There is a little catch-22 here.
I would be happy to fund a student to spend a summer cleaning up the error
messages and making it more userfriendly, but the catch-22 problem is that I
don’t have much confidence that this effort would be adopted into the basic R
distribution — and the only
Hello everybody,
I have experience coding with R, but am brand new to R Shiny. I am trying
to produce an application that will allow users to upload their own
dataset, select columns they want an ANOVA analysis run on, and generate
graphs that will allow users to view their results. However, I am
[lvadd %in% names(wageszm14)]
>
> > On 5 Jun 2019, at 06:46 , Rolando I. Valdez via R-help <
> r-help@r-project.org> wrote:
> >
> > Hello,
> >
> > I have almost 40 variables that I am trying to include in a formula.
> >
> > I tried to includ
Hello,
I have almost 40 variables that I am trying to include in a formula.
I tried to include them using as.formula(), however the variables do not
follow a patter in the name. e.g. These variables are named like: lvacb23
lvacb30 lvacb300 lvacb40 . lvacb81.
> lvadd <- paste0("
Thank you very much, Sarah, for your help. The colvar argument was
certainly what I needed. I will follow your suggestion.
Cheers,
Jorge.-
On Thu, Jul 14, 2016 at 1:29 PM, Sarah Goslee
wrote:
> I assume you want some variant of:
> with(d0, scatter3D(X, Z, Y, bty = "b2", co
Dear R-help,
I am using the plot3D package to produce 3D spheres along with 95% CIs
distinguishing each sphere with a predefined colour (see the reproducible
example at the end).
I have been successful in producing a similar plot using a different data
set (kindly see
https://www.dropbox.com/s
Dear Mohammad,
What's wrong with the result?
Best,
Jorge.-
On Monday, March 7, 2016, Mohammad Tanvir Ahamed via R-help <
r-help@r-project.org> wrote:
> Hi,
>
> a <- c(1:5)b <- c(1:3)
> c <- 1
> d <- 5
> e <- list(a,b,c,d)
>
> # To extract every 1st element
> lapply(e,"[[",1)
>
> ## Out-put
> [[1]
This has been eating at me for a very long time. From everything I understand,
the plm package does not yet allow robust standard errors to be calculated for
the Hausman-Taylor using the coeftest command you would for other estimators
(i.e. pooled, within, random). This is of course the error
Dear Dr. Steipe,
Thank you for the code. It does exactly what I needed.
Best regards,
Jorge.-
On Wed, Oct 28, 2015 at 1:18 PM, Boris Steipe
wrote:
> Your code does not produce the matrix in your image.
> The first three rows contain all-zeros and the last row is missing.
> The
Dear all,
I thought I would better send an image illustrating that the problem is
(hope the file gets through). In the picture, the matrix "m" is given by
## input
m <- structure(c(0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 2, 2,
2, 2, 2, 2, 2, 2, 2, 0, 0, 0, 3, 3, 3, 3, 3, 3,
Thank you all for your solutions and comments.
As Dr. Carlson mentioned, we leave rows 1 to 3 out as they are all zeroes.
Then, the entries I need to select from m are
entry value
4,1 ---> 1
5,2 ---> 2
6,3 ---> 3
7,1 ---> 1
8,2 ---> 2
9
Dear Jim,
Thank you very much for your quick reply.
I am sorry for the confusion it may have caused, but I messed up the
indexes in my example. I would like, from the following matrix "m"
## input
m <- structure(c(0L, 0L, 0L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 0L,
0L, 0L, 2L, 2
Dear R-help,
I am working with a matrix "m" from which I would like to extract some
elements. An toy example is as follows:
## input matrix
m <- structure(c(0L, 0L, 0L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 0L, 0L, 0L, 2L, 2L, 2L, 2L,
2L, 2L, 2
-0.58 -0.50 0.30 0.60
Best regards,
Jorge.-
On Mon, Oct 26, 2015 at 11:43 AM, Judson wrote:
> How do I control the number of digits to display,
> say, in a matrix, without rounding or losing accuracy
> in subsequent calculations?
> round() of course reduces accuracy.
>
>
Dear R-help,
I am working on what it seems to be a simple problem, but after several
hours trying to come up with a solution, unfortunately I have not been able
to.
I would like to go from "datain" to "dataout", that is, create the NEWREF
variable according with some restrict
hafizuddinarsha...@gmail.com> wrote:
> Dear R users,
>
> Could someone help me on this? I have this kind of data set:
>
> structure(list(Year = c(1971L, 1971L, 1971L, 1971L, 1971L, 1971L,
> 1971L, 1971L, 1971L, 1971L, 1971L, 1971L, 1971L, 1971L, 1971L,
> 1971L, 1971L, 1971L,
Dear R-help,
I am interested in plotting some pedigrees and came across the kinship2
package. What follows is an example of the pedigrees I am working with.
Now, when running
## check package availability
if(!require(kinship2)) install.packages('kinship2')
require(kinship2)
## data
Hi Hui,
I have used the source_url function in the devtools package with good
results. Give it a shot!
Best,
Jorge.-
On Tue, Mar 10, 2015 at 9:39 AM, Hui Du wrote:
> Hi All,
>
> I am trying to parse some information from website, say, a linkedin page.
> The linkedin url was
>
on(data, TitleGraph){
>
>
> pdf("TitleGraph.pdf",width=7,height=5)
> plot(data)
> dev.off()
> }
>
> test(cars <- c(1, 3, 6, 4, 9),TitleGraph="etc")
>
> My problem is that I want graph pdf being saved as etc and not as
> Titlegraph.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 cal
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 wrote:
> Hello,
>
> I have tried the solve the non-negativity constraint "if else function" in
> R. But I have done something wrong becaus
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 wrote:
>
> Hello dear R-helpers,
>
> I have a small problem in my algorithm. I have sequences of "0&
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 wrote:
> R 3.1.1
> OS X
>
> Colleagu
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 wrote:
>
>
> Hi all,
>
> I have a matrix with consecutive and non-consecutive numbers
> in columns. For example, the first 2 columns have consecutive
Perhaps I am missing something but I do not get the same result:
x <- read.table(textConnection("Family.ID Sample.ID Relationship
2702 349 mother
2702 3456 sibling
2702 9980 sibling
3064 3 father
3064 4 mother
3064 5sibling
3064 86 sibling
3064 87 sibling&q
$PID[l$Relationship == 'sibling'] <- l$Sample.ID[father]
if(sum(mother) == 0)
l$MID[l$Relationship == 'sibling'] <- 0
else l$MID[l$Relationship == 'sibling'] <- l$Sample.ID[mother]
l
}))
It is assumed that when either parent is not available the M/PID is 0.
[l$Relationship == 'sibling'] <- l$Sample.ID[father]
l$MID[l$Relationship == 'sibling'] <- l$Sample.ID[mother]
l
}))
res
HTH,
Jorge.-
Best regards,
Jorge.-
On Sun, Aug 17, 2014 at 5:42 AM, Kate Ignatius
wrote:
> Hi,
>
> I have a data.table question (
nal output is a list.
> How do I covert it back a dataframe?
> -Sohail
>
>
> On Fri, Aug 15, 2014 at 5:37 AM, Jorge I Velez
> wrote:
>
>> Dear Sohail,
>>
>> Using Jim's data set skdat, two more options would be
>>
>> # first option
>
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 optio
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 wrote:
> Hi,
>
> Let say I have 2
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
wrote:
> Dear all,
> I need some help with plotting boxplots in groups. I have a
> fi
Hi Arun,
Thank you very much for your suggestion.
While running some tests, I came across the following:
# sample data
n <- 2000
p <- 1000
x2 <- data.frame(variable = rep(paste0('x', 1:p), each = n), id =
rep(paste0('p', 1:p), n), outcome = sample(0:2, n*p, TRUE), r
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(v
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
wrote:
> I have 4 columns, and about 300K plus rows with 0s and 1s.
>
> I'm trying to count how many rows satisfy a certa
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 wrote:
> Dear list,
>
> Please consider the following example:
>
> library(plotrix)
> barp(c(2,3,4,5,6,7
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 iterati
der q1 q2 q3 q4
>
> 11 f 1 1 5 1
>
> 22 f 2 1 4 1
>
> 31 f 2 2 4 3
>
> 51 m 4 5 2 4
>
> 62 m 5 4 5 5
>
> 82 m 4 5 5 5
>
>
>
> Next I changed the o
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
Hi Nevil,
Try
apply(A, 2, function(x) x == B)
HTH,
Jorge.-
On Fri, May 2, 2014 at 6:46 PM, nevil amos 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
Hi Kristi,
Try
out1970$smoot
HTH,
Jorge.-
On Fri, May 2, 2014 at 10:00 AM, Kristi Glover wrote:
> 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
d a fast way to fix the following question?
>
> Given a verctor of length N, for example bo = [3 8 4 6 1 5],
> I want to drive a vector whose elements are 1, 2, ..., N and the order of
> elements is the same as that in verctor bo.
> In this example, the result is supposed to be
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
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 t
25, 2014 at 7:51 PM, Luigi Marongiu wrote:
> Dear all,
> I would like to calculate the confidence intervals on aggregate data. I
> know how to do this using the t test, but it did not work together with the
> aggregate function.
> Is there a function that can be applied to the agg
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, eliz
Hi Eliza,
Perhaps the following?
matpoints(t(dat), type = 'l')
HTH,
Jorge.-
On Sat, Mar 22, 2014 at 10:18 PM, eliza botto wrote:
>
> 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 &quo
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 wrote:
> Dear all!
>
> Is there a possibility to replace all duplicates values in data frame with
> 0?
>
>
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 t
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 wrote:
> Hi,
> I want to install "languageR" but is doesn'
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 wrote:
> 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 context:
>
2
leftTranspose <- function(x) t(x[, ncol(x):1])
leftTranspose(x)
HTH.
Jorge.-
On Tue, Oct 22, 2013 at 12:52 PM, Vokey, John <> wrote:
> useRs,
> I frequently require the following transform of a matrix that I call a
> leftTranspose:
>
> -- transposes x such that the la
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 wrote:
> Hi everyone
>
> I have a problem with replacing the NA values with the mean of the column
> whic
T
hank you very much, Bert. Point well taken.
Regards,
Jorge.-
On Wed, Jun 19, 2013 at 12:07 AM, Bert Gunter wrote:
> Jorge: No.
>
> > a <-c(1,5,8,15,32,33.5,69)
> > b <-c(8.5,33)
> > a[findInterval(b, a)]
> [1] 8 32 ##should be 8 33.5
>
> I believe
Dear Andras,
Try
> a[findInterval(b, a)]
[1] 8 32
HTH,
Jorge.-
On Tue, Jun 18, 2013 at 10:34 PM, Andras Farkas 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)
>
&
Hi Miao,
Try
attributes(test1)[[1]]
HTH,
Jorge.-
On Tue, Jun 11, 2013 at 3:49 PM, jpm miao 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
t1* 0.4069613 0.005465687 0.07355997
See http://www.mayin.org/ajayshah/KB/R/documents/boot.html for more
information and examples.
HTH,
Jorge.-
On Fri, Jun 7, 2013 at 6:03 PM, Rguy wrote:
> I am getting started with the boot package and boot command. As a first
> step I tried the
"Y" | sample==1 | !is.na(glb_ind))
> subset2<-subset(dframe, cwar_ind="Y" |sample==2 | !is.na(cwar_ind))
> subset3<-subset(dframe, reg_ind="Y" | sample==3 | !is.na(reg_ind))
>
>
> On Wed, Jun 5, 2013 at 9:33 AM, Daniel Tucker
> wrote:
>
>> I am
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" wrote:
> Dear R Helpers,
>
> I have a fairly complicated list of data frames. To give you an idea of
> the structure, the to
Try
rep(1:length(v), v)
HTH,
Jorge.-
On Fri, May 17, 2013 at 8:53 PM, Stefan Petersson 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
[f<0]
> [1] -2 -8 -32 -128
> > f[f>0]
> [1] 4 16 64
> > f[f=4]
> [1] 16
>
>
> 2013/5/8 Jorge I Velez
>
>> f [ f < 0 ]
>>
>>
>> On Wed, May 8, 2013 at 11:54 AM, jpm miao wrote:
>>
>>> Hi,
>>>
>&g
f [ f < 0 ]
On Wed, May 8, 2013 at 11:54 AM, jpm miao 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
&
company, which is the primary output
> result. How can I access the first column, rowname, 4, 5, 6, 7, 8, 9, 11?
> How can I build a dataframe consisting of two columns, the company name
> (code) and the efficiency?
>as.data.frame(a) produce only a column with company names, wh
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
wrote:
> Hi everyone,
>
> I know there have been several requests regarding subsetting before, but
> n
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 wrote:
> Christofer,
>
> The following should get you started:
>
> r <- Mat[match(rownames(M
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 h
Dear Dr. Harding,
Try
sapply(L, "[", 1)
sapply(L, "[", 2)
HTH,
Jorge.-
On Thu, Apr 25, 2013 at 8:16 PM, Ted Harding wrote:
> Greetings!
> For some reason I am not managing to work out how to do this
> (in principle) simple task!
>
> As a result of
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
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 wrote:
> I can't understand what is happening. Th
nchar(x) # digits in x
b <- 5 # period
tapply(strsplit(x, "")[[1]], rep(1:(nchar(x)/b), each = b), function(x)
prod(as.numeric(x)))
HTH,
Jorge.-
On Thu, Apr 18, 2013 at 6:47 PM, Janesh Devkota wrote:
> Hello,
>
> I have a big number lets say of around hundred digits. I w
.-
On Wed, Apr 17, 2013 at 6:08 PM, S'dumo Masango wrote:
> I have installed the arm package and its dependents (e.g MATRIX, etc), but
> cannot use the functions "invlogit" and "bayesglm" because it gives me the
> error message "Error: could not find function
Please do not forget to CC the list to increase your chances of getting
help. --JIV
On Wed, Apr 10, 2013 at 9:31 PM, catalin roibu wrote:
> Hello!
> I try to compute the double exponential function this code:
> f<-function(cls,a,b,c,d)a*exp(cls*b)+c*exp(cls*d)
>
> n2<-n
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, p
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
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 interva
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, s
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
Try
ifelse(ABS ==1 | DEFF == 1, 1, 0)
HTH,
Jorge.-
On Fri, Mar 22, 2013 at 12:02 AM, Tasnuva Tabassum wrote:
> 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 take val
ad idea -- dangerous -- confusing statefulness, etc. (See
> explanations in the archives as to why)
>
> >
> > which results in adding your column names to the search path of R for
> name
> > resolving.
> >
> > Pierrick Bruneau
> > CRP Gabriel Lippmann
> &g
hanks arun!!
>
>
> On Mon, Mar 18, 2013 at 10:06 AM, arun wrote:
>
> Hi,
> >Try this:
> >set.seed(25)
> >mat1<-
> matrix(cbind(sample(1:15,20,replace=TRUE),sample(16:30,20,replace=TRUE)),ncol=2)
> > nrow(mat1[sapply(seq_len(nrow(mat1)),function(i)
>
run <> wrote:
> Hi,
> Try this:
> set.seed(25)
> mat1<-
> matrix(cbind(sample(1:15,20,replace=TRUE),sample(16:30,20,replace=TRUE)),ncol=2)
> nrow(mat1[sapply(seq_len(nrow(mat1)),function(i)
> any(seq(mat1[i,1],mat1[i,2])==12)),])
> #[1] 17
>
>
> set.seed
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,
&g
Try
x <- c("Tom Cruiser", "Bread Pett", "Arnold Schwiezer")
sapply(strsplit(x, " "), function(r) paste0(substr(r[1], 1, 3),
substr(r[2], 1, 3)))
[1] "TomCru" "BrePet" "ArnSch"
HTH,
Jorge.-
On Thu, Mar 14, 2013 at 1:21
Try
substr(tempdf$abb 4, 6)
--JIV
On Thu, Mar 14, 2013 at 1:15 AM, SH wrote:
> Dear Jorge,
>
> I gave me this result (below) since it defines starting from the forth
> letter and ending 6th letter from the first element.
>
> > substr(tempdf$name, 4, 6)
> [1] "
Dear SH,
Hmmm... what about
substr(tempdf$name, 4, 6))
?
HTH,
Jorge.-
On Thu, Mar 14, 2013 at 1:06 AM, SH 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,
>
&g
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
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 like:
> c("a1b1","a2b2","a1b2") which I want to spilt into two parts like:
> c(
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 &l
3.44
>
> [[2]]
> [1] 0.1 3.0
>
> [[3]]
> [1] 12.0 33.1
>
> [[4]]
> [1] 1.1 23.0
>
> lapply(x, t) doesn't do the job, I think.
>
> ishida
>
>
> 2013/3/8 PIKAL Petr
>
> > Hi
> >
> > > -Original Message-
> > &g
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
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 wrote:
> Hi Bert,
>
> I re
Dear Julien,
Check
citation('stats')
HTH,
Jorge.-
On Wed, Mar 6, 2013 at 12:05 AM, Julien Mvdb 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,
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,
>
>
Nico,
Check https://stat.ethz.ch/pipermail/r-help/2008-May/161725.html for some
alternatives. You might have to change the structure of your data, though.
HTH,
Jorge.-
On Wed, Feb 13, 2013 at 10:13 PM, Nico Met <> wrote:
> Dear all,
>
> I have a data (bellow) and I want to mak
e 2nd element of x is at the 3rd position
> in y
> 3rd element of z is NA, because the 3rd element of x is not in y
>
> Of course I can write the function findIndexIn() using a for loop, but
> in 80% of cases when I felt the urge to use "for" in R it turned out
> that there
Dear Wim,
You could try
1*(df > 0)
HTH,
Jorge.-
On Tue, Feb 5, 2013 at 5:33 AM, Wim Kreinen <> wrote:
> Hello,
>
> I have a dataframe with positive integers and
> for every value > 0
> I would like to have 1 and the rest should be zero.
>
> For instance
Dear Gundala,
Try
as.data.frame.table(foo)
HTH,
Jorge.-
On Fri, Feb 1, 2013 at 5:00 PM, Gundala Viswanath <> wrote:
> I have the following data frame:
>
> > foo
>w x y z
> n 1.51550092 1.4337572 1.2791624 1.1771230
> q 0.0
Hi David,
Check
?"%in%"
for a simpler approach.
Regards,
Jorge.-
On Wed, Jan 30, 2013 at 8:42 PM, David Studer <> wrote:
> Hello everybody!
>
> I have again a rather simple question concerning recoding of variables:
>
> I have a variable/data-frame column BIRT
;
HTH,
Jorge.-
On Wed, Jan 30, 2013 at 8:05 PM, Mat <> wrote:
> Hello together,
>
> i have a question for "substring".
> I know i can filter a number like this one:
> bill$No<-substring(bill$Customer,2,4)
>
> in this case i get the 2nd, 3rd and 4th number
Hi m p,
You can use either
apply(fhours, 2, function(x) ifelse(x < 0, x + 24, x)
or
shours <- fhours
shours[shours < 0 ] <- shours[shours < 0 ] + 24
shours
HTH,
Jorge.-
On Sat, Jan 19, 2013 at 4:17 PM, m p <> wrote:
> Hello,
> It should be easu but I cannot
Dear Kaveh,
Take a look at http://www.statmethods.net/stats/power.html
HTH,
Jorge.-
On Thu, Jan 10, 2013 at 3:21 PM, Kaveh Zakeri <> wrote:
> Hello,
>
> I am trying to get the power function to report the sample size rather
> than the power. My goal is to input a variety o
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