))
# I assume you would like to add up the values with na.rm = TRUE
meanFn <- function(x) mean(x, na.rm = TRUE)
# see ?aggregate
aggregate(dat[, c("PO1M", "PO1T", "PO2M")],
by = dat["STUDENT_ID"],
FUN = meanFn)
# if you have largis
2, 2018 1:30:37 PM MDT, "Tóth Dénes" <toth.de...@kogentum.hu> wrote:
On 05/02/2018 07:11 PM, Kevin E. Thorpe wrote:
I suspect this is pretty easy, but I'm having trouble figuring it
out.
Basically, I have a list of data frames such as the following
example:
list(A=data.fr
On 05/02/2018 07:11 PM, Kevin E. Thorpe wrote:
I suspect this is pretty easy, but I'm having trouble figuring it out.
Basically, I have a list of data frames such as the following example:
list(A=data.frame(x=1:2, y=3:4),B=data.frame(x=5:6,y=7:8))
I would like to turn this into data frame
Hi Martin,
On 11/29/2017 10:46 PM, Martin Morgan wrote:
On 11/29/2017 04:15 PM, Tóth Dénes wrote:
Hi,
A benchmarking study with an additional (data.table-based) solution.
I don't think speed is the right benchmark (I do agree that correctness
is!).
Well, agree, and sorry for the wording
d more, see
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posti
ng-guide.html
and provide commented, minimal, self-contained, reproducible code.
[[alternative HTML version deleted]]
__
On 09/29/2017 12:02 AM, Tóth Dénes wrote:
On 09/28/2017 10:25 PM, Dan Abner wrote:
Hi all,
I have a large number of text strings to search for enumerated items.
However, I am receiving this error message even though I thought that I
properly escaped the special character closed parenthesis
t.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
Dr. Tóth Dénes ügyvezető
Kogentum Kft.
Tel.: 06-30-2583723
Web: www.kogentum.hu
___
Dear friends,
How do I stop partial matching of list names?
e.g.,
x - list(=a, B=b)
is.null(x$A) #returns FALSE even though there is no element A.
if(is.null(x$A)) {result - x$} else {result - x$A}
result #is even though there is no x$A element
x -
You could use ?unlist:
structure(data.frame(
matrix(unlist(strsplit(beatles, )),length(beatles),2,T)),
names=c(FirstName,LastName))
Note that this compact code does not guard you against typos, that is
names with 2 or 2 elements.
Hope that helps,
Denes
I have a vector of character
Hi there,
Since you failed to provide us with data and sessionInfo(), I can only
guess that for some reason you call the rtmvt.rejection function instead
of rtmvt.gibbs.
Just look at the code of rtvmt by typing:
rtmvt
There you can see that it is a wrapper for rtmvt.rejection or rtmvt.gibbs.
Hi Wendy,
try this:
Z - matrix(runif(1000)0.5,10,100)
rowSums(Z)
HTH,
Denes
Hi all,
I have a huge matrix of TRUE/FALSE table like following, and I want to
count
the number of TRUEs in each row. Instead of looping through each row and
do
length(Z[Z==TRUE]), I am wondering if there is
I do not know the bioconductor packages you mentioned, but the corr.test
function in the psych package, or the rcorr function in the Hmisc package
should do the work.
Also note that the c() in method=c(pearson) is redundant. Just write
method=pearson instead (or nothing, since this is the default
Dear Dirk,
You should avoid indexing in the glm call so that the name of the terms
will not contain the indexing part. (Check str(lg) in your example.)
A more preferred solution uses predefined data frames in the original calls:
n - 250
x - rnorm(n)
noise - rnorm(n,0,0.3)
y -
Just have a look at ?quantile and the probs argument.
tapply(x, l.c.1, quantile,probs=0.75)
Anyway, quantiles and quartiles are not the same. I guess you meant the
3rd quartile.
All -
I have an example data frame
x l.c.1
43.38812035 085
47.55710661 085
47.55710661 085
You might also consider the Deducer package. You can build up a plot by
point and click and then have a look at (and amend) the code and learn the
syntax of ggplot2, which is a nice alternative to the lattice package.
The website of the Deducer package (www.deducer.org) is a good start.
--
to vacillate between lattice and
ggplot2. I should probably settle on one or the other and learn it
better. I'll admit I like the default look of lattice plots better, but
so far custom panel functions still baffle me.
--Chris
Tóth Dénes wrote:
You might also consider the Deducer package. You can
Hi,
I guess you have commas as decimals in your data. Replace it to decimal
points.
Best,
Denes
On Mar 21, 2011, at 1:57 PM, pat...@gmx.de wrote:
Hi list,
I have problems with the as.numeric function. I have imported
probabilities from external data, but they are classified as
On Mar 21, 2011, at 2:59 PM, Tóth Dénes wrote:
Hi,
I guess you have commas as decimals in your data. Replace it to
decimal
points.
If that is true then the easiest fix would be to set the proper
decimal argument in read.table
In this particular case, sure. If patsko happens to work
Hi,
predictAll should do what you want. See ?predict.gamlss.
HTH,
Denes
Dear All:
I have succeeded in fitting a GAMLSS.dist model to growth data I am
working
with it.
My aim is to create a matrix of predicted percentiles and the
corresponding
the fitted model's sigma mu nu by
Hi!
Sorry, I made an error in the previous e-mail.
So try this:
by(df[,-1],df$id,function(x) apply(x,2,tabulate))
This gives you a list. You can rearrange it into a data frame or a 3d
array if you wish.
Regards,
Denes
Hello everybody,
I have a data frame in R which is similar to the
t(matrix(a,3,4))
for more complex arrays, see ?aperm
Hi again,
I'd like to ask you a question again.
I have a matrix like this:
a -matrix(c(1,2,3,4,5,6,7,8,9,10,11,12))
a
[,1]
[1,]1
[2,]2
[3,]3
[4,]4
[5,]5
[6,]6
[7,]7
[8,]8
[9,]
?unlist
quantile(unlist(data))
Hi dear all,
It may be a simple question, i have a list output with different number of
elements as following;
[[1]]
[1] 0.86801402 -0.82974691 0.3974 -0.98566707 -4.96576856
-1.32056754
[7] -5.54093319 -0.07600462 -1.34457280 -1.04080125
Hi!
Not an elegant solution, but seems to work:
date - c(1:300)
flow - sin(2*pi/53*c(1:300))
levels - factor(rep(c(high,med,low),100))
data - cbind.data.frame(date, flow, levels)
colours - as.numeric(levels)+1
# interpolate
resolution - 0.001
appres -
Indeed, I forgot about the segments function.
with(d,plot(date,flow,type=n))
with(d,segments(start,start.y,end,end.y,col=colour))
Hi,
because each colour is defined on non-consecutive points, you'll
probably need to cut the intervals to define segments around each
point. One approach
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