Hi Bert, The “status” at the end of the study does exist in the original
dataset, what was missing was the time between events. And there exists so many
events that fall between the first and last day to be explored in this work.
The suggestion I received then, was to compute time between the in
Hi Jeff, thanks and I will explore your suggestions too..
Regards
---
Kevin Wame
On 7/4/16, 12:43 AM, "Jeff Newmiller" wrote:
There are a great many hits when I search on the keywords "kaplan meier plot
R"... so my
A kaplan-meier plot requires for each individual (in each treatment
group, if there are more than one):
1. Survival time,which in your case appears to mean time without disease;
2. Status at end of time on study: whether the individual was censored
(still without disease) or died (in your case, wa
Dear R experts,
I compare two diagnostic tests. Therfore, I collected patient data from
several studies. The dataframe is similar to this one:
set.seed(10)
data = data.frame( test1 = rbinom(1000, 1, 0.6),
test2 = rbinom(1000, 1, 0.4),
reference = rbinom(1000,
There are a great many hits when I search on the keywords "kaplan meier plot
R"... so my first reaction is that you should be referring to some of the
existing packages for doing this type of analysis. I do not do this type of
analysis normally, so am probably not your best helper... perhaps som
Hi Bert, my first task is to make a Kaplan Meier Plot to evaluate the risk of
developing disease in the treated vs the non-treated individuals. I therefore
figured it might be easier to compute dates first as any further analysis will
be based on time, in this case days. I keep getting recommend
I haven't followed this thread closely, but if it's not too late, I
might suggest that you stop worrying about how you want your data
frame to look and start worrying about you want to display/analyze
your data. As Jeff suggested, you and your supervisor are probably
being driven by paradigms from
Do the elements in 'locs' hahabe an _ somewhere? If not the search and
replace find nothing.
Bert's suggestion of taking a substring is better if you are just
interested in characters on fixed positions.
Bert also suggested that you could maybe benefit from reading a few
tutorials and I agree.
B
Hi Jeff, It works on well on a dataset with 10 rows and I figure it will
work well with the “real” dataset. You’ve been of great help and I am starting
to make headway.
It creates a new dataframe (result), as shown below that doesn’t quite have the
result as I would want it.
ID admin
Thanks Jeff, let me try it on the larger dataset.
Regards
---
Kevin Wame
On 7/3/16, 10:09 PM, "Jeff Newmiller" wrote:
result <- ( result0
%>% select( -admin_period1 )
%>% inner_join( result0 %>
Typo on the second line
result <- ( result0
%>% select( -admin_period1 )
%>% inner_join( result0 %>% select( ID, admin_period1, end=start )
, by = c( ID="ID", admin_period ="admin_period1" )
)
%>% mutate( ddays = end -
Hi Lily,
My suggestion should remove the underscore and everything after it, leaving
just aClim and bClim in the ID column.
Best
Ulrik
On Sun, 3 Jul 2016, 20:34 lily li, wrote:
> Hi Ulrik,
>
> Thanks. This is for one group, but how to do for several groups? I tried
> gsub(c(),c(),df$ID), but i
I strongly suspect that you do not need to do this. What I think you
do need to do is to create a new column (which will be a factor)
identifying the climate ("a" or "b"), which can then be used to group
climates in plots, used as a covariate in statistical analyses, etc.
Moreover, there is probabl
I still get the impression from your mixing of information types that you are
thinking like this is Excel.
Perhaps something like
drug_study$admin_period <- ave( "Y" == drug_study$drug_admin, drug_study$ID,
FUN=cumsum )
library(dplyr)
result0 <- ( drug_study
%>% filter( 0 != admin_
Hi Ulrik,
Thanks. This is for one group, but how to do for several groups? I tried
gsub(c(),c(),df$ID), but it does not work.
On Sun, Jul 3, 2016 at 12:24 PM, Ulrik Stervbo
wrote:
> Hi Lily,
>
> you can use gsub:
>
> df$ID <- gsub("_.*", "", df$ID)
>
> HTH
> Ulrik
>
> On Sun, 3 Jul 2016 at 20:
Hi Lily,
you can use gsub:
df$ID <- gsub("_.*", "", df$ID)
HTH
Ulrik
On Sun, 3 Jul 2016 at 20:16 lily li wrote:
> I have a problem in changing row names in a dataframe in R. The first
> column is ID, such as aClim_st02, aClim_st03, aClim_st 05, bClim_st01,
> bClim_st02, etc. How to rename the
I have a problem in changing row names in a dataframe in R. The first
column is ID, such as aClim_st02, aClim_st03, aClim_st 05, bClim_st01,
bClim_st02, etc. How to rename the names, so that aClim_ all grouped to
aClim, while bClim_ all grouped to bClim? Thanks for your help.
df
ID
HI Jeff, it’s been an uphill task working with the dataset and I am not the
first to complain. Nonetheless, data-cleaning is ongoing and since I cannot
wait for that to get done, I decided to make the most of what the dataset looks
like at this time. It appears the process may take a while.
Tha
> On 03 Jul 2016, at 13:47 , varin sacha via R-help
> wrote:
>
> Dear R-experts,
>
> I am trying to calculate the bootstrapped (BCa) regression coefficients for a
> robust regression using MM-type estimator (lmrob function from robustbase
> package).
>
> My R code here below is showing a wa
Your goal of putting character representations of dates in certain rows of a
column is hard to imagine a use for. Your goal of identifying start and end
dates seems reasonable enough. It can be accomplished using aggregate from base
R (less external dependency) or summarise from dplyr (faster,
Well, yes, ... but no: there is no need to pre-define the matrix.
The following is still a (interpreted) loop, but it is fast and short.
## ex is the downloaded array, here filled with random numbers
reqX = c(35,35,40,65,95)
reqY = c(2,5,10,112,120)
out <-sapply(seq_along(reqX), function(i)ex[r
Thank you both.
Yes, this is basically the issue of able to subset an array rather than
extracting from the netCDF file. The dX = ncvar_get(nc=myNC,
varid="myVar")command already results in the array. And one can subset that
array using indices.
In turn the problem can be stated as follows:Let u
The data set did not show up. The R-help list tends to strip out most file
types as a safety precaution. Try renaming the file from xxx.csv to xxx.txt
and it should come through alright.
John Kane
Kingston ON Canada
> -Original Message-
> From: kwa...@kemri-wellcome.org
> Sent: Sun,
Dear R-experts,
I am trying to calculate the bootstrapped (BCa) regression coefficients for a
robust regression using MM-type estimator (lmrob function from robustbase
package).
My R code here below is showing a warning message ([1] "All values of t are
equal to
22.2073014256803\n Can not cal
Hi Marietta,
You may not be aware that the variable k is doing nothing in your
example except running the random variable generation 2 or 3 times for
each cycle of the outer loop as each successive run just overwrites
the one before. If you want to include all two or three lots of values
you will h
Hi Jeff, pardon me, I was surely not making it easy. I hope this time I will ☺
Attached is snippet of the dataset in csv format and below is the R.script I
have managed so far.
---
Hi Clemence,
I don't have sciplot installed, but the help page suggests that the
"xaxt" argument is available. This will prevent the x axis from being
displayed and you can then specify the x axis you want. Assume that
you want an x axis from 0 to 300 by 50:
axis(1,at=seq(0,300,by=50))
Jim
On T
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