As I noted in my earlier post, it does - had checked that ;-)
It works by taking corresponding pair fo the input vectors (after possible
recycling, as eluded by Helmut in his remark on working on only one vector) as
needed for outer.
Thanks for the reminder, though,
Benno
> On 29. Sep
I would assume the following snippet does what you want - note the use of outer
with anonymous function wrapping powerTOST:
z <- outer(xs, ys, function(x, y)power.TOST(CV = y, theta0 = x, design =
"2x2x4", method = "central", n = res[1]))
contour(xs, ys, z)
Benno
> On 29. Sep 2020, at
Is this a for a homework assignment? We try to avoid doing others' homework
on this list.
To learn more about how to do simulations in R, search on "how to simulate
data in R" or similar. This brought up many tutorials that should be
helpful.
If this is not homework, maybe someone else can make s
Hi R User,
I was trying to create a data matrix with four columns: height, locations
(three locations), temperature (three levels) and slope (three classes) and
used the following code, but it did not work. I basically wanted to have
20 rows for each class (I have a total of 9 classes). Can you he
Jim,
I tried a few things, I found that clip() works if I just do some regular
graphing tasks. But as long as I run lines(fit) with "fit" object is a survfit
object, this would reset to default plot region. See the ovarian example below:
library(survival)
ovarian1<-ovarian
ovarian1$fustat[ovari
Hi Jim,
I tried points(-1,-1) before lines() and before clip(), but either way, it
still shows everything, :-(
It's interesting that the examples with hist() provided by the R help of clip
function works nicely.
I also tried a simple linear regression plots below, clip() works, too.
dat<-data
Hi John,
I should have remembered this. For some reason, the clip() function
doesn't operate until you have issued a graphics command. Try:
points(-1,-1)
before calling lines()
Jim
On Wed, Sep 30, 2020 at 12:26 PM array chip wrote:
>
> Hi Jim,
>
> I tried the clip() function below, surprisingl
Hi Jim,
I tried the clip() function below, surprisingly it did not work! I read the R
help file and feel your script should work. To have a workable example, I used
the ovarian dataset in the survival package as an example:
ovarian1<-ovarian
ovarian1$fustat[ovarian$futime>450]<-0
ovarian1$futim
Hi, Sarah,
There is a mistake in the column name and the desired output is pretty much
the same as the input data frames except for the replacement of
'some_string' by the word 'alcohol' in the column I specified above.
Yet, your code is what I am looking for and helped me to get started and
figur
Your desired output seems to be the same as your desired input in your
example, and your data frames have different column names.
Nonetheless, this bit of code will find rows with "alcohol" in column
3, and for those rows replace the contents of column 4 with column 3.
That may not be exactly what
Dear all,
I have had difficulties copying the word "alcohol" in the "vehicle" column
to replace the string in the column "accident_type". It is a huge list but
I have prepared a workable and simple example below and the desired
output. I am sure you guys can give me some advice on how to deal wit
A simpler, cleaner, and maybe faster approach is to use outer():
nm <- unique(dat$PLATE)
dat <- cbind(dat, 1+outer(dat$PLATE,nm, "=="))
names(dat)[-(1:3)] <- nm
Bert
Bert Gunter
"The trouble with having an open mind is that people keep coming along and
sticking things into it."
-- Opus (aka Be
Hello,
Sorry, I didn't understand that 1 and 2 are the final values, I thought
they would be counts of PLATE. I have changed the auxiliary column
'counts' to 'flag'.
mc %>%
group_by(PLATE) %>%
mutate(flag = 2) %>%
pivot_wider(
id_cols = c("FID", "IID"),
names_from = "PLATE",
Hello,
Something like this?
mc <- read.table(text = "
FID IID PLATE
1 fam0110 G110 4RWG569
2 fam0113 G113 cherry
3 fam0114 G114 cherry
4 fam0117 G117 4RWG569
5 fam0118 G118 5XAV049
6 fam0119 G119 cherry
", header = TRUE)
library(dplyr)
library(tidyr)
mc %>%
group_by(PLATE) %>%
You should be able to figure this out yourself!
The number of new columns you add **must agree with the number of unique
id's PLATE** !
Here is the general approach, slightly more efficient, probably:
PL = dat$PLATE
nm<- unique(PL)
len <- length(nm)
m <- matrix(0, ncol = len, nrow = nrow(dat), d
oh it seems that I can just use your last line of code and solve my problem:
m2=tapply(mc$IID, list(FID=mc$FID, PLATE=mc$PLATE), mean)
m2=as.data.frame(m2)
library(data.table)
m3=setDT(m2, keep.rownames = TRUE)[]
colnames(m3)[1] <- "FID"
mt=merge(mc,m3,by="FID"
for(i in 4:ncol(mt)) mt[,i] <- 1 + (n
HI Bert,
thank you for getting back to me.
I tried this:
> dat <- cbind(mc, matrix(0,ncol = 34))
> head(dat)
FID IID PLATE 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22
1 fam0110 G110 4RWG569 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
2 fam0113 G113 cherry 0 0 0
Thank you Marc as well! I'll try both ways! Yes this is an oncology study with
time set at 5
John
On Tuesday, September 29, 2020, 07:08:39 AM PDT, Marc Schwartz
wrote:
Hi John,
>From the looks of the first plot, it would appear that perhaps you are engaged
>in a landmark analysis in a
Thank you very much Jim, this is great!!
John
On Tuesday, September 29, 2020, 01:35:48 AM PDT, Jim Lemon
wrote:
Hi John,
Perhaps the most direct way would be:
plot(fit1, col=1:2)
xylim<-par("usr")
clip(4,xylim[2],xylim[3],xylim[4])
lines(fit2,col=1:2)
Remember that the new clipping rec
I am not sure reshape2 is appropriate for this task, but, assuming I
understand correctly, it's quite easy without it. The following is one way,
which probably can be done more elegantly and efficiently, but I think it
does what you want.
"dat" is your example data frame, in which the columns were
Hello,
I have a data frame like this:
> head(mc)
FID IID PLATE
1 fam0110 G110 4RWG569
2 fam0113 G113 cherry
3 fam0114 G114 cherry
4 fam0117 G117 4RWG569
5 fam0118 G118 5XAV049
6 fam0119 G119 cherry
...
> dim(mc)
[1] 16254
> length(unique(mc$PLATE))
[1] 34
I am trying to make a ne
Dear Duncan and Benno,
Duncan Murdoch wrote on 2020-09-29 15:21:
On 29/09/2020 9:16 a.m., Puetz, Benno wrote:
As I noted in my earlier post, it does - had checked that ;-)
Great!
The result in all of its "beauty":
https://forum.bebac.at/mix_entry.php?id=21930#top21939
THX again!
Helmut
Hi John,
>From the looks of the first plot, it would appear that perhaps you are engaged
>in a landmark analysis in an oncology setting, with the landmark time set at 5
>years. On the off chance that you are not familiar with the pros and cons of
>that methodology, Google "landmark analysis", w
On 29/09/2020 9:16 a.m., Puetz, Benno wrote:
As I noted in my earlier post, it does - had checked that ;-)
Great!
Duncan Murdoch
It works by taking corresponding pair fo the input vectors (after possible
recycling, as eluded by Helmut in his remark on working on only one vector) as
needed
That won't work unless power.TOST is vectorized. outer() will pass it
vectors of x and y values.
Duncan Murdoch
On 29/09/2020 8:11 a.m., Helmut Schütz wrote:
Dear Benno,
THX, you made my day! Case closed.
Helmut
Puetz, Benno wrote on 2020-09-29 13:14:
I would assume the following snippet
Dear Benno,
THX, you made my day! Case closed.
Helmut
Puetz, Benno wrote on 2020-09-29 13:14:
I would assume the following snippet does what you want - note the use
of outer with anonymous function wrapping powerTOST:
z <- outer(xs, ys, function(x, y)power.TOST(CV = y, theta0 = x, design
=
Dear Duncan,
Duncan Murdoch wrote on 2020-09-29 11:57:
On 29/09/2020 5:37 a.m., Helmut Schütz wrote:
Here I'm lost. power.TOST(theta0, CV, ...) vectorizes properly for
theta0 _or_ CV but no _both_. Hence
library(PowerTOST)
power.TOST(theta0 = c(0.9, 0.95, 1), CV = 0.25, n = 28)
and
power.TOST(t
On 29/09/2020 5:37 a.m., Helmut Schütz wrote:
Dear Duncan,
Duncan Murdoch wrote on 2020-09-28 21:47:
You're doing a lot of manipulation of the z matrix; I haven't followed
all of it, but that's where I'd look for problems. Generally if you
keep your calculation of the z matrix very simple you
Dear Duncan,
Duncan Murdoch wrote on 2020-09-28 21:47:
You're doing a lot of manipulation of the z matrix; I haven't followed
all of it, but that's where I'd look for problems. Generally if you
keep your calculation of the z matrix very simple you are better off.
For example, once you have xs
Hi All,
I recreated a new "sintetic" df1 based on df2 and I could merge:
> day_1 <- as.POSIXct("2005-11-14-00-00", format="%Y-%m-%d-%H-%M",
tz="Etc/GMT-1")
> day_2 <- as.POSIXct("2005-11-14-12-00", format="%Y-%m-%d-%H-%M",
tz="Etc/GMT-1")
> df2 <- data.frame(data_POSIX=seq(day_1, day_2, by="30 mi
Hi John,
Perhaps the most direct way would be:
plot(fit1, col=1:2)
xylim<-par("usr")
clip(4,xylim[2],xylim[3],xylim[4])
lines(fit2,col=1:2)
Remember that the new clipping rectangle will persist until you or
something else resets it.
Jim
On Tue, Sep 29, 2020 at 10:34 AM array chip via R-help
wr
Hi Stefano,
For the merge() part, I'll leave it to more expert users (I rarely use
merge(), and every time I need it, it's painful...).
To know why instead of NA, check the results with str(df3); I guess
it is not the mode you expected.
For more details, you should provide the file, or bet
Dear R users,
I'm struggling with a simple "merge".
I have an external file called df.txt like
data_POSIX, event
2005-11-14 02:30:00, "start"
2005-11-14 11:30:00, "end"
I load it with
df1 <- read.table(file="df.txt", header=TRUE, sep=",", dec = ".",
stringsAsFactors=FALSE)
df1$data_POSIX <- as
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