ht simplify this?
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On 27 Dec 2015, at 08:00, "Polwart Calum (COUNTY DURHAM AND DARLINGTON NHS
FOUNDATION TRUST)" mailto:calum.polw...@nhs.net>> wrote:
***
m, to basically say... has anyone come across a
function in R that might simplify this?
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On 27 Dec 2015, at 08:00, "Polwart Calum (COUNTY DURHAM AND DARLINGTON NHS
FOUNDATION TRUST)"
mailto:calum.polw...@nhs.net>&l
I'm far from an expert on stats but what I think you are saying is if you try
and compare Baseline with Version 3 you don't think your p-value is as good as
version 1 and 2. I'm not 100% sure you are meant to do that with p-values but
I'll let someone else comment on that!.
tot
I've been tweaking code for several days on and off in R, cut and pasting in
from a text editor (I just leave them open all the time). I think I got
something that was usable but then a powersurge tripped the fuses and
unfortunately the machine I was working on doesn't have a UPS.
Does R hold
Rolf Write:
> If you've been cut-and-pasting from a text editor, then your commands
> *might* be in the file .Rhistory. Unfortunately this history gets saved
> only when you exit R (and by default only when you also say ``yes'' to
> saving the workspace) or if you explicitly savehistory().
Was ho
OK - I've been doing some work on getting a forest plot or two together for a
sub-group analaysis. Thats the crucial thing - because its a sub-group
analysis rather than a meta-analsysis and all the forest (meta) and forestplot
(rmeta) instructions assume you are doing a meta-analysis.
I found
>Example,
>data("GlaucomaM", package = "ipred") is accepted. Now instead of GlaucomaM,
I> need to give my own data. But the data format for GlaucomaM is not given.
>So how can I know that?
Not familliar with the packages at all but if you simply enter:
> data ("GlaucomaM", package="ipred")
> Glau
Hi,
I'm sure this should be simple but I can't figure it out! I want to get the
median survival calculated by the survfit function and use the value rather
than just be able to print it. Something like this:
library(survival)
data(lung)
lung.byPS = survfit(Surv (time, status) ~ ph.ecog, data=
>
> # I tried defining a function like this
> myplot <- function(...)plot(..., pch=19, col=c("blue","red")[treatment])
>
> # So i can call it like this:
> with(mydfr, myplot(Xmeas, Ymeas))
>
> # but:
> Error in plot.xy(xy, type, ...) : object 'treatment' not found
>
basically that is something like
> I have got 27 graphs to export (not a lot...I know!). How can I fit all of
> them into a single file like PNG without adjusting the size of the graphs?
> What's in my mind is like pasting graphs into Word, in which I can just
> scroll down to view the graphs.
Pretty sure PNG can only cope with s
>> col=c("blue","red")mydfr$[treatment]
>
> Yes, but I would like to use the function for lots of other dataframes
> as well, so embedding 'mydfr' in the function is not the ideal
> solution...
In that case I'd try something like:
myplot <- function(..., tmnt) {
plot(...,
pch=19,
Sorry I'm having one of those moments where I can't find the answer but I bet
its obvious...
I'm outputting my results to a file using sink()
Is there a command simillar to php's echo command that would allow me to add
some text to that file ie:
dataFr$a = 1:10
dataFr$b = 2*1:10
sink ("filepat
Hi guys I'm hoping someone can help me with this. It should be easy but it
seems to get stuck for no obvious reason! I am trying to set a report up in
odfWeave so that we can re-run the same analysis at 3 time points as the data
matures and provide an 'instant' report.
To simplify the situati
Solved my own problem by using:
odfTable.matrix(
as.matrix (
with (mydata, table (site_id, reaction))
)
)
This message may contain confidential information. If yo...{{drop
I'm hoping I'm missing some (probably fundamental basic process) which might
make my life easier!
Lets assume I have a 3 column table summarizing results from a trial from three
arms (Arm A, B and C).
For each arm there will be a number of pieces of information to report. The
simplest example
I'm anything but an expert in R however if I'm labeling a graph axis with a
superscript I have tended to use:
> plot (x , y , xlab = expression ("label"^2))
But when you try to have more than one superscript it fails. Assuming you are
in a UTF8 location (Western Europe) you could try:
> plot
I tried to post this a few times last week and it seems to have got stuck
somehow so I'm trying from a different email in the hope that works. If
somehow this has appeared on the list 20 tiems and I never saw any of them I
apologize ;-)
I'm basically an R-newbie. But I am VERY computer liter
Slightly confused because if I try:
> newdata.yaxis = c(2.473, 3.123456, 3.23456, 2.67890, 1.56789)
> newdata.yaxis_4 = round (newdata.yaxis, digits = 4)
> newdata.yaxis
[1] 2.47 3.123456 3.234560 2.678900 1.567890
> newdata.yaxis_4
[1] 2. 3.1235 3.2346 2.6789 1.5679
As you see - I ge
So the issue is something to do with the [['xxx']] construction of your data.
Can you explain what thats' all about - as it errors all over the shop when I
try using that...
You've set me on a mission to find the answer! So I'd really like to recreate
a little bit of your data here, and play..
> Are n.FD and n.RD the number of people who received the full/reduced dose
Yes - but I don't have the data structured like that YET - thats what I want to
get to because thats what forest plot seems to be wanting.
> and surv.FD and surv.RD the number of people that survived?
Mmm... was more thin
>> What I want to do is do a forrest (forest) plot for subgroups within my
>> single dataset as a test of heterogeniety. I have a dataset who received
>> either full dose(FD) or reduced dose(RD) treatment, and a number of
>> characteristics about those subjects: age, sex, renal function, weight,
>
> library(RODBC)
> library(HYDAT)
> You will need to install HYDAT (the zip file) from
> http://www.geog.ubc.ca/~rdmoore/Rcode.htm
>
> Below is my current code - which works. The [[]] is the way i am accessing
> the columns from the data frame.
>
> thanks again for all your help
>
> # load HY
Duh! did it again! the variables need str in front of them don't they!!
sub = sprintf('Seasonal station with natural streamflow - Lat: %s Lon: %s
Gross Area %s km\UB2 - Effective Area %s km\UB2 ',
round( str[['metadata']][['latitude']], digits=4 ),
round( str[['metadata']][['longitude']], digits =
>Ah, I think I see what you want. Try this on each pair of exclusive sets:
>
n_total<-dim(mydataset)[1]
under65<-mydataset$age <= 65
n_under65<-sum(under65)
under65row<-c(sum(mydataset$dose[under65] == "FD"),
sum(mydataset$dose[under65] == "RD"),
sum(mydataset$vitalstatus[under65] == "dead" &
>Ah, I think I see what you want. Try this on each pair of exclusive sets:
>Then under65row and over65row should be the first two rows of your result.
>Can't test this at the moment, but I don't think it's too far wrong.
>
I knew this shouldn't need so much work ;-)
Not cracked it yet - because
I may be doing this wrong! but I have a function which I have simplified a lot
below. I want to pass some 'field names' of a data-frame to the function for
it to then do some manipulation of.
Here's my code:
#build a simple dataset
mydataset = data.frame (
ages=c('40-49','40-49','40-49','30-
> I run the script and it exports a PDF called "version 1".
> I want it to check if "version 1" already exists. If so,
> then I want the new graphs to be exported as
> "version 2", and so on.
>
> Is it possible to do it in R?
Someone may know a way. However its certainly possible to execute a co
>LinZhongjun wrote:
>>
> >I ran Winbugs under R. I could get the results, but I kept getting the error
> >messages:
>>
> >Error in
> > file(con, "wb") : cannot open the connection
> >In addition: Warning messages:
> >1: In file.c
Sorry this may well be defined as Off Topic. I apologize in advance.
I am interested in performing what I think would be a probabilistic sensitivity
simulation. I've done some crude ones before in excel but I'm wondering if R
can help me do it more effectively?
I have a set of theoretical va
I have a dataset which for the sake of simplicity has two endpoints. We would
like to test if two different end-points have the same eventual meaning. To
try and take an example that people might understand better:
Lets assume we had a group of subjects who all received a treatment. The could
se, absence of evidence is not evidence of absence. Failing to
> reject the null hypothesis that the
> distributions are different is not proof that the distributions are equal.
Yes absolutely - however I'm half expecting to detect a difference and so then
dismiss using A as a surrog
Some colleagues nationally have developed a system which means they can pick
the optimal sets of doses for a drug. The system could apply to a number of
drugs. But the actual doses might vary. To try and explain this in terms that
the average Joe on the street might understand if you have som
I might be missing something really obvious, but is there an easy way to locate
all non-unique values in a data frame?
Example
mydata <- numeric()
mydata$id <- 0:8
mydata$unique <- c(1:5, 1:4)
mydata$result <- c(1:3, 1:3, 1:3)
> mydata
$id
[1] 0 1 2 3 4 5 6 7 8
$unique
[1] 1 2 3 4 5 1 2 3 4
$r
This may be a really obvious question but I just can't figure out how to do it.
I have a small dataset that I am trying to compare to some controls. It is
essential that the controls are matched on Cancer Stage (a numerical factor
between 1 and 4), and then ideally on Age (integer), Gender (fac
I have some data which is censored and I want to determine the median. Its
actually cost data for a cohort of patients, many of whom are still on
treatment and so are censored.
I can do the same sort of analysis for a survival curve and get the median
survival... ...but can I just use the surv
ert.lancas...@orbitz.com]
Sent: 03 November 2011 19:55
To: Polwart Calum (COUNTY DURHAM AND DARLINGTON NHS FOUNDATION TRUST);
r-help@r-project.org
Subject: RE: Kaplan Meier - not for dates
I think it really depends on what your event of interest is. If your event is
that the patient got better
> 2. The answer will be wrong. The reason is that the censoring occurs on a
> time scale, not a $ scale: you don't stop observing someone because
> total cost hits a threshold, but because calendar time does. The KM routines
> assume that the censoring process and the event process are on the
> 2. The answer will be wrong. The reason is that the censoring occurs on a
> time scale, not a $ scale: you don't stop observing someone because
> total cost hits a threshold, but because calendar time does. The KM routines
> assume that the censoring process and the event process are on the
Before I go and do this another way - can I check if anyone has a way of
looping through data in odfWeave (or possibly sweave) to do a repeating
analysis on subsets of data?
For simplicity lets use mtcars dataset in R to explain. Dataset looks like
this:
> mtcars
mpg cyl disp
I am using seq with the expression seq(1.4, 2.1, by=0.001) to create a sequence
of references from 1.4 to 2.1 in 0.001 increments. They appear to be created
correctly. They have a related pair of data which for the purposes of this we
will call val. I'm interested in the content on the row wi
R-help On Behalf Of Ben Tupper
> Sent: Thursday, January 17, 2019 2:43 PM
> To: POLWART, Calum (COUNTY DURHAM AND DARLINGTON NHS FOUNDATION TRUST)
>
> Cc: r-help@r-project.org
> Subject: Re: [R] I can't get seq to behave how I think it should
>
> Hi,
>
> This lo
about it from FAQ 7.31.
Cheers
Petr
> -Original Message-
> From: POLWART, Calum (COUNTY DURHAM AND DARLINGTON NHS FOUNDATION
> TRUST)
> Sent: Thursday, January 17, 2019 2:56 PM
> To: PIKAL Petr ; Ben Tupper
>
> Cc: r-help@r-project.org
> Subject: RE: [R] I can&#x
I'm writing a quite large document in Rmarkdown which has financial data in it.
I format that data using scales::dollar() currently something like this:
>
> require (scales)
> x = 10
> cat (dollar (x, prefix ="£", big.mark=","))
£100,000
But actually, I'd quite like to get £100k out in tha
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