500 77.51 12 0.3724666
Thanks,
Alfredo
_
Hi,
I've two dataframe:
> snag_totale
AREA snag_ha
12 1.628128
23 10.274249
34 2.778503
45 73.764307
57 12.015985
> log_totale
AREAlog_ha
11 22.29846
22 17.16889
33 48.80377
44 144.18996
55 70.30962
66 61.81850
77 13.24876
>
How c
?
Hi,
The following script will run without asking me anything.
$ Rscript plot.R
> x=1:10
> y=1:10
> X11()
> plot(x,y)
> par(ask=T)
>
Regards,
Peng
On Sun, Sep 6, 2009 at 11:55 AM, RIOS,ALFREDO ARTURO wrote:
> Hi Peng
>
> I think this is what you are looking for
t(x,y)
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--
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Hi Peng
I think this is what you are looking for
par(ask=T)
Alfredo
On Sun Sep 06 12:52:31 EDT 2009, Peng Yu
wrote:
Hi,
In 'example(barplot)' running in R, I see 'Hit to see
next
plot:', then R waits for my input. I am wondering how to wait for
a
user response
I think should work
rep(c(1,2,3),10)
Alfredo
On Sat Aug 29 15:14:15 EDT 2009, njhuang86
wrote:
Hey guys,
I was wondering how to create this sequence: 1, 2, 3, 1, 2, 3, 1,
2, 3, 1,
2, 3... with the '1, 2, 3' repeated over 10 times.
Also, is there a simple method to genera
You might want to check vif in the Design package.
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Tomas Zelinsky
Sent: Tuesday, August 11, 2009 3:20 PM
To: r-help@r-project.org
Subject: [R] Heteroscedasticity in binary logit models
sample(c("a","b"))[1]
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of damien landais
Sent: Thursday, August 06, 2009 11:02 AM
To: R-help@r-project.org
Subject: [R] random between two values
Hi,
I would obtain a random value between
nc[154]
[1] NA
I must obtain only the first value, in this case: inc[82].
Regards,
Alfredo
2009/6/23 Henrique Dallazuanna
>
> Try this:
>
> which(diff(is.na(inc)) < 0)
>
> On Tue, Jun 23, 2009 at 11:00 AM, Alfredo Alessandrini
> wrote:
>>
>> Hi,
&
Now is ok...thanks.
Alfredo
2009/6/23 Alfredo Alessandrini :
> I've the NA value also between the value of the vector:
>
>> inc
> [1] NA NA NA NA NA NA NA
> [8] NA NA NA NA NA
tain the position of first value of the vector...
In this case is 82.
> inc[82]
[1] 13.09550
Regards,
Alfredo
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I extract a vector without the NA value?
Regards,
Alfredo
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A simple question :-)
I'm writing a R scritp(#!/usr/bin/Rscript)
How can execute a bash command inside the script, like the command for
change the directory (cd)?
Thanks in advance,
Alfredo
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NA
X1006402F X1006406F X1006407F X1006408F X1006421F X1006605F
1NA65NANANA45
>
How can I count the field =! NA??
Alfredo
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Hi,
How can I join two string?
frequency = 15
I want join the number frequency with a string.
Alfredo
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>for (i in files_rwl) {
> thisfile <- get(i)
> thisfile[is.na(thisfile)] <- 0
> assign(i, thisfile)
> }
>
> It's likely that you could condense it, but using the long form makes it
> clear what's happenin
> Don't use a loop for this. Do this.
I need to use a loop...
I've many data.
Alfredo
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i)] <- 0)
> ls()
[1] "cimfasy_rwl" "files_rwl" "files.rwl" "i" "rocquce_rwl"
>
> cimfasy_rwl
[1] 0
>
Thanks in advance,
Alfredo
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d column.
This isn't a problem, I utilise only the data that start from a
definite start date.
Alfredo
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gt; files_rwl
[1] "cimfasy_rwl" "rocquce_rwl"
> for (i in files_rwl) assign(gsub("\\_rwl$", "", i), i[is.na(i) ] <- 0)
> ls()
[1] "cimfasy" "cimfasy_rwl" "files_rwl" "files.rwl" "i"
[6] "roc
gt; class(a)
[1] "data.frame"
This loop import all the files rwl in a single data.frame ( a ).
Can I import a single files to a single data.frame? like this:
for (i in files) {cimfasy <- read.rwl(i,header=0)}
for (i in files) {rocquce <- read.rwl(i,header=0)}
Thanks in advance
> have you read the docs?
Yes, I'm reading the manual "An Introduction to R"..
> start with ?`[`
Ok...I haven't read it.
Thanks.
Alfredo
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quot;0028002F" "0028013F" "0028032F"
Ok
> colnames(rwl[,1])
NULL
why?? I expect: "0028002F"
Thanks in advance,
Alfredo
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PLE
3 1.78
1995 1.15 0.92 1.50 0.97 0.60 0.00
1996 0.00 0.00 0.00 0.00 0.00 0.00
thanks...
Alfredo
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2008/6/18 Alfredo Alessandrini <[EMAIL PROTECTED]>:
> I've a matrix like this:
>
> 1985 1.38 1.27 1.84 2.10 0.59 3.47
> 1986 1.05 1.13 1.21 1.54 0.21 2.14
> 1987 1.33 1.21 1.77 1.44 0.27 2.85
> 19
1.89 0.60 0.87
1995 1.13 1.04 1.19 1.52 1.13 1.78
Can I utilise a cumsum inverse? from 1995 to 1985?
Thank you,
Alfredo
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PLEASE do rea
>
> # set X's colnames to match d's
> colnames(X) <- colnames(d)
>
> # Now rbind()
> rbind(d, X)
thanks,
it's work...
Alfredo
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NA NANANANA
If I try to add X to d, I've an error like this:
rbind(X,d)
Errore in match.names(clabs, names(xi)) : names do not match previous names
Why?
Best wishes,
Alfredo
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0 0
> length (a)
[1] 9
I've a questions...
There is a function for show the lenght of single observation?
> (a)
[1] 12
Best Wishes,
Alfredo
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ed?
print (b)
a b
1 14 21
2 21 45
3 14 23
4 4 11
Best Wishes,
Alfredo
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wishes,
Alfredo
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Hi,
If I have this values:
21
23
14
58
26
How can I sum the values by a progression like this:
(21)
(21 + 23)
(21 + 23 + 14)
(21 + 23 + 14 + 58)
(21 + 23 + 14 + 58 + 26)
(21 + 23 + 14 + 58 + 26)
I've try with the function "loop"
Best Wishes,
Alfredo
[[a
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