David,
No hay mucho detalle, pero aqui van algunas ideas:
1. Construye una funcion que tome como argumentos los parametros con
los que generas tus muestras y construye una lista con los resultados.
Cada componente de la lista contendra las *B *muestras sobre las cuales
vas a realizar los
Gracias, Javier. Alguien tiene experiencia con Mac? Que tal la
comunicacion en OS X?
Saludos,
Jorge Velez.-
2015-02-26 0:54 GMT+11:00 Javier Marcuzzi javier.ruben.marcu...@gmail.com:
Estimados
No es una pregunta, pero si una buena noticia, le� que R y mysql en windows
ahora se llevan
O solo
R sum(combn(x, 2, prod))
[1] 14121
Saludos,
Jorge.-
2015-02-24 21:00 GMT+11:00 Carlos Ortega c...@qualityexcellence.es:
Hola,
Otra forma de hacerlo, es as�:
#---
x- c(24,12,45,68,45)
sum(apply(combn(x,2),2,prod))
[1] 14121
#---
Y te ahorras los l�os
Fernando,
Podrias intentar
R a - rep('a', 5)
R b - rep('b', 5)
R a
[1] a a a a a
R b
[1] b b b b b
R c(rbind(a, b))
[1] a b a b a b a b a b
Saludos,
Jorge.-
2015-02-24 23:49 GMT+11:00 Fernando Macedo ferm...@gmail.com:
Buenas a todos.
Relato el problema:
- tengo un archivo de 316
I Velez escribi�:
Fernando,
Podrias intentar
R a - rep('a', 5)
R b - rep('b', 5)
R a
[1] a a a a a
R b
[1] b b b b b
R c(rbind(a, b))
[1] a b a b a b a b a b
Saludos,
Jorge.-
2015-02-24 23:49 GMT+11:00 Fernando Macedo ferm...@gmail.com:
Buenas a todos.
Relato el
Hi Evgenia,
Try
test2 - function(data, TitleGraph){
pdf(paste0(TitleGraph, .pdf), width = 7, height = 5)
plot(data)
dev.off()
}
instead. Take a look at ?paste0 for more information.
HTH,
Jorge.-
On Tue, Feb 10, 2015 at 12:14 AM, Evgenia ev...@aueb.gr wrote:
test-function(data,
Hola Jesus,
Intenta lo siguiente, donde x es tu w.list:
R unlist(sapply(x, '[', 'a'))
a a
1 11
R unlist(sapply(x, '[', 'b'))
b1 b2 b3 b4 b5 b1 b2 b3 b4 b5
41 42 43 44 45 71 72 73 74 75
R unlist(sapply(x, '[', 'c'))
c c
X Z
Saludos,
Jorge.-
2015-01-29 22:37 GMT+11:00 Jesus Herranz
Hola Javier,
La ayuda de ?aictabe establece que la implementacion del AIC no funciona
para objetos de clase lmer (a la que pertenecen los modelos ajustados).
De ahi el error.
La idea es simplemente comparar los modelos basados en el AIC? O tienes
pensado algo mas? Si es lo primero, los objetos
Hola Hector,
No soy experto, pero en
http://r.789695.n4.nabble.com/Fitting-weibull-exponential-and-lognormal-distributions-to-left-truncated-data-td869977.html
http://www.r-bloggers.com/r-help-follow-up-truncated-exponential/
http://www.jstatsoft.org/v16/c02/paper
hay algunas ideas.Espero
Gracias Emilio por el codigo en R.
Hace poco estuve revisando una situacion similar a la que ilustras en tu
mensaje (i.e., seleccion de modelos e inferencia). Dale una mirada a estos
dos articulos:
http://www-stat.wharton.upenn.edu/~berkr/PoSI-submit.pdf
Dear all,
Given vectors x and y, I would like to compute the proportion of
entries that are equal, that is, mean(x == y).
Now, suppose I have the following matrix:
n - 1e2
m - 1e4
X - matrix(sample(0:2, m*n, replace = TRUE), ncol = m)
I am interested in calculating the above proportion for
Hola Juan Carlos,
Quizas lo siguiente pueda serte util:
# test
R s - Merluccius merluccius
R strsplit(s, )
[[1]]
[1] Merluccius merluccius
R strsplit(s, )[[1]]
[1] Merluccius merluccius
R s - strsplit(s, )[[1]]
R paste0(substr(s[1], 1, 1), ., s[2])
[1] M.merluccius
# funcion
convertir -
What about
ifelse(w 0, 0, w)
See ?ifelse for more information.
Best,
Jorge.-
On Sat, Dec 20, 2014 at 3:26 PM, Esra Ulasan esra_ula...@icloud.com wrote:
Hello,
I have tried the solve the non-negativity constraint if else function in
R. But I have done something wrong because it still
Dear jeff6868,
Here is one way:
ifelse(with(data, c(0, diff(mydata))) != 1, 0, 1)
You could also take a look at ?rle
HTH,
Jorge.-
On Mon, Dec 15, 2014 at 9:33 PM, jeff6868 geoffrey_kl...@etu.u-bourgogne.fr
wrote:
Hello dear R-helpers,
I have a small problem in my algorithm. I have
Dear Dennis,
Assuming that your data.frame() is called dd, the following should get you
started:
colnames(dd[,-1])[apply(dd[,-1], 1, function(x) which(x == 'Yes'))]
HTH,
Jorge.-
On Sat, Nov 1, 2014 at 12:32 PM, Fisher Dennis fis...@plessthan.com wrote:
R 3.1.1
OS X
Colleagues,
I have a
Carlos,
Una forma de resolverlo es usando la funcion ?rmultinom
Saludos,
Jorge.-
2014-10-19 20:32 GMT+11:00 Carlos Hern�ndez-Castellano
carlos.hernandezcastell...@gmail.com:
Saludos compa�eros/as.
Consideren una matriz de n columnas y n filas.
Quiero distribuir aleatoriamente 6 n�meros
Hola Eric,
Revisa
?%in%
Saludos,
Jorge.-
2014-09-04 8:41 GMT+10:00 eric ericconchamu...@gmail.com:
Estimados, tengo un data.frame con una columna que tiene tres diferentes
niveles (aunque la columna no es propiamente de un factor, son solo tres
letras diferentes), por ejemplo c, t y s, y
Hola Angela,
Una forma de resolver tu problema es utilizando la funcion difftime. En
?difftime hay varios ejemplos. Observa el argumento units en
as.numeric(..., units = 'seconds').
Saludos,
Jorge.-
2014-08-26 8:16 GMT+10:00 Angela Andrea Camargo Sanabria
angela.andrea.cama...@gmail.com:
Hi James,
Try
mat[, apply(mat, 2, function(x) any(diff(x) == 1))]
HTH,
Jorge.-
On Fri, Aug 22, 2014 at 10:18 PM, James Wei zwei0...@hotmail.com wrote:
Hi all,
I have a matrix with consecutive and non-consecutive numbers
in columns. For example, the first 2 columns have consecutive
Buenas noches Javier y Jos�,
Estoy en contra de usar attach(), asi que propongo la siguiente alternativa
con with():
# paquete
require(epicalc)
# los argumentos en ... pasan de epicalc:::cc
# ver ?cc para mas informacion
foo - function(var1, var2, var3, ...){
or1 - cc(var1, var2, ...)
or2 -
Hola Ernesto,
Te sugiero trabajar con el paquete texreg:
Philip Leifeld (2013). texreg: Conversion of Statistical Model Output in
R to
LaTeX and HTML Tables. Journal of Statistical Software, 55(8), 1-24. URL
http://www.jstatsoft.org/v55/i08/.
Saludos,
Jorge.-
2014-08-19 9:26 GMT+10:00
, Aug 16, 2014 at 6:48 PM, Jorge I Velez jorgeivanve...@gmail.com
wrote:
Dear Kate,
Assuming you have nuclear families, one option would be:
x - read.table(textConnection(Family.ID Sample.ID Relationship
14 62 sibling
14 94 father
14 63 sibling
14
2702 3456 sibling 0 842
2702 9980 sibling 0 842
3064 3 father 0 0
3064 4 mother 0 0
3064 5sibling 879 880
3064 86 sibling 879 880
3064 87 sibling 879 880
On Sat, Aug 16, 2014 at 9:31 PM, Jorge I Velez jorgeivanve...@gmail.com
wrote:
Dear Kate,
Try
Dear Kate,
Assuming you have nuclear families, one option would be:
x - read.table(textConnection(Family.ID Sample.ID Relationship
14 62 sibling
14 94 father
14 63 sibling
14 59 mother
17 6004 father
17 6003 mother
17 6005
Dear Sohail,
Using Jim's data set skdat, two more options would be
# first option
d - with(skdat, table(ID, lettertag))
names - colnames(d)
d - c(list(rownames(d)), lapply(1:ncol(d), function(i) as.numeric(d[,i])))
names(d) - c('ID', names)
d
# second option
d - with(skdat, table(ID,
is a list.
How do I covert it back a dataframe?
-Sohail
On Fri, Aug 15, 2014 at 5:37 AM, Jorge I Velez jorgeivanve...@gmail.com
wrote:
Dear Sohail,
Using Jim's data set skdat, two more options would be
# first option
d - with(skdat, table(ID, lettertag))
names - colnames(d)
d - c
Hola Miguel,
A que te refieres con y nada? Obtienes algun error? Algun mensaje? Has
probado con scan() y/o readLines()?
Saludos,
Jorge.-
2014-08-15 7:38 GMT+10:00 Miguel Fiandor Guti�rrez
miguel.fiandor.gutier...@gmail.com:
Hola,
Pens� que esto iba a ser trivial en R, pero me estoy
Hola Alfredo,
Algunos comentarios/observaciones:
1. No uses attach. Mejor, explora la funcion with() y/o within(). attach
es muy peligroso.
2. Solo por curiosidad, como hiciste para crear la tabla usando
latabla$ciudad de origen? Supongo que deberia ser latabla$ciudad de
origen
3. Lo que
Dear Ron,
What about this?
set.seed(123)
d - 4
x1 - sample(0:1, d, TRUE)
x2 - sample(0:1, d, TRUE)
x1
x2
expand.grid(x1 = x1, x2 = x2)
See ?expand.grid for more information.
Best,
Jorge.-
On Sat, Aug 9, 2014 at 7:46 PM, Ron Michael ron_michae...@yahoo.com wrote:
Hi,
Let say I have 2
Estimado Prof. Di Rienzo,
Creo que lo que busca puede hacerlo con la función download.packages()
Saludos cordiales,
Jorge.-
2014-07-22 14:03 GMT+10:00 Julio Alejandro Di Rienzo
dirienzo.ju...@gmail.com:
Hola
Alguien sabe como descargar una lista de librerías de R en formato zipeado.
Por
Hola Juan,
Una forma es la siguiente:
1. Debes decirle a R donde estan los ficheros. Para ello usa setwd()
2. Determina el nombre de los ficheros. Usa list.files()
3. Toma el nombre de cada fichero, leelo y genera el data.frame() que
necesitas. Al exportar el data.frame() usando
Dear Anupam,
Try
boxplot(DISO ~ POS * NODE_CAT, data = yourdata)
Another option would be the last example in ?boxplot
HTH,
Jorge.-
On Fri, Jul 11, 2014 at 4:38 PM, anupam sinha anupam.cont...@gmail.com
wrote:
Dear all,
I need some help with plotting boxplots in groups. I have a
julio de 2014, 14:34, Jorge I Velez jorgeivanve...@gmail.com
escribió:
Hola Alberto,
Necesitas
as.Date(as.numeric(as.Date(Sys.time())), origin = '1970-01-01')
Esta parte
as.numeric(as.Date(Sys.time()))
# 16260
te da el numero de dias que han transcurrido desde Ene 1, 1970. Luego
Hola Juan Antonio,
Has pensado considerar una aproximacion diferente? De ser asi, explora
?cut y
?car:::recode.
Saludos,
Jorge.-
2014-07-10 16:58 GMT+10:00 juan(uned) j...@edu.uned.es:
Estimados compañeros, hoy me ha surgido una duda, quizás trivial, pero que
no encuentro sentido. Tengo
Hola Alberto,
Necesitas
as.Date(as.numeric(as.Date(Sys.time())), origin = '1970-01-01')
Esta parte
as.numeric(as.Date(Sys.time()))
# 16260
te da el numero de dias que han transcurrido desde Ene 1, 1970. Luego,
utilizando ese dia/año como origen, determinas la fecha actual.
Saludos,
Jorge.-
Estimado Alejandro,
Lo mejor es trabajar con listas, sea creadas antes de o despues de leer los
datos (esto ultimo automaticamente desde R). En cuanto a los nombres de
las variables, creo que ahorras tiempo y problemas si los incluyes.
A continuacion un ejemplo (necesitas el paquete mets):
#
fuente
permanente de dolor de cabeza. Mi recomendaciones usar siempre que podáis
UTF-8.
Fran
El 03/07/2014 9:57, Jorge I Velez escribió:
Hola Eric,
Me incliniaria mas por un problema de enconding. Intenta agregando
enconding = 'latin1' al final de read.csv()
A lo mejor enviandonos
Hola Eric,
Me incliniaria mas por un problema de enconding. Intenta agregando
enconding = 'latin1' al final de read.csv()
A lo mejor enviandonos tu sessionInfo() podriamos ayudarte un poco mas.
Saludos,
Jorge.-
2014-07-03 5:32 GMT+10:00 neo ericconchamu...@gmail.com:
Estimada comunidad,
] - do.call(rbind, r)
X
})
# user system elapsed
# 0.125 0.011 0.074
wide2 - wide1
wide2$id - as.character(wide2$id)
wide$id - as.character(wide$id)
all.equal(wide, wide2, check.attributes=F)
#[1] TRUE
A.K.
On Sunday, June 29, 2014 11:48 PM, Jorge I Velez jorgeivanve
Dear R-help,
I am working with some data stored as filename.txt.gz in my working
directory.
After reading the data in using read.table(), I can see that each of them
has four columns (variable, id, outcome, and rate) and the following
structure:
# sample data
x2 - data.frame(variable =
Hi Kate,
You could try
sum(X[, 1] == 1 X[, 2] == 1)
where X is your data set.
HTH,
Jorge.-
On Sun, Jun 22, 2014 at 12:57 AM, Kate Ignatius kate.ignat...@gmail.com
wrote:
I have 4 columns, and about 300K plus rows with 0s and 1s.
I'm trying to count how many rows satisfy a certain
Hola Jose,
Me funciona perfectamente:
install.packages('ergm')
#--- Please select a CRAN mirror for use in this session ---
# also installing the dependencies 'statnet.common', 'trust'
#snipped
require(ergm)
#snipped
sessionInfo()
#R version 3.0.2 Patched (2013-12-11 r64449)
#Platform:
Hi Pascal,
Perhaps I am missing something, but what about changing passing ylim = c(0,
10) to barp()?
Best,
Jorge.-
On Fri, Jun 13, 2014 at 7:50 PM, Pascal Oettli kri...@ymail.com wrote:
Dear list,
Please consider the following example:
library(plotrix)
barp(c(2,3,4,5,6,7,8),
Dear Juan,
Perhaps the last example in
http://stat.ethz.ch/R-manual/R-devel/library/utils/html/txtProgressBar.html
is what you are looking for.
Best,
Jorge.-
On Thu, Jun 12, 2014 at 8:49 PM, Juan Andres Hernandez
jhernandezcabr...@gmail.com wrote:
Hi I need to print the iteration number of
Hi Bill,
You need
require(plyr)
?rbind.fill
and then the rest of the code you already tried.
Best,
Jorge.-
On Mon, Jun 2, 2014 at 3:49 AM, Bill Bentley valuetr...@gmail.com wrote:
The following works as it should...
both-rbind(females,males)
both
workshop gender q1 q2 q3 q4
1
Try
options(digits = 22)
168988580159 * 36662978
# [1] 6195624596620653568
HTH,
Jorge.-
On Sun, May 4, 2014 at 10:44 PM, ARTENTOR Diego Tentor
diegotento...@gmail.com wrote:
Trying algorithm for products with large numbers i encountered a difference
between result of 168988580159 *
Hi Nevil,
Try
apply(A, 2, function(x) x == B)
HTH,
Jorge.-
On Fri, May 2, 2014 at 6:46 PM, nevil amos nevil.a...@gmail.com wrote:
I wish to return True in a matrix for only the first match of a value
per row where the value equals that in a vector with the same number of
values as rosw
Hi Kristi,
Try
out1970$smoot
HTH,
Jorge.-
On Fri, May 2, 2014 at 10:00 AM, Kristi Glover kristi.glo...@hotmail.comwrote:
Hi R User,
I am wonedring how I can extract a part of objects from list.
For example
str(out1970)
List of 8
$ comm : num [1:16, 1:57] 1 1 1 1 1 1 1 1 1 1 ...
Hi Xueming,
Try
(1:length(bo))[rank(bo)]
In a function the above would be
f - function(x){
N - length(x)
(1:N)[rank(x)]
}
f(bo)
# [1] 2 6 3 5 1 4
HTH,
Jorge.-
On Sat, Apr 26, 2014 at 7:54 PM, xmliu1...@gmail.com xmliu1...@gmail.comwrote:
Hi,
could anybody help me to find a
Hi Beatriz,
Try
paste(val_mapped_petpe_, 1976:1981, 01.txt, sep=)
Best,
Jorge.-
On Mon, Apr 21, 2014 at 6:43 PM, Beatriz R. Gonzalez Dominguez
aguitatie...@hotmail.com wrote:
Dear all,
I'm trying to create a loop to select a series of files into my computer
but I haven't been successful
Hi there,
Try
X[X %% 3 == 0]
HTH,
Jorge.-
On Thu, Mar 27, 2014 at 6:46 PM, Prabhakar Ghorpade
dr.prabhaka...@gmail.com wrote:
Hi,
here's my code
X - 1:100
I want to select number divisible by 3 out of them how can I select it?
( I tried following
X - 1:100
DIV - Y - X/3
But I
Hi Luigi,
Thanks for sending the data in reproducible format. Perhaps something
like this?
aggregate(my.data[,3], list(my.data[,2]), FUN = function(x) t.test(x)$
conf.int[1:2])
#Group.1 x.1 x.2
#1 Unstimulated 5.296492e+02 2.410510e+03
#2ESAT6 9.105338e+00
Hi Eliza,
Perhaps the following?
matpoints(t(dat), type = 'l')
HTH,
Jorge.-
On Sat, Mar 22, 2014 at 10:18 PM, eliza botto eliza_bo...@hotmail.comwrote:
Dear useRs,
I have two column vectors of different lengths say x=1,2,3,4,5,6,7,8 and
y=1,2,3,4,5. I wanted to plot them by using
You are welcome, Eliza.
If I understand correctly, the following will do:
x - 1:8
y - 1:5
matrix(apply(expand.grid(x = y, y = x), 1, function(r) paste0((, r[1],
,, r[2], ))), ncol = length(x))
Best,
Jorge.-
On Sat, Mar 22, 2014 at 10:37 PM, eliza botto eliza_bo...@hotmail.comwrote:
Thankyou
Hi Catalin,
The following should give you some ideas:
set.seed(123)
x - rpois(50, 2)
x
idx - duplicated(x)
x[idx] - 0
x
Best,
Jorge.-
On Thu, Mar 13, 2014 at 11:35 PM, catalin roibu catalinro...@gmail.comwrote:
Dear all!
Is there a possibility to replace all duplicates values in data
Hi Berry,
What about using
NROW(input)
?
Best,
Jorge.-
On Sat, Feb 15, 2014 at 2:26 AM, Berry Boessenkool
berryboessenk...@hotmail.com wrote:
Hi,
In my function, I want to allow input to be a vector or a data.frame.
Certain operations need to be done if the length or nrows exceeds one,
Dear Eve,
See http://cran.r-project.org/web/packages/languageR/index.html The name
of the package is languageR, not LanguageR.
Best,
Jorge.-
On Thu, Feb 13, 2014 at 3:39 PM, Eve Dupierrix evedupier...@gmail.comwrote:
Hi,
I want to install languageR but is doesn't work.
I tried it by two
Try
R plot(1:10)
R text(1,3, expression((x, *hat(y)*)), pos=3)
Best,
Jorge.-
On Sun, Nov 24, 2013 at 10:51 AM, David Arnold dwarnol...@suddenlink.netwrote:
Hi,
I'd like to do this:
text(1,3,(x,yhat),pos=3)
But using (x,hat(y)). Any suggestions?
D.
--
View this message in
Dear Dr. Vokey,
Here is one approach, although may not be the more efficient:
x - matrix(1:8, ncol = 4)
x
# [,1] [,2] [,3] [,4]
#[1,]1357
#[2,]2468
t(x[, ncol(x):1])
# [,1] [,2]
#[1,]78
#[2,]56
#[3,]34
#[4,]12
Consider the following:
f - function(x){
m - mean(x, na.rm = TRUE)
x[is.na(x)] - m
x
}
apply(de, 2, f)
HTH,
Jorge.-
On Tue, Jul 30, 2013 at 2:39 AM, iza.ch1 iza@op.pl wrote:
Hi everyone
I have a problem with replacing the NA values with the mean of the column
which contains them. If
Dear Andras,
Try
a[findInterval(b, a)]
[1] 8 32
HTH,
Jorge.-
On Tue, Jun 18, 2013 at 10:34 PM, Andras Farkas motyoc...@yahoo.com wrote:
Dear All,
would you please provide your thoughts on the following:
let us say I have:
a -c(1,5,8,15,32,69)
b -c(8.5,33)
and I would like to
they may not
be minimum?
a - c(1, 8, 9)
b - c(2,3)
Then what are the 2 closest values of a to b?
-- Bert
On Tue, Jun 18, 2013 at 5:43 AM, Jorge I Velez jorgeivanve...@gmail.com
wrote:
Dear Andras,
Try
a[findInterval(b, a)]
[1] 8 32
HTH,
Jorge.-
On Tue, Jun 18, 2013
Hi Miao,
Try
attributes(test1)[[1]]
HTH,
Jorge.-
On Tue, Jun 11, 2013 at 3:49 PM, jpm miao miao...@gmail.com wrote:
Hi,
I have a structure, which is the result of a function
How can I access the elements in the gradient?
dput(test1)
structure(-1.17782911684913, gradient =
Hi there,
You need a function for your statistic:
boot(x, function(x, index) mean(x[index]), R = 1000)
ORDINARY NONPARAMETRIC BOOTSTRAP
Call:
boot(data = x, statistic = function(x, index) mean(x[index]),
R = 1000)
Bootstrap Statistics :
original biasstd. error
t1*
Daniel,
You need == instead of =.
HTH,
Jorge.-
Sent from my phone. Please excuse my brevity and misspelling.
On Jun 6, 2013, at 10:36 AM, Daniel Tucker dtuck...@u.rochester.edu wrote:
Also tried this but results werent any different
subset1- subset(dframe, glb_ind=Y | sample==1 |
Try
names(ResList)
HTH,
Jorge.-
Sent from my phone. Please excuse my brevity and misspelling.
On Jun 5, 2013, at 12:34 AM, Sparks, John James jspa...@uic.edu wrote:
Dear R Helpers,
I have a fairly complicated list of data frames. To give you an idea of
the structure, the top of the str
Try
rep(1:length(v), v)
HTH,
Jorge.-
On Fri, May 17, 2013 at 8:53 PM, Stefan Petersson ste...@inizio.se wrote:
I want to create a sequence, repeating each element according to a vector.
I have this:
v - c(4, 4, 4, 3, 3, 2)
And want to create this:
1 1 1 1 2 2 2 2 3 3 3 3 4 4 4 5 5 5
-32 -128
f[f0]
[1] 4 16 64
f[f=4]
[1] 16
2013/5/8 Jorge I Velez jorgeivanve...@gmail.com
f [ f 0 ]
On Wed, May 8, 2013 at 11:54 AM, jpm miao miao...@gmail.com wrote:
Hi,
I have a vector f with some negative columns. I remember that there is
an easy expression that can find out
f [ f 0 ]
On Wed, May 8, 2013 at 11:54 AM, jpm miao miao...@gmail.com wrote:
Hi,
I have a vector f with some negative columns. I remember that there is
an easy expression that can find out negative items. Can someone tell me
how I can do it?
It seems to be
f[i such that f[i]0
rownames(a)
perhaps?
HTH,
Jorge.-
On Mon, May 6, 2013 at 6:03 PM, jpm miao miao...@gmail.com wrote:
Hi,
Below is the output from an R package. The first column (4, 5, 6, 7,
which is unnamed) is the company name (code), while the second column
efficiency is the performance of each
Hi Kasia,
You need
subset(REC2, INFECTION==Infected )
(note the space after Infected).
HTH,
Jorge.-
On Fri, May 3, 2013 at 7:48 PM, Katarzyna Kulma
katarzyna.ku...@gmail.comwrote:
Hi everyone,
I know there have been several requests regarding subsetting before, but
none of them really
Christofer,
The following should get you started:
r - Mat[match(rownames(Mat), Subscript_Vec),]
rownames(r) - Subscript_Vec
r
HTH,
Jorge.-
On Mon, Apr 29, 2013 at 11:38 PM, Christofer Bogaso
bogaso.christo...@gmail.com wrote:
Hello again,
Let say I have 1 matrix:
Mat - matrix(1:12, 4,
Sorry, the first line should have been
Mat[match( Subscript_Vec, rownames(Mat)),]
and the rest remains the same.
Best,
Jorge.-
On Mon, Apr 29, 2013 at 11:45 PM, Jorge I Velez jorgeivanve...@gmail.comwrote:
Christofer,
The following should get you started:
r - Mat[match(rownames(Mat
Dear Dr. Harding,
Try
sapply(L, [, 1)
sapply(L, [, 2)
HTH,
Jorge.-
On Thu, Apr 25, 2013 at 8:16 PM, Ted Harding ted.hard...@wlandres.netwrote:
Greetings!
For some reason I am not managing to work out how to do this
(in principle) simple task!
As a result of applying strsplit() to a
Try
subset(Dat, AA == A | (AA == B BB == b))
HTH,
Jorge.-
On Wed, Apr 24, 2013 at 8:21 PM, Christofer Bogaso
bogaso.christo...@gmail.com wrote:
Hello again,
Let say I have following data:
Dat - structure(list(AA = structure(c(3L, 1L, 2L, 1L, 2L, 3L, 3L, 2L,
3L, 1L, 1L, 3L, 3L, 2L, 2L,
Mike,
You need
subset(agoa, agoa$X.1 == AGOA )
instead of
subset(agoa, agoa$X.1 == AGOA)
(note the space after the last A in AGOA.
HTH,
Jorge.-
On Tue, Apr 23, 2013 at 7:14 AM, Mihai Nica mihain...@yahoo.com wrote:
I can't understand what is happening. This is the code and results:
Dear Janesh,
Here is one way:
# note x is a character
x -
73167176531330624919225119674426574742355349194934969835203127745063262395783180169848018694788518438586156078911294949545950173795833195285320880551112540698747158523863050715693290963295227443043557
k - nchar(x) # digits in x
b - 5 #
Hi Carrington,
You also need the boot package (see
http://stat.ethz.ch/R-manual/R-patched/library/boot/html/inv.logit.html )
As for the other function, please load the arm package, e.g.,
require(arm)
require(boot)
and then you will be able to use the functions mentioned below.
HTH,
Jorge.-
Dear Catalin,
You can look at ?nls.
Alternatively, you could also consider a linear model as follows, where d
is your data:
# plot your data
with(d, plot(cls, proc, las = 1))
# linear model
fit - lm(proc ~ I(1/cls) + I((1/cls)^2), data = d)
summary(fit)
# plotting
with(d, plot(cls, proc, las
the regression coefficients.
On 10 April 2013 12:19, Jorge I Velez jorgeivanve...@gmail.com wrote:
Dear Catalin,
You can look at ?nls.
Alternatively, you could also consider a linear model as follows, where
d is your data:
# plot your data
with(d, plot(cls, proc, las = 1))
# linear
Dear Miao,
Check
require(MASS)
?mvrnorm
for some ideas.
HTH,
Jorge.-
On Wed, Apr 3, 2013 at 4:57 PM, jpm miao wrote:
Hi,
I conduct a panel data estimation and obtain estimators for two of the
coefficients beta1 and beta2. R tells me the mean and covariance of the
distribution of
Pablo,
Check the qqPlot function in car:
require(car)
qqPlot(x, dist = gamma, shape = 1.7918012, rate = 0.9458022)
Best,
Jorge.-
On Tue, Apr 2, 2013 at 4:41 AM, pablo.castano wrote:
Hi,
I want to create upper and lower 95% confidence intervals for a p-p plot of
an empirical
Hi Dimitri,
If I understood correctly, the following will do:
system.time(sum1 - apply(mycombos, 1, function(x) sum(values1[x])))
system.time(sum2 - apply(mycombos, 1, function(x) sum(values2[x])))
system.time(sum3 - apply(mycombos, 1, function(x) sum(values3[x])))
cbind(sum1, sum2, sum3)
HTH,
Sahana,
The notation
df[a,b)]
is plain wrong. I think you meant (but I may be mistaken)
df[a, b]
and I am not still sure if that would work in your example. Have you
instead considered subset()? E.g.,
subset(df, a = 10 b = 10)
See ?subset for more details.
Also, df is a very bad name
Or simply
subset(dat, a 0)
HTH,
Jorge.-
On Thu, Mar 21, 2013 at 6:58 PM, Michael Weylandt wrote:
On Mar 21, 2013, at 7:39, Pierrick Bruneau pbrun...@gmail.com wrote:
Hi Borja,
You may issue:
attach(data)
No -- bad idea -- dangerous -- confusing statefulness, etc. (See
Try
ifelse(ABS ==1 | DEFF == 1, 1, 0)
HTH,
Jorge.-
On Fri, Mar 22, 2013 at 12:02 AM, Tasnuva Tabassum t.tasn...@gmail.comwrote:
I have two indicator variables ABS and DEFF. I want to create another
indicator variable which will take value 1 if either ABS=1 or DEFF=1.
Otherwise, it will
Hi Jim,
Try either of the following (untested):
sum( x[1, ] 12 x[2, ] 12)
sum(apply(x, 2, function(x) x[1] 12 x[2] 12))
where x is your 2x1000 matrix.
HTH,
Jorge.-
On Tue, Mar 19, 2013 at 12:03 AM, Jim Silverton wrote:
Hi,
I have a 2 x 1 matrix of confidence intervals. The
Thats cumbersome, Arun.
sum(mat1[,1] 12 mat1[,2] 12)
[1] 17
will do the job and even faster:
system.time(replicate(1, sum(mat1[,1] 12 mat1[,2] 12)))
# user system elapsed
# 0.067 0.001 0.078
HTH,
Jorge.-
On Tue, Mar 19, 2013 at 1:06 AM, arun wrote:
Hi,
Try this:
If you don't use apply() it would be even faster:
system.time(sum(mat2[,1] 12 mat2[,2] 12))
user system elapsed
0.004 0.000 0.003
Regards,
Jorge.-
On Tue, Mar 19, 2013 at 1:21 AM, arun wrote:
Hi,
Jorge's method will be faster.
#system.time(res1-sum(apply(mat2,1,function(x)
Dear Johannes,
May not be the best way, but this looks like what you described:
x - c(a1b1,a2b2,a1b2)
x
[1] a1b1 a2b2 a1b2
substr(x, 1, 2)
[1] a1 a2 a1
substr(x, 3, 4)
[1] b1 b2 b2
HTH,
Jorge.-
On Wed, Mar 13, 2013 at 7:37 PM, Johannes Radinger wrote:
Hi,
I have a vector of strings
Dear Catalun,
If I understood your description, please see ?%in% and try
subset(x, names(x) %in% c(1834,1876,1901,1928,2006) )
where x is your data.
HTH,
Jorge.-
On Wed, Mar 13, 2013 at 9:25 PM, catalin roibu wrote:
Hello all!
I have a problem with R. I try to merge data like this:
Dear SH,
Hmmm... what about
substr(tempdf$name, 4, 6))
?
HTH,
Jorge.-
On Thu, Mar 14, 2013 at 1:06 AM, SH empti...@gmail.com wrote:
Dear list:
I would like to extract three letters from first and second elements
in one column and make a new column.
For example below,
tempdf =
like to have letters from first and second elements if possible.
Thanks for replying,
Steve
On Wed, Mar 13, 2013 at 10:10 AM, Jorge I Velez
jorgeivanve...@gmail.com wrote:
Dear SH,
Hmmm... what about
substr(tempdf$name, 4, 6))
?
HTH,
Jorge.-
On Thu, Mar 14, 2013 at 1
and last name
and to combine them to make another column 'abb'. The column 'abb' is
to be a my final product. I can make column 'abb' using 'paste'
function once I have two parts from the first column 'name'.
Thanks,
Steve
On Wed, Mar 13, 2013 at 10:17 AM, Jorge I Velez
jorgeivanve
Is the following that you are looking for?
unlist(lapply(x.list, [, 2))
HTH,
Jorge.-
On Mon, Mar 11, 2013 at 9:52 PM, ishi soichi wrote:
say I have a matrix and lists like
x - matrix(c(12.1, 3.44, 0.1, 3, 12, 33.1, 1.1, 23), nrow=2)
x.list - lapply(seq_len(nrow(x)), function(i) x[i,])
One option would be
x - list(c(12.1, 0.1, 12, 1.1), c(3.44, 3, 33.1, 23))
do.call(c, apply(do.call(rbind, x), 2, list))
HTH,
Jorge.-
On Fri, Mar 8, 2013 at 9:06 PM, ishi soichi soichi...@gmail.com wrote:
Thanks. The result should be a list of lists like...
x
[[1]]
[1] 12.10 3.44
Dear Marin,
May be not the cleanest way to do it, but the following seems to work:
write.table(as.character(round(pi, 10)), pi.txt, row.names = FALSE,
col.names = FALSE, quote = FALSE)
Best,
Jorge.-
On Fri, Mar 8, 2013 at 11:24 AM, Marino David davidmarino...@gmail.comwrote:
Hi Bert,
I
If I understood correctly,
lapply(x, [, 1:3)
will do what you want.
HTH,
Jorge.-
On Fri, Mar 8, 2013 at 5:05 PM, ishi soichi wrote:
hi. I have a list like
x - list(1:10,11:20,21:30)
It's a sort of a 3 x 10 matrix in list form.
I would like to reduce the dimension of this list.
it
Dear Julien,
Check
citation('stats')
HTH,
Jorge.-
On Wed, Mar 6, 2013 at 12:05 AM, Julien Mvdb julien.m...@gmail.com wrote:
The question is in the title.
Then, I would like to know how I should refer to the documentation
regarding the use of each functions.
Thanks,
Julien Mehl
Hi Alain,
The following should get you started:
apply(df[,-1], 2, function(x) cut(x, breaks = quantile(x), include.lowest =
TRUE, labels = 1:4))
Check ?cut and ?apply for more information.
HTH,
Jorge.-
On Tue, Feb 19, 2013 at 9:01 PM, D. Alain wrote:
Dear R-List,
I would like to recode
101 - 200 of 388 matches
Mail list logo
