I am writing my master thesis in which I compared two cultures . So for my
statistics I need to compare Age,Sex,Culture as well as have a look at the
tasks scores .
Anyone familiar with this ?
I’d love to share my script so you guide me where I did wrong .
Regards
___
Hi all,
I have a question about calculating a P for trend on my data. Let�s give an
example that is similar to my own situation first: I have a continuous outcome,
namely BMI. I want to investigate the effect of a specific medicine, let�s call
it MedA on BMI. MedA is a variable that is catego
Dear John and Peter,
Thank you both for your answers. I am going to try the solutions you gave me!
Thanks again,
Lisa
From: Fox, John
Sent: 23 November 2018 16:54:49
To: Lisa van der Burgh
Cc: r-help@R-project.org; peter dalgaard
Subject: RE: [R] Question
work. I know I am probably doing it completely wrong,
but I do not know how to solve it. Besides that, I do not know what to fill in
at the �x�.
Can somebody help me?
Thank you in advance.
Lisa
[[alternative HTML version deleted]]
ce again, I have tried using y and x lim as well as
max functions, but have not managed to attain any results.
I realize these are amateur questions but I have been struggling with this
for quite some time and would really appreciate any help I can get.
Thanks and best regards,
Lisa
__
Thank you all for your help, it worked!
Op 23 mei 2018 om 19:27 heeft marta azores
mailto:martazo...@gmail.com>> het volgende geschreven:
Try that code
NewDF<-DF[!DF$Anxiolytics==1,]
2018-05-23 10:14 GMT+00:00 Lisa van der Burgh
mailto:lisavdbu...@hotmail.com>>:
Hi all,
in a way that 2 becomes 1, such as:
> summary(DF$Anxiolytics)
0 1 NA's
1102 20 440
Can you help me with the code for doing this?
Thank you in advance!
Best, Lisa
[[alternative HTML version deleted]]
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> Hello,
>
> I would like to Know if i can install r on my iPad Air 2? Is this possible?
>
> Regards Lisa-Marie Kindler
Von meinem iPhone gesendet
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Don,
Thank you for your helpful response. At this point, I do believe it is a
package error and I have contacted the developer.
Thanks,
Lisa
-Original Message-
From: MacQueen, Don [mailto:macque...@llnl.gov]
Sent: Thursday, May 19, 2016 4:27 PM
To: Rees, Lisa Marie (MU-Student); r
Michael,
Thanks for your response.
I tried table(table(Bmat)) and it gave me the following error:
[ Error in table(Bmat) : object 'Bmat' not found]
FYI--
"values" contains 16,383 observations ranging from 0 to less than 1.
Lisa
-Original Message-
From: Mich
I'm using the "GameTheory" package--- DefineGame(14,values) and values is equal
to 16,383 observations.
I keep getting the following error-
[Error in data.frame(rep(0, 2^n - 1), row.names = Bmat) :
duplicate row.names: 1, 11, 111, 12, 112, 1112, 2, 13, 113, 1113, 3,
1213, 11213, 111213
I am considering using RUSBoost (https://github.com/SteveOhh/RUSBoost) and
was wondering if anyone has used this package, and could give me some
insight and help. The help would be on more of the machine learning side,
I just have a few questions about implementation.
[[alternative HTML v
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in resampled performance measures.3: In
train.default(d, classData, method = "J48", model = FALSE) :
missing values found in aggregated results
What am I doing wrong? Also note I'm using model=FALSE in training to
conserve memory, as this has been a problem
--
Lisa Gandy, PhD
As
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Try xtabs()
>> df2 = data.frame(ID = c(10,10,10,10,10,11,11,12),Group =
c(1,2,3,4,5,3,4,4),Value = c(10,20,30,40,50,60,70,80))
>> xtabs(df2$Value~df2$ID + df2$Group)
I think this is exactly what you want.
On Fri, Mar 28, 2014 at 4:51 PM, Mat wrote:
> Hello togehter,
>
> i have a litte problem.
He wants a[1] b[1] a[2] b[2] a[3] b[3]
I think you can do:
x = as.vector(rbind(a, b))
On Mon, Mar 24, 2014 at 2:39 PM, Frans Marcelissen <
fransiepansiekever...@gmail.com> wrote:
> Why not simply
>
> > a<-1:3
> > b<-4:5
> > c(a,b)
> [1] 1 2 3 4 5
>
>
> 2014-03-22 23:22 GMT+01:00 Tham Tran :
>
But I don't think a list can contain that much data - we dont know how big
each txt file is.
So I'd like to do:
city.set <- c('Asheville','Charlotte')
month.set <- c('Dec', 'Jan', 'Feb')
comb.set <- expand.grid(city.set,month.set)
comb.set$filename = paste(comb.set[,1],comb.set[,2],sep = '_')
fo
In this calculation:
(dg0%*%t(dx0))
is
[,1] [,2]
[1,]0 -0.75
[2,]0 0.50
So: exp2 = NaN
It is extremely easy to find out the issue if you go back line by line,
this is a basic procedure of debugging. Please do it yourself next time.
On Mon, Jan 13, 2014 at 3:09 PM, IZHAK shabso
uot;
Then, I installed xfonts-100dpi, xfonts-75dpi (Ubuntu-based).
xset -q #output
Font Path:
/usr/share/fonts/X11/misc,/usr/share/fonts/X11/Type1,built-ins
Regards,
Lisa.
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d0 | persons), family=binomial,
data=tmp1, nAGQ=3)
a[ii,1]=lm1@fixef[1]
a[ii,2]=lm1@fixef[2]
a[ii,3]=vcov(lm1)[1,2]/prod(sqrt(diag(vcov(lm1
a[ii,4:5]=sqrt(diag(vcov(lm1)))
}
return(a)
}
##
what i want is for the function to go on to the next data
3]){
tmp<-mydata[,,ii]
tmp1<-as.data.frame(tmp)
names(tmp1)=c("persons", "d1", "tp", "fn", "fp", "fn", "detect", "d0",
"outcome")
lm1<-tryCatch(lmer(outco
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> C
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Thanks for the illustrative example. In my project actually my supervisor
wanted to estimate the probabilities using a "conditional MLE" approach,
which happens to be the case that *uses clogit() while trying to achieve
aim b in your words*.
I learned that clogit() is based on the sufficient stati
sorry
for the misunderstandings.
Thanks,
Lisa
On 1 Mar, 2013, at 10:10 PM, David Winsemius wrote:
>>>> I still don't think the exp(lp)/(1+exp(lp)) gonna work. Since this is
>>>> conditional logit model, while this formula is only used in unconditional
>>>> on
I do appreciate this answer. I heard that in SAS, conditional logistic model do
predictions in the same way. However, this formula can only deal with in-sample
predictions. How about the out-of-sample one? Is it like one of the former
responses by Thomas, say, it's impossible to do the out-of-sa
more = TRUE)
j=j+1
print(xy, pos = c(0.33, 0.0, 0.66, 0.45), more = TRUE)
j=j+1
counts <- table(mtcars$gear)
barplot(counts, main="Car Distribution", #this part is not correct
xlab="Number of Gears")
Thanks in advance,
Lisa
___
TRUE)
j=j+1
print(xy, pos = c(0.66, 0.50, 0.99, 0.95), more = TRUE)
j=j+1
print(xy, pos = c(0.0, 0.0, 0.33, 0.45), more = TRUE)
j=j+1
print(xy, pos = c(0.33, 0.0, 0.66, 0.45), more = TRUE)
j=j+1
print(xy, pos = c(0.66, 0.0, 0.99, 0.45), more = FALSE)
grid.text("Upper row", x=0.5, y=0
5), more = TRUE)
j=j+1
print(xy, pos = c(0.33, 0.0, 0.66, 0.5), more = TRUE)
j=j+1
print(xy, pos = c(0.66, 0.0, 0.99, 0.5), more = FALSE)
Many thanks,
Lisa
- Original Message -
From: Lisa Daniel
To: "r-help@r-project.org"
Cc:
Sent: Thursday, February 14, 2013 12:14 PM
Sub
7;E' , ' ','G','
'))),
If possible, I would like a heading for each rows (justified to center) above
the individual heading.
Please help.
Many thanks,
Lisa.
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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.
1.0), more = TRUE)
j=j+1
print(xy, pos = c(0.66, 0.5, 0.99, 1.0), more = TRUE)
j=j+1
print(xy, pos = c(0.0, 0.0, 0.33, 0.5), more = TRUE)
j=j+1
print(xy, pos = c(0.33, 0.0, 0.66, 0.5), more = TRUE)
j=j+1
print(xy, pos = c(0.66, 0.0, 0.99, 0.5), more = FALSE)
Pascal:
I would prefer not to use p
(xy, pos = c(0.33, 0.0, 0.66, 0.5), more = TRUE)
j=j+1
print(xy, pos = c(0.66, 0.0, 0.99, 0.5), more = FALSE)
Also, I would like to create a main title for the first row and second row of
figures.
Many thanks in advance,
Lisa
- Original Message -
From: Duncan Mackay
To: Lisa Daniel ;
c(0.66, 0.5, 0.99, 1.0), more = TRUE)
print(xy, pos = c(0.0, 0.0, 0.33, 0.5), more = TRUE)
print(xy, pos = c(0.33, 0.0, 0.66, 0.5), more = TRUE)
print(xy, pos = c(0.66, 0.0, 0.99, 0.5), more = FALSE)
Thanks in advance,
Lisa
- Original Message -
From: Lisa Daniel
To: "r-help@r-projec
frow=c(2,2))
hist(a,main="")
title(main="A",adj=0)
hist(b,main="")
title(main="B",adj=0)
hist(c,main="")
title(main="C",adj=0)
hist(d,main="")
title(main="D",adj=0)
I tried this to xyplot, but it was not wor
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Understand major shortcomings of using only decision trees and how tree
ensembles can help overcome these challenges
clusters while the next 5 - the code is exactly the same.
I assume there is some randomization that I don't see. Is there a way
to get a consistent answer. Thanks.
-Lisa
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?
Thank you very much in advance,
Lisa-Marie
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and
nsform them somehow but I don't know how.
ANY help would be really appreciated, along with comments about the rest of
the work done.
Thanks in advance,
Lisa
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dy helped me to think about
this problem a little differently.
Sincerely,
Lisa
On Tue, Nov 8, 2011 at 9:32 PM, John Fox wrote:
> Dear Lisa,
>
> There doesn't seem to be anything logically wrong with your model.
>
> I don't have much time today to look into it, but
can i be taken off of this mailing list please?
is there another way that you can access this without having to get all the
emails??
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Thank you. I will try this.
Lisa
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Thanks you for your reply.
I consider only the performance of “vars” (v1 to v10), so I just want to
indicate then on the y-axis at the point they should be located.
Lisa
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Sent from
"",
ylab = "", axes = F)
abline(v = 0, lty = 2)
axis(1)
mtext(side = 2, text = c(vars), at = sid, las = 2, line = 0.8)
axis(3)
But the text annotation can not be displayed correctly, i.e., some of them
stick together.
Can anybody help me with this particular pr
Thanks. I have installed PBAT on my computer.
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I consider a model with the disease as a phenotype, and two covariates
(age, weigth) and two SNPs as predictors as bellow:
pbat.m(AffectionStatus ~ age + weight, y1, x1, fbat="gee")
The function had been running for a very long time and no output was
returned until I had to stop it.
Any help wo
HI everyone,
I'm trying to assign colors to multiple lines in a graph. Problem is I don't
want to type in as many colors as there are linesis there a way around
this? In brief, I'm plotting the logratio for up to 60 samples and want a
different color for each sample. Here is the code I'm us
Thank you!
LT
On 7/7/11 3:46 PM, "Joshua Wiley" wrote:
Hi lt2,
I would use the ggplot2 or lattice package. It strikes me as more
effort to do in traditional graphics. Anyway, here are some examples.
Lattice is a very nice package, but I am not quite as familiar with
it, so my examples
question here to see if there is some R
functions or some idea that can help me solve this problem. Thanks.
Lisa
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This is not a homework. I just want to see if there are some R functions or
some ideas I can borrow to solve my problem.
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there is another way to get the similar
result, i.e., first take the derivative of y with respect to each element of
b, and then take its reciprocal. But it is not what I want. Could someone
please tell me how to solve this problem? Thank you in advance.
Lisa
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Thank you very much, Bill. Your script works very well.
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Thank you so much, Filipe. Your R script is what I am looking for.
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Thank you, Duncan,
Here “a” has the length of 24, and “b” has the length of 20 with numbers
from 1 to 20 uniquely. I just want encode “a” from 1 to 20 based on “a”
current order using “b”. So, a1[1] = b[1] = 1
a1[2] = b[2] = 5
a1[3] = a1[4] = b[3] = 8 (since third and fourth numbers are the same i
Thank you for your help, Pete. I tried b[a], but it is not a1.
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, 8, 9, 9, 14, 20, 3, 10, 10, 12, 6, 16, 7, 11, 13, 13, 17,
18, 2, 4, 15, 19)
Does anyone have a suggestion how to deal with this? Thank you in advance.
Lisa
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Hi, Phil,
Yes. That's what I am looking for. Thank you so much.
Lisa
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reatly appreciated.
Lisa
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Thanks for your comments and suggestion. I didn’t show all my own function
here because it has many lines. “x” is the results of another function. I am
calling summary because I want to extract some values from the results.
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and
doesn’t need showing.
Does someone have any idea or any suggestion? Thank you in advance.
Lisa
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Hi, Bill,
Thank you for your help. Your R script works very well.
Lisa
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uld appreciate any help on this question. Thanks a lot.
Lisa
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Hi, Duncan,
On your and William Dunlap’s suggestion, I tried this:
cat(sep="\n", strwrap("my long character string")),
and it works very well. Thanks so much for your help.
Lisa
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unity for students to learn about ongoing
> researches in the field of mathematics, computer science, physics,
> chemistry, a$
Any help will be appreciated!
Lisa
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I am trying to write a function to check unusual values in my datasets and
correct them. As some R users suggested, I try to use readline() and scan()
in my function. Suppose there are several unusual values in a dataset. I
want to change the line numbers in scan() to something like: unusual value
9
Any help will be appreciated.
Lisa
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don’t know how to change the source code. Could you please help how to get
this done? Thanks.
Lisa
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Thanks. I will try them.
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continues to execute.
Can anybody please help how to get this done? Thanks a lot in advance
Lisa
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Yes. That is what I want. Thank you very much.
Lisa
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Thanks a lot.
Lisa
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Dear all,
I just want to determine if the characters in a character string are the
same or not. For example,
temp <- c("aa", "aA", "ab")
How do I determine the first one have the two same “a”, and the second and
third have the different characters? Thanks i
Thank you so much!
Lisa
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Thank you for your help. Your R code works well.
Lisa
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Thank you for your reply again. I really know that NA is not "NA". I just
want to figure out how to remove "NA" from the levels. Thanks again.
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Did you see the data frame "d"? Thanks.
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=NA)
[1] xx yy NA
Levels: NA xx yy
But “NA” is still listed in the levels. How can I solve this problem? Thanks
in advance.
Lisa
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Thank you for your kind help. Your R code works very well.
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, the new dataset is:
group subject result v4 v5
8 3 1 01 0
9 3 2 00 1
10 3 3 10 0
11 3 4 01 0
Can anybody please help how to get this done? Your help would be greatly
appreciated.
Lisa
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rial but a different probability, for example, rbinom(1, 1, 0.7).
Repeat these processes…
How can I do this efficiently if there are more than thousand records
(rows)? Can anybody please help how to get this done? Thanks a lot in
advance
Lisa
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Thank you very much!
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Hi, everybody,
I just want to pass arguments to a function as below:
range <- c(0.1, 0.5)
runif(1, range)
But it doesn’t work.
Does anyone have any suggestions to offer?
Thanks.
Lisa
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Your R script works very well. Thank you very much.
Lisa
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0.0 0.00.2
30.0 0.0 0.00.0 0.00.0 0.80.7
Can anybody please help how to get this done? Thanks a lot in advance
Lisa
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Thank you so much.
Lisa
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ne? Your help would be greatly
appreciated.
Lisa
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It works well. Thank you very much.
Lisa
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How to delete, say, the first columns of the list or the second rows? Thank
in advance.
Lisa
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That's what I want. Thank you so much.
Lisa
Henrique Dallazuanna wrote:
>
> Try this:
>
> my.f <- function(a, b) {
> x1 <- 2 * 3
> x2 <- 3 / 6
> x3 <- 4 * 4 / 5 - sqrt(2)
> y <- get(deparse(substitute(a))) + get(deparse(sub
und”.
Can anybody help me solve this problem? Thanks in advance.
Lisa
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Thank you for your kind help. Your R script works well.
Lisa
Dieter Menne wrote:
>
>
>
> Lisa wrote:
>>
>> I have a dataset that looks like this:
>>
>>> data
>> idcode1code2
>> 1 114
>&g
rds, I want to change 3 to 4 and 4 to 6.
Can anybody please help how to get this done? Thanks a lot in advance
Lisa
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Your script works well. Thank you very mcuh.
Lisa
Nutter, Benjamin wrote:
>
> It seems from your example that you're assuming all of the vectors have
> the same length. If this is the case, then a data.frame might be your
> friend.
>
>> df <- data.frame(
>
Thanks. I also need column name "v5".
Lisa
jholtman wrote:
>
> ?get
>
> something like:
>
>> v1 <- c(0, 1, 0)
>> v2 <- c(1, 1, 0)
>> v3 <- c(2, 1, 2)
>> v4 <- c(2, 2, 1)
>> v5 <- c(0, 1, 1)
>> x <- 5
>>
t; "v5"
[3,] "0" "0" "v5"
This is not what I want. I want to get this:
v1 v2 v5
[1,] 0 1 0
[2,] 1 1 1
[3,] 0 0 1
Can you give me further suggestions or comments? Thanks a lot.
Lisa
Barry Rowlingson wrote:
>
> On Tue,
Thank you for your reply. But in the following case, “cat()” or “print()”
doesn’t work.
data.frame(cbind(variable 1, variable 2, cat(paste("variable", x), "\n"))),
where x is a random number generated by other R script.
Lisa
Duncan Murdoch wrote:
>
> On 29/1
Dear All,
I am not sure how to remove double quotation marks in a string, e.g.,
paste("variable", 1). Can anybody please help me solve it? Thank you in
advance.
Lisa
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