it.
Or see the class of those objects.
class(myBiomodModelEval_55["ROC","Testing.data",,,])
class(myBiomodModelEval_55["TSS","Testing.data",,,])
You do not apply(), simply do
mean(myBiomodModelEval_55["ROC","Testing.data",,,])
An
Hello,
After the plot just do
abline(v = median(A))
As for how to plot points, see, well, ?points().
Hope this helps,
Rui Barradas
Em 22-10-2017 16:33, varin sacha via R-help escreveu:
Dear R-experts,
Here below is my code,
I would like to add a vertical line on my plot, showing the
Hello,
R-Help answers questions on R code, your question is about statistics.
You should try posting the question to
https://stats.stackexchange.com/
Hope this helps,
Rui Barradas
Em 23-10-2017 18:54, kende jan via R-help escreveu:
Dear all, I am trying to fit a multiple linear regression
Hello,
I think that your code is simple enough to be considered "nice". If you
are worried about the for loop, don't, were loops worrying they wouldn't
exist.
Hope this helps,
Rui Barradas
Em 23-10-2017 22:09, Ed Siefker escreveu:
I have a list of file names, and a
ols_3.4.2
[13] stringr_1.2.0munsell_0.4.3compiler_3.4.2 colorspace_1.3-2
[17] tibble_1.3.4
Hope this helps,
Rui Barradas
Em 26-10-2017 21:09, Lara Dutra Silva escreveu:
Boa noite,
Não sei se português.
Estou a ter algumas dificuldades na alteração do tamanho, letra de um
plot, ou seja, as
# total non zero
totals
# V1 V2 V3 V4 V5 V6 V7 V8 V9 V10
# 6 8 8 8 8 7 7 8 6 10
grand_total
#[1] 76
# another way
prod(dim(data)) - sum(zero + na)
#[1] 76
Hope this helps,
Rui Barradas
Em 29-10-2017 10:25, Engin YILMAZ escreveu:
Dear R Staff
You can see my data.csv
Hello,
This is cross-posted from
https://stackoverflow.com/questions/47042591/how-to-resolve-nested-variables-inside-a-loop-in-r
And you already have the answer there. See my comment.
Rui Barradas
Em 31-10-2017 19:00, Edward Guda via R-help escreveu:
How do i resolve this?
symbol <-
Hello,
If cat is giving you an error try print(attr <- ...etc...)
Hope this helps,
Rui Barradas
Em 01-11-2017 17:21, Priya Arasu via R-help escreveu:
Hi David,Thank you for the example.When I try to use the cat function, I get an
error
cat(attr<-getAttractors(net, type="a
Hello,
Also
tail(test_df$Movie, 10)
Hope this helps,
Rui Barradas
Em 05-11-2017 19:18, Ulrik Stervbo escreveu:
R can have a bit of a learning curve... There are several ways to achieve
your goal - depending on what you want:
test_df <- data.frame(Movie = letters, some.value = rnorm
that the column in question holds "levels" do you mean it's
a factor? (factors are R's categorical variables.)
Hope this helps,
Rui Barradas
Em 06-11-2017 19:26, Matti Viljamaa escreveu:
It’s sometimes faster to ask from someone who has already learnt the syntax.
In this ca
Hello,
Try
print(head(...))
Hope this helps,
Rui Barradas
Em 07-11-2017 20:01, Tom Backer Johnsen escreveu:
Dear R-help,
I am running a Mac under Sierra, with R version 3.4.2 and RStudio 1.1.383. When
running head () or tail () on an object in a script using source (
Hello,
Using base R only, the following seems to do what you want.
with(mydf, ave(speed, date_time, type, FUN = weighted.mean, w = n_vehicles))
Hope this helps,
Rui Barradas
Em 09-11-2017 13:16, Massimo Bressan escreveu:
Hello
an update about my question: I worked out the following
Sorry, I messed up. Only checked the final result after sending the
previous mail. The solution is wrong.
Rui Barradas
Em 09-11-2017 13:27, Rui Barradas escreveu:
Hello,
Using base R only, the following seems to do what you want.
with(mydf, ave(speed, date_time, type, FUN = weighted.mean, w
5
#5 5 2 0 NA
Hope this helps,
Rui Barradas
On 11/22/2017 10:34 AM, Massimo Bressan wrote:
Given this data frame (a simplified, essential reproducible example)
A<-c(8,7,10,1,5)
A_flag<-c(10,0,1,0,2)
B<-c(5,6,2,1,0)
B_flag<-c(12,9,0,5,0)
mydf<-data.frame(A, A
Hello,
Sorry, I obviously read in a hurry.
icol <- grepl("flag", names(mydf))
is.na(mydf[!icol]) <- mydf[icol] == 0
mydf
# A A_flag B B_flag
#1 8 10 5 12
#2 NA 0 6 9
#3 10 1 NA 0
#4 NA 0 1 5
#5 5 2 NA 0
Hope this help
21
#6 B 47 512
#9 C 61 521
#10 C 68 235
Hope this helps,
Rui Barradas
On 12/9/2017 12:48 AM, Ashta wrote:
Hi David, Ista and all,
I have one related question Within one group I want to keep records
conditionally.
example within
group A I want keep rows that have " x" values ran
ex){
mine(data[index, ])$MIC
}
results=boot(data = cbind(C,D), statistic = myCor, R = 2000)
boot.ci(results,type="all")
Hope this helps,
Rui Barradas
On 12/10/2017 3:19 PM, varin sacha via R-help wrote:
Dear R-Experts,
Here below is my R code (reproducible example) to calculate the
Hello,
Here is one way.
tdat$D <- ifelse(tdat$B %in% tdat$A, tdat$A[tdat$B], 0)
tdat$E <- ifelse(tdat$B %in% tdat$A, tdat$A[tdat$C], 0)
Hope this helps,
Rui Barradas
On 12/13/2017 9:36 PM, Val wrote:
Hi all,
I have a data frame
tdat <- read.table(textConnection("A B C Y
A1
Example , sum)
Happy holidays,
Rui Barradas
On 12/29/2017 12:03 AM, PABLO ORTIZ PINEDA wrote:
Hello there. Happy new year for everyone!
I need help with a table. This table contains 300 rows and 192 columns.
Being the first column the ID of my samples that can have several
observations.
I nee
Hello,
You have to create the vector 'd' outside the loop before using it.
d <- numeric(nrow(data))
Only then comes the loop.
Hope this helps,
Rui Barradas
On 12/29/2017 2:31 PM, Luca Meyer wrote:
Hello,
I am trying to run the following syntax for all cases within the datafra
Hello,
I believe the following regex will do it.
x <- "\":\"03-JAN-2018 16:00:00\""
sub('^.*(\\d{2}-\\w{3}-\\d{4} \\d{2}:\\d{2}:\\d{2})[:"]', '\\1', x)
Hope this helps,
Rui Barradas
On 1/3/2018 2:26 PM, Christofer Bogaso wrote:
Hi,
- "H04Q007/32"
select_many <- c("H04Q007/32", "H04M001/275")
oecd2 <- subset(oecd, IPC == select_one)
oecd3 <- subset(oecd, IPC %in% select_many)
Hope this helps,
Rui Barradas
On 1/3/2018 7:53 PM, Saptorshee Kanto Chakraborty wrote:
Hello,
I have a data of P
IBM&f=d1t1l1c1p2ohgv'
In addition: Warning message:
In download.file(paste("https://finance.yahoo.com/d/quotes.csv?s=";, :
InternetOpenUrl failed: 'Não foi possível processar o nome ou o
endereço do servidor'
So it's a server error. Note that getSymbols(&quo
Hello,
I didn't like the video but that has nothing to do with the language, I
just happen to prefer other type(s) of music.
Rui Barradas
On 1/9/2018 8:45 PM, Rolf Turner wrote:
On 10/01/18 09:31, Doran, Harold wrote:
It would be better for you to instead read the blog post that
Hello,
Can you please post the output of
dput(all) # if all is small
dput(head(all, 30)) # if all is big
in a mail?
Hope this helps,
Rui Barradas
On 1/24/2018 4:28 PM, Francisca R. Souza Pereira wrote:
Hi,
I'm not a programmer, so I have a question about R functions,
I run the R
Hello,
I believe the following is simpler.
It changes the OP's code a bit and uses lapply, not apply.
res2 <- lapply(C, fitdist, "gamma")
do.call(rbind, lapply(res2, `[[`, "estimate"))
# shape rate
#A 3.702253 1.234126
#B 31.300800 3.912649
Hope this h
)
Hope this helps,
Rui Barradas
On 1/30/2018 8:58 AM, Daniel Nordlund wrote:
On 1/29/2018 9:03 PM, smart hendsome via R-help wrote:
Hello everyone,
I have a question regarding simulating based on runif. Let say I have
generated matrix A and B based on runif. Then I find mean for each
matrix A
Hello,
First of all, your question is about 'predict' but you include graphic
instructions that have nothing to do with it. They do not hurt, but the
reproducible example should also be minimal.
Second, whenever you use RNG's, you should start it with set.seed().
Now, I have edited your code
ountry + IDNum, mydata)
addmargins(tbl2)
Hope this helps,
Rui Barradas
On 2/3/2018 3:00 AM, Val wrote:
Hi all,
I have a data set need to be summarized by unique ID (count and sum of a
variable)
A unique individual ID (country name Abbreviation followed by an integer
numbers) may hav
l(x != 0))
tbl1b <- addmargins(tbl1[inx, ])
tbl1b
Hope this helps,
Rui Barradas
On 2/3/2018 4:42 PM, Val wrote:
Thank you so much Rui.
1. How do I export this table to excel file?
I used this
tbl1 <- table(Country, IDNum)
tbl2=addmargins(tbl1)
write.xlsx(tbl2,"tt
way around, call dyn.unload before modifying the C code.)
Hope this helps,
Rui Barradas
On 3/1/2018 8:52 AM, Rolf Turner wrote:
I am working with a function "foo" that explicitly dynamically loads a
shared object library or "DLL", doing something like dyn.load("bar.so&qu
lt;- paste(dynlib, .Platform$dynlib.ext, sep = "")
dyn.unload(dynlib)
}
These have saved me lots and lots of typing along the years. (Note that
they still call paste/sep = "")
Hope this helps,
Rui Barradas
Eric
On Thu, Mar 1, 2018 at 2:21 PM, Rui Barradas <mailt
se you have a matrix (or data.frame)
dat <- cbind(dv, iv)
apply(dat[, -1], 2, cor, dat[, 1])
Hope this helps,
Rui Barradas
On 3/6/2018 12:03 PM, faiz rasool wrote:
Dear list, I have the following how-to-do it in R, questions.
Suppose I have ten independent variables, and one dependent va
Hello,
Maybe you want
while(x[i] < 5)
not <=
Hope this helps,
Rui Barradas
On 3/31/2018 2:45 PM, Henri Moolman wrote:
Could you please provide help with something from R that I find rather
puzzling? In the small program below x[1]=1, . . . , x[5]=5. R also
finds that x[1]<=
Hello,
Though Bert's and David's answers are what you should do, note that some
R functions that need factors will coerce their input variables when
necessary.
Have you tried to run the code you haven't posted without coercing to
factor? It might run...
Hope this helps,
Rui
t;dir1","dir2","dir3")
for(directory in directories) {
old_dir <- setwd(directory)
# do whatever you want to do
# then return to the original directory
setwd(old_dir)
}
Hope this helps,
Rui Barradas
This will allow you to start and finish in the same dire
t;sn7", i)
DF
})
seg <- do.call(rbind, seg)
row.names(seg) <- NULL
Hope this helps,
Rui Barradas
On 4/16/2018 9:54 PM, Ding, Yuan Chun wrote:
Hi All..,
I need to do the following repetitive jobs:
seg71 <-
read.csv("C:/Awork/geneAssociation/removed8samples/neuhausen71/se
Hello,
Nor do I, no gmail, also got spam.
Rui Barradas
On 4/17/2018 8:34 PM, Ding, Yuan Chun wrote:
No, I do not use gmail, still got dirty spam email twice.
-Original Message-
From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of Fowler, Mark
Sent: Tuesday, April 17, 2018
n try is
https://regex101.com
Hope this helps,
Rui Barradas
On 4/18/2018 11:19 PM, Jim Lemon wrote:
Hi Farnoosh,
Perhaps this will help:
drop_dollar<-function(x) return(as.numeric(as.character(gsub("\\$","",x
sapply(My.Data,drop_dollar)
Jim
On Thu, Apr 19, 2018 a
frames in the list. Then save the results as csv files.
stats_list <- lapply(txt_list, summary_stats_function)
csv_files <- sub("txt", "csv", txt_files)
lapply(seq_along(csv_files), function(i) write.csv(stats_list[[i]],
csv_files[i]))
setwd(old_dir)# reset, if nee
Hello,
When programming it is better to use dat[["variable"]] than dat$variable.
So your code could be
pfas.pheno[[cat.var]] <- NA
pfas.pheno[[cat.var]][pfas.pheno[,i] <= quantile(pfas.pheno[,i],0.25,
na.rm =T)] <- 0
etc.
Untested.
Hope this helps,
Rui Barradas
O
are mutually exclusive (i.e. the user
should define only one of these). How do I code this within the function?
See
?missing
See also the first example in help("switch")
Hope this helps,
Rui Barradas
HTH,
Chuck
__
R-help@r-project.org maili
Hello,
instead of ifelse, the following is exactly the same and much more
efficient.
d0[[nm]] <- as.integer(regexpr(d1[i,1], d0$X0) > 0)
Hope this helps,
Rui Barradas
On 4/28/2018 8:45 PM, Luca Meyer wrote:
Thanks Don,
for (i in 1:10){
nm <- paste0("V"
ifelse creates both vectors, the true part and the false part, and then
indexes those vectors in order to return the appropriate values. This is
the double of the trouble and a great deal of memory used.
Rui Barradas
On 4/28/2018 10:12 PM, Rui Barradas wrote:
Hello,
instead of ifelse, the
Hello,
Another thing to note is that regexpr is likely to take (much) more time
than ifelse or as.integer.
And the code will therefore not be very optimizable.
Rui Barradas
On 4/30/2018 4:25 PM, MacQueen, Don wrote:
Luca,
If speed is important, you might improve performance by making d0
Hello,
Is it something like this that you want?
x <- data.frame(a = c(1:3, 5, 5:10), b = c(1:7, 7, 9:10))
y <- data.frame(a = 1:10, b = 1:10)
which(x != y, arr.ind = TRUE)
Hope this helps,
Rui Barradas
On 5/1/2018 11:35 AM, Chintanu wrote:
Hi,
May I please ask how I do the follow
, though.
Hope this helps,
Rui Barradas
On 5/4/2018 5:26 PM, David Winsemius wrote:
On May 4, 2018, at 12:04 AM, sunyeping wrote:
--
From:David Winsemius
Send Time:2018 May 4 (Fri) 13:25
To:孙业平
Cc:R Help Mailing List
20
#8 820
#9 920
#10 1020
So your code works: df$newcolumn <- number
Hope this helps,
Rui Barradas
On 5/10/2018 2:04 PM, Marcelo Mariano Silva wrote:
Hi,
I am a begginer in R programming.
I am traying to create a a column in my data frame filled down with
Hello,
This is cross posted from StackOverflow:
https://stackoverflow.com/questions/50314015/is-there-any-method-to-defer-the-execution-of-code-in-r
Cross posting is discouraged in r-help.
Rui Barradas
On 5/13/2018 8:59 AM, akshay kulkarni wrote:
dear members,
I have created the following
Hello,
I don't understand.
It *is* the same question. Same code, same words. And same 'AKSHAY M
KULKARNI' (the OP here) and 'AKshayKulkarni' (SO).
Exactly the same.
Rui Barradas
On 5/13/2018 2:07 PM, Jeff Newmiller wrote:
I am puzzled by the use of the term "
se unrelated to R-Help, I will stop it now.
Rui Barradas
On 5/13/2018 5:41 PM, Berry, Charles wrote:
On May 13, 2018, at 9:24 AM, Jeff Newmiller wrote:
Not when I click on that link.
Nor me, but what I get is actually
https://stackoverflow.com/questions/1174799/how-to-make-executi
Hello,
I downloaded the file "polycor_0.7-8.tar.gz" from CRAN with Mozilla
Firefox and then ran
gzFile <- file.choose()
install.packages(gzFile, repos=NULL, type="source")
and it worked at the first try.
Hope this helps,
Rui Barradas
On 5/16/2018 6:26 PM, William
", split_str)
Hope this helps,
Rui Barradas
On 5/19/2018 12:07 PM, Christofer Bogaso wrote:
Hi,
I am struggling to split a data.frame as will below scheme :
DF = data.frame(name = c('a', 'v', 'c'), val = 0); DF
split_str = c('a', 'c')
Try
plot(1:10, xlab = expression(NO[3]^{'-'}~(mg/L)))
Hope this helps,
Rui Barradas
On 5/21/2018 1:09 PM, Jinsong Zhao wrote:
hi there,
I find the following codes produce strange output.
plot(1:10, xlab = expression(NO[3]^-~(mg/L)))
you will notice that the unit, mg/L is in s
Hello,
Try this.
ss1 <- "z:f(5, a=3, b=4, c='1:4', d=2)"
ss2 <- "f(5, a=3, b=4, c=\"1:4\", d=2)*z"
fun <- function(s) sub("(\\().*(\\))", "\\1\\2", s)
fun(ss1)
#[1] "z:f()"
fun(ss2)
#[1] "f()*z"
Hope
oups, just \\( and \\) with .* between
them, replaced by ()
Hope this helps,
Rui Barradas
On 5/21/2018 3:00 PM, Ulrik Stervbo wrote:
I would use
sub("\\(.*\\)", "()", s)
It is essentially the same as Rui's suggestion, but I find the purpose
to be more clear. It mi
ypred <- predict(fit)
y <- d$crp
median(y - ypred)^2
}
dat <- data.frame(crp, bmi, glucose)
nboot <- 100
medse <- boot(dat, bootMedianSE, R = nboot)
medse$t0
mean(medse$t)# This is the value you want
Hope this helps,
Rui Barradas
On 5/22/2018 12:19 AM, vari
I forgot, you should also set.seed() before calling boot() to make the
results reproducible.
Rui Barradas
On 5/22/2018 10:00 AM, Rui Barradas wrote:
Hello,
If you want to bootstrap a statistic, I suggest you use base package boot.
You would need the data in a data.frame, see how you could do
- ypred)^2)
}
Sorry,
rui Barradas
On 5/22/2018 11:32 AM, Daniel Nordlund wrote:
On 5/22/2018 2:32 AM, Rui Barradas wrote:
bootMedianSE <- function(data, indices){
d <- data[indices, ]
fit <- rq(crp ~ bmi + glucose, tau = 0.5, data = d)
ypred <- predict(fit)
y &l
"73\\.", row.names(mat))
new_mat <- mat[-inx, ]
new_mat
Hope this helps,
Rui Barradas
On 5/22/2018 11:48 AM, Ahmed Serag wrote:
Dear R-experts,
How can I remove a certain feature or observation by a part of its name. To be
clear, I have a matrix with 766 observations as a rows. T
a special
character.)
Hope this helps,
Rui Barradas
On 5/22/2018 12:50 PM, Ahmed Serag wrote:
Thank you Mr. Barradas. The code works great. Unfortunately I have also
some labeles with
173.1
273.1
the grep script remove them also ?
Any ideas Plz, Thanks again
*A
uot;dd"))
y <- c("D","E", "F")[i]
y
#[1] "E" "D" "F"
# This will give you the inverse,
# just apply order() to the output of order(),
# function order() is its own inverse
y[ order(i) ]
#[1] "D" "E" "F&q
)
summary(DF$Anxiolytics)
Hope this helps,
Rui Barradas
On 5/23/2018 11:14 AM, Lisa van der Burgh wrote:
Hi all,
I have a very general question and I think for you maybe very easy, but I am
not able to solve it.
I have a dataset and that dataset contains the variable Anxiolytics. This
va
)
R <- ifelse(Race = "black", 1.159, 1)
141 * pmin(SCr/k, 1)^alpha * pmax(SCr/k, 1)^(- 1.209) * 0.993^Age *
R * S
}
I really don't know what this is, my knoledge of biology is, to be nice,
sketchy.
[1] https://www.kidney.org/content/ckd-epi-creatinine-equation-2009
Hope t
Hello,
I am not sure I understand the question, but see if the following is
what you want.
split(Imputed, cumsum(c(0, diff(Imputed$Y) != 1)))
Hope this helps,
Rui Barradas
On 5/24/2018 3:46 PM, Ioanna Ioannou wrote:
Hello everyone,
I want to transform a data.frame into an array (lets
l be the other imputation.
If this is not what you want, can you please post an example output
mydata[[1]] from the database you have posted?
Rui Barradas
On 5/24/2018 4:14 PM, Ioanna Ioannou wrote:
Hello everyone,
Thank you for this. Nonetheless it is not exactly want i need.
I need mydata
Hello,
See if this is it:
priceStore_Grps$StoreID <- paste("Store",
seq_len(nrow(priceStore_Grps)), sep = "_")
Hope this helps,
Rui Barradas
On 5/26/2018 2:03 PM, Jeff Reichman wrote:
ALCON
I'm trying to figure out how to rename groups in a data fr
})
res <- do.call(rbind, res)
row.names(res) <- NULL
Hope this helps,
Rui Barradas
On 5/26/2018 2:22 PM, Rui Barradas wrote:
Hello,
See if this is it:
priceStore_Grps$StoreID <- paste("Store",
seq_len(nrow(priceStore_Grps)), sep = "_")
Hope this helps,
Rui Barrada
e(dat$A == 1 & dat$B == 0, 1, 0)
dat$A3 <- ifelse(dat$A == 0 & dat$B == 1, 1, 0)
dat$A4 <- ifelse(dat$A == 0 & dat$B == 0, 1, 0)
dat
Hope this helps,
Rui Barradas
On 5/27/2018 3:13 PM, smart hendsome via R-help wrote:
Hi everyone,
I have two columns:
A B
’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
#
#Residual standard error: 286.1 on 29 degrees of freedom
#Multiple R-squared: 0.2426, Adjusted R-squared: 0.1381
#F-statistic: 2.322 on 4 and 29 DF, p-value: 0.08039
Hope this helps,
Rui Barradas
Às 16:54 de 01-06-2018, nguy2952 University of Minnesota esc
Hello,
I don't believe what you want is possible because:
axis.ticks.x and axis.ticks.y change the width of the tick marks
axis.ticks.length changes the length but there is no x and y axis
versions, just a general purpose one.
Sorry I couldn't be of much help,
Rui Barradas
Às 1
Hello,
What type of data do you have? A vector? Or is it a matrix, a
data.frame, a list, etc?
Suppose it is a vector named x. Then you could do something like
rate <- 0.2
is.na(x) <- sample(length(x), rate*length(x))
At an R prompt type
?is.na
?sample
Hope this helps,
Rui Barradas
Hello,
Please repost in plain text, NO HTML formating.
Also, you are missing an open parenthesis right after while:
while( sum(abs(Sb-D-Sc-t(Pi))>1E-5)){
Hope this helps,
Rui Barradas
Às 14:25 de 15-07-2018, Atanasio Alberto Tembe Tembe escreveu:
Hello!
Is there anyone who can help me
Hello,
Maybe the following is not the recommended way but it works
(and I believe makes sense).
f <- function(){}
formals(f) <- formals(fc)
body(f) <- body(fc)
f
#function (x)
#{
# x <- x + 1
# pi * x
#}
f(1)
#[1] 6.283185
Hope this helps,
Rui Barradas
Às 03:25 de 16-07-20
(head(data, 30))
Hope this helps,
Rui Barradas
Às 17:47 de 18-07-2018, Francesca escreveu:
Dear R help,
I am new to ggplot so I apologize if my question is a bit obvious.
I would like to create a plot where a compare the fraction of the values of a
variable called PASP out of the number of
Hello,
The first argument of mean is a vector, the dots argument is to be
"passed to or from other methods." (from ?mean)
Try instead
mean(c(dd1, dd2))
Hope this helps,
Rui Barradas
Às 17:39 de 21-07-2018, John Kane via R-help escreveu:
Either I am doing something very stup
]=1/(K*A[i,j]*c[1,1]+K*A[i,j]*c[1,2]+K*A[i,j]*c[1,3])
the denominator is zero because A[i,j] and c[] are multiplied by it.
3) What are the i,j in A[i,j] and P[i,j]?
Hope this helps,
Rui Barradas
Às 11:51 de 21-07-2018, Atanasio Alberto Tembe Tembe escreveu:
Dear Mr. Barradas,
Thank you for
pipe the output of
kable() through sub().
do.call("rbind", efas) %>%
kable() %>%
sub("^\\|[^|]+(\\|.*)", "\\1", .)
Hope this helps,
Rui Barradas
Às 04:21 de 24-07-2018, michael matta escreveu:
I have been trying to write a function in Rstudio that
anges you do to it will only have effect on the copy. The original
is left as it were.
So run the same code (with the obvious bug yhatResponse$yhat corrected)
but eliminating the 'attach' instruction. And see if it reorders Response.
Hope this helps,
Rui Barradas
Em 01-04-2017
print(deal[[tranche]]$CashFlow$BeginBal)
}
deal
}
result <- lapply(1:1, function(i) BeginBal(Deal, i))
result
Hope this helps,
Rui Barradas
Em 01-04-2017 18:05, Glenn Schultz escreveu:
All,
I am working on structuring a FHLMC credit risk transfer deal. I have
the dea
Hello,
Try the following.
GENDER$Gender_male <- as.integer(GENDER$Gender == "male")
GENDER$Gender_female <- as.integer(GENDER$Gender == "female")
Hope this helps,
Rui Barradas
Em 02-04-2017 19:48, BR_email escreveu:
Hi R'ers:
I need a jump start to obtain my o
Hello,
I don't use ESS but I can use google. The first hit on "ESS help mailing
list" was
https://stat.ethz.ch/mailman/listinfo/ess-help
Hope this helps,
Rui Barradas
Em 03-04-2017 17:46, MacQueen, Don escreveu:
Strikes me as a good question for the ESS help mailing list (a
Hello,
Unless you have strong reasons why not, use the most recent one, R 3.3.
Hope this helps,
Rui barradas
Em 05-04-2017 03:47, Bogdan Tanasa escreveu:
Dear all,
please could you advise me on the following :
on a server, in a folder "x86_64-redhat-linux-gnu-library", i have 2
v
Hello,
There's no need to send the same question twice, we've got it at the
first try.
Maybe I don't understand but is this it?
kk1 <- list(a_c = union(kk$a, kk$c), b = kk$b)
kk1
$a_c
[1] 1 2 3 4 5 6 7 8 9 10 11
$b
[1] 6 7 8 9 10
Hope this helps,
Rui Barra
it's the first time you've came across this function, I can guarantee
you that it is really, really usefull.
Hope this helps,
Rui Barradas
Em 15-04-2017 00:58, Carl Sutton via R-help escreveu:
Hi Jeff
I have seen the seq_along function but never knew the what or why of it. Your
Hello,
Maybe package psych, function multi.hist is what you want.
https://cran.r-project.org/web/packages/psych/index.html
And don't post in HTML, your data is unreadable.
Hope this helps,
Rui Barradas
Em 19-04-2017 14:05, prateek pande escreveu:
Hi,
I have a data as mentioned bel
Hello,
A closure is, like you say, a function.
At an R prompt try:
> typeof(time)
[1] "closure"
So like Duncan suggested rename 'time', for instance capitalize it
'Time'. That should do it.
Hope this helps,
Rui Barradas
Em 11-05-2017 21:20, Tobias Christ
Hello,
I have never used plm but the standard way of adding a quadratic term is
I(time(year)^2)
Hope this helps,
Rui Barradas
Em 12-05-2017 15:40, Tobias Christoph escreveu:
Hey guys,
thanks a lot for your tips. The regression is finally running. As you
said, I had to integrate the column
Hello,
It seems to be right.
You have Chi-squared = 23.724, df = 8, p-value = 0.002549. So try the R
function ?pchisq:
pchisq(23.724, df = 8, lower = FALSE)
[1] 0.002549054
Hope this helps,
Rui Barradas
Em 23-05-2017 13:08, Dhivya Narayanasamy escreveu:
Hi,
I am working with bivariate
you need.
Hope this helps,
Rui Barradas
Em 28-05-2017 15:51, Glenn Schultz escreveu:
If is specify a spline basis as follows
knots <- c(6, 12, 22, 30, 35)
x <- c(0.0, .25, 1.0, 2.0, 3.0)
SCurve <- bs(x = x, knots = knots, intercept = FALSE, Boundary.knots =
c(0,3.5))
I would like to now
d base packages:
[1] splines stats graphics grDevices utils datasets methods
[8] base
loaded via a namespace (and not attached):
[1] compiler_3.4.0 tools_3.4.0
Rui Barradas
Em 28-05-2017 17:37, Rui Barradas escreveu:
Hello,
Since function splines::bs returns an object of class bs
Hello,
Em 29-05-2017 10:43, Duncan Murdoch escreveu:
On May 28, 2017 1:53:29 PM PDT, Rui Barradas wrote:
> predict.bs(SCurve, xnew = 40:45)
Error in predict.bs(SCurve, xnew = 40:45) :
could not find function "predict.bs"
You should call it using the generic, i.e.
predict
Hello,
In order for us to help we need to know how you've imported your data.
What was the file type? What instructions have you used to import it?
Did you use base R or a package?
Give us a minimal but complete code example that can reproduce your
situation.
Hope this helps,
Rui Bar
] stats graphics grDevices utils datasets methods base
loaded via a namespace (and not attached):
[1] compiler_3.4.0
Hope this helps,
Rui Barradas
Em 05-06-2017 19:14, Omar André Gonzáles Díaz escreveu:
Hi,
I want to reporte some strange behaviour with the "months" function
Hello,
See FAQ 7.31
And try ?all.equal.
Hope this helps,
Rui Barradas
Em 07-06-2017 15:32, li li escreveu:
Hi all,
In checking my R codes, I encountered the following problem. Is there a
way to fix this?
I tried to specify options(digits=). I did not fix the problem.
Thanks so much
al Beliefs`, Attitude, `Normative
Beliefs`, `Subjective Norm`, `Control Beliefs`, `Perceived Behavioural
Control`, Intention, Behaviour)
Error in rbind(`Behavioural Beliefs`, Attitude, `Normative Beliefs`,
`Subjective Norm`, :
object 'Behavioural Beliefs' not found
Please correct this
s, modes = TPB_modes)
Error in is_tabular(x) : object 'TPBDATA' not found
So we need to know what 'TPBDATA' is.
Rui Barradas
Em 12-06-2017 00:19, Sarah Sinasac escreveu:
Hello Rui,
I must have missed that line when I copied and pasted my code.
Behavioural Beliefs is:
"Be
Hello,
You have to be aware that mylist[1] and mylist[[1]] are different things.
class(mylist[1])
[1] "list"
class(mylist[[1]])
[1] "NULL"
Apparently you want/need the latter:
is.null(mylist[[1]])
[1] TRUE
Hope this helps,
Rui Barradas
Em 15-06-2017 16:33, ce esc
code. Note that I've made up some data, since we don't have
the .csv file.
x <- "Jun-11"
Dataset <- data.frame(x)
str(Dataset)
tmp <- paste("01", as.character(Dataset$x), sep = "-")
Dataset$x <- as.Date(tmp, format = "%d-%b-%y")
str(
Hello,
You don't need dev.new, but you need to tell R that the plot is finished
by calling dev.off after the plot command.
Hope this helps,
Rui Barradas
Em 20-06-2017 06:18, Yogesh Gupta escreveu:
Dear All,
I am learning R so it's a very simple problem but I do not understand
1 - 100 of 3007 matches
Mail list logo
