Hi,
I just noticed that you also have all is.na() or !is.na() subsets. To do that:
mydatatest - structure(list(CaseID = structure(1:8, .Label = c(1605928,
1605943, 1605945, 1605947, 1605949, 1605951, 1605952,
1605953), class = factor), Structure = structure(c(1L, 1L,
3L, 2L, 2L, 3L, 3L, 4L),
On Wed, Mar 19, 2014 at 4:03 AM, Wolf, Steven wolfs...@msu.edu wrote:
Hi R users,
This isn't a R-specific issue, per-se, but I thought that this list would
have some helpful input on this topic. First, a bit of background. I am
working on a project which is interested in following approx
This package provides meta-analysis of significance values.
Although the preferred way of performing meta-analysis uses effect
sizes sometimes these are unavailable and only p-values can be
obtained from the primary studies. The metap package provides several
ways of synthesising p-values and
Wolf, Steven wolfs...@msu.edu [2014-03-18 21:05]:
I am looking for a verification protocol which can look at a data file and
see if it has been modified. Ideally, this should be something that I can
check the file with to see if the file has been changed or corrupted and
incorporate into
Hi everybody,
I am sorry if this is a silly question (I have tried to search it at
http://tolstoy.newcastle.edu.au/R/ but I get a 404 page).
I have a script writing some data.frames to csv files.
I write the csv files using the write.csv function and the option
row.names=TRUE.
Everything works
The argument is col.names, not colnames.
Best,
Ista
On Mar 19, 2014 6:14 AM, Luca Cerone luca.cer...@gmail.com wrote:
Hi everybody,
I am sorry if this is a silly question (I have tried to search it at
http://tolstoy.newcastle.edu.au/R/ but I get a 404 page).
I have a script writing some
Thanks Ista,
sorry it was a typo in my email, but in the code I have the right option.
On Wed, Mar 19, 2014 at 11:58 AM, Ista Zahn istaz...@gmail.com wrote:
The argument is col.names, not colnames.
Best,
Ista
On Mar 19, 2014 6:14 AM, Luca Cerone luca.cer...@gmail.com wrote:
Hi everybody,
Thanks Duncan,
I knew that, I wanted to know if there was some way not involving
having to recreate the df.
It is not a big issue in this case (they are not very big), but it
might be problematic for big data frames.
Thanks for advice on using col.names=NA.
Cheers,
Luca
Luca Cerone
Tel: +34 692
Hello,
write.csv(df, row.names=TRUE, col.names=c(ID, colnames(df)))
,a,b
A,1,3
B,2,4
C,3,5
Warning message:
In write.csv(df, row.names = TRUE, col.names = c(ID, colnames(df))) :
attempt to set 'col.names' ignored
This gives a warning, not an error message. I've also tried setting
col.names
Hi Weiwei,
I found your post on Mahalanobis distances five years ago.
/ Error in solve.default(cov, ...) : system is computationally singular:
// reciprocal condition number = 1.09501e-25
/I have the same problem now - and I don't know how to deal with it. I would be
thankful if you could
Topic: importing data from a web page
Dear Sir/ Madam
Hi. First I apologize for this question. But this also shows I'm
inexperienced in using R.
I need to import some tables from a web page (https) into R and analyze
them through the readHTMLTable function. After loading XML package, the
HI all
How to add row names at the end of the record after transposing a matrix
Sample code
Data = read.table(set_a_2.txt,header = TRUE, sep = \t)
Data.T - t(Data[,0:ncol(fooData)])
write.table(Data.T, file =set_a_2_format.tab,row.names = TRUE, col.names
= TRUE, quote = FALSE)
example of
I would like to announce the release of version 1.2.2 of the RSelenium
package on CRAN. RSelenium is an R package that provides a set of R
bindings for the Selenium 2.0 webdriver. Selenium automates browsers. Using
RSelenium you can automate browsers locally or remotely. The set of
supported
Hi
Not sure what you exactly want but some comments in line
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
project.org] On Behalf Of Ankush Sharma
Sent: Wednesday, March 19, 2014 11:36 AM
To: r-help@r-project.org
Subject: [R] Row names at the end of
Hi
Although I am not experienced with XML package the warning seems pretty
straightforward to me. The object you want to load is not recognised as XML.
Most probably readHTMLTable requires some format as input and your url is not
accepted.
Regards
Petr
-Original Message-
From:
On 19 Mar 2014, at 11:10 , Mehrshad Koleini mehrkole...@gmail.com wrote:
I need to import some tables from a web page (https) into R and analyze
them through the readHTMLTable function. After loading XML package, the
following commands are run:
url -
Is this what you're trying to do?
mtrx - structure(c(BB, AB, BB, BB, BB, BB, AA, BB,
BB,
BB, BB, BB, AB, AB, BB, AA, BB,
BB, BB, BB,
BB, AB, BB, BB, BB), .Dim = c(5L, 5L),
.Dimnames = list(
c(rs1, rs2, rs3, rs4, rs5), c(LG_1,
HI,
It seems like you used ?attach(). Better would be to use ?with.
set.seed(45)
dat1 - as.data.frame(matrix(sample(LETTERS,80,replace=TRUE),20,4))
lapply(dat1,levels) ##get the levels of all the dataset variables in a list
with(dat1,levels(V1))
A.K.
I have another question, at the
My question is related to a cox model with time-dependent variable.
When I think about it more, I get a little confused about
non-increasing assumption for survival probability for an individual.
For example, for a time-dependent ,say x, assuming increasing x
increases the risk of event.
given the data below
x1-c(-1,0,1,-1,0,1,-1,0,1,-1,0,1,-1,0,1,-1,0,1,-1,0,1,-1,0,1,-1,0,1)
y-c(-1,-1,-1,0,0,0,1,1,1,-1,-1,-1,0,0,0,1,1,1,-1,-1,-1,0,0,0,1,1,1)
z-c(-1,-1,-1,-1,-1,-1,-1,-1,-1,0,0,0,0,0,0,0,0,0,1,1,1,1,1,1,1,1,1)
using maple i can fit as below and get (1)
#PolynomialFit(3, x1,
I need to plot and analyze irregular time series. Having read the JSS
article on zoo (2005), ?zoo, ?plot.zoo, and the quickref vignette I'm still
not doing this correctly.
My work flow is to read the *.csv file to create a dataframe, convert the
df to a matrix using as.matrix() (see
Hi Satish,
Pascal's suggestion is better.
Please use ?dput() to show the example data.
op - options(digits.secs=6)
library(zoo)
##added a couple more rows
dat - read.zoo(text=V1 V2 V3 V4 V5 V6
'2014-03-14 22:41:46.988804' 10 2 8 3 14
'2014-03-14 22:41:46.991126' 13 4 9 5 15
As the author of that post, I must respond...
# - first comment ---
Ducan: The article that Henrik cited gives a reasonable description up
until near the end
Suraj: Given the complexity of the topics at hand, the many accolades I've
received for the post, and how thoroughly I've
Hi Everyone: Suraj will respond to Duncan's comments below promptly. Suraj
doesn't have the original thread so I am just helping out by commenting
here so that he can respond and the thread can be kept continuous.
Mark
On Sun, Mar 9, 2014 at 9:09 AM, Duncan Murdoch
Hi
Isn't melt just what you want?
melt(mydt)
Petr
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
project.org] On Behalf Of Abraham Mathew
Sent: Tuesday, March 18, 2014 1:59 AM
To: r-help@r-project.org (r-help@r-project.org)
Subject: [R] Reshaping
Dear all,
I have obtained the averages (means) and standard deviations (SD) of
different variable using the aggregate function, which I have found very
useful for these kind of computation.
However I would like to calculate the lower and upper ends of the
data (that is mean - SD and mean + SD)
I need to plot and analyze irregular time series. Having read the JSS
article on zoo (2005), ?zoo, ?plot.zoo, and the quickref vignette I'm still
not doing this correctly. This did not post to the list with the dput()
outputs attached; they are at the end of this message.
My work flow is to
Dear R-Family,
I have a data-set of the following format. I am only presenting a part of it.
DateA B C
D
1-Jan-61
0.00
1.27
8.128
0.25
2-Jan-61
6.10
9.144
94.742
Dear Arun and Jim,
Thanks for your replies. I will be extremely careful next time.Anyway, here is
the dput format of dat
dput(dat)
structure(list(Date = c(1-Jan-61, 2-Jan-61, 3-Jan-61, 4-Jan-61,
5-Jan-61, 6-Jan-61, 7-Jan-61, 8-Jan-61, 9-Jan-61, 10-Jan-61), A =
c(0, 6.1, 0, 0, 0, 0, 0, 0, 0,
Hi,
dat - read.table(text=Date A B C D
1-Jan-61 0.00 1.27 8.128 0.25
2-Jan-61 6.10 9.144 94.742 15.49
3-Jan-61 0.00 0.508 1.27 0.00
4-Jan-61 0.00 0 NA 0.00
5-Jan-61 0.00 0 0 0.00
6-Jan-61 0.00 NA 0 0.00
7-Jan-61 0.00 0 0 0.00
8-Jan-61 0.00 NA 0 0.00
9-Jan-61
Hello,
Maybe something like the following.
res - merge(AVG, SD, by = stimulation)
names(res)[2:3] - c(AVG, SD)
res$lower - res$AVG - res$SD
res$upper - res$AVG + res$SD
res
Hope this helps,
Rui Barradas
Em 19-03-2014 18:15, Luigi Marongiu escreveu:
Dear all,
I have obtained the averages
Don't really understand your equations but.
If X is the (random variable of) time of event, then the survival
S(x) = 1 - Pr( X x )
has to be non increasing since:
If t1 t2. then
S(t1) - S(t2) = Pr(X t2) - Pr( X t1 ) = Pr( t1 X t2) 0
which means that
S(t1) S(t2).
But you must not
Hi all,
I've been using the randomForest package and I'm trying to make the switch
over to party. My problem is that I have an extremely unbalanced outcome
(only 1% of the data has a positive outcome) which makes resampling methods
necessary.
randomForest has a very useful argument that is
You might look at the 'bag' function in the caret package. It will not
do the subsampling of variables at each split but you can bag a tree
and down-sample the data at each iteration. The help page has an
examples bagging ctree (although you might want to play with the tree
depth a little).
Max
On 14-03-19 1:18 PM, Suraj Gupta wrote:
As the author of that post, I must respond...
# - first comment ---
Ducan: The article that Henrik cited gives a reasonable description up
until near the end
Suraj: Given the complexity of the topics at hand, the many accolades
I've received
I need to plot and analyze irregular time series. Despite reading the 2005
JSS article on zoo, ?zoo, ?plot.zoo, and the quickref vignette I'm still not
doing this correctly. dput() output of a matrix and the zoo object are at
the end of this message.
My work flow is to read the *.csv file to
MaxUpdated_row-NULL
MaxUpdated_val- 2013-06-28
rbind(MaxUpdated_row, as.Date(MaxUpdated_val, %Y-%m-%d))
[,1]
[1,] 15884
c(MaxUpdated_row, as.Date(MaxUpdated_val, %Y-%m-%d))
[1] 15884
c(MaxUpdated_row, MaxUpdated_val)
[1] 15884
Evidently, I'm again missing something simple, as I would
On 03/20/2014 08:46 AM, Jason Rupert wrote:
MaxUpdated_row-NULL
MaxUpdated_val- 2013-06-28
rbind(MaxUpdated_row, as.Date(MaxUpdated_val, %Y-%m-%d))
[,1]
[1,] 15884
c(MaxUpdated_row, as.Date(MaxUpdated_val, %Y-%m-%d))
[1] 15884
c(MaxUpdated_row, MaxUpdated_val)
[1] 15884
Evidently,
Not sure I follow.
I would like the date formatting information preserved as is implemented when
using as.character:
MaxUpdated_row-NULL
MaxUpdated_val- 2013-06-28
c(MaxUpdated_row, as.character(MaxUpdated_val))
[1] 2013-06-28
However, when I try to used as.Date(...) I loose the date
Instead of making your initial value of MaxUpdated_row NULL,
make it a 0-long Date object. Then c() will work as you wish
(but not rbind). E.g.,
MaxUpdated_row - as.Date(character(0))
MaxUpdated_row - c(MaxUpdated_row, as.Date(2013-06-28, %Y-%m-%d))
MaxUpdated_row - c(MaxUpdated_row,
Hi Duncan: If you think there is anything incorrect in the document, I'd
be interested because I refer to that document whenever I'm confused about
namespaces, importing etc. Thanks.
On Wed, Mar 19, 2014 at 4:31 PM, Duncan Murdoch murdoch.dun...@gmail.comwrote:
On 14-03-19 1:18 PM, Suraj
I'm trying to create a function to return the date x months in the past.
With the code below I'm getting the warning message:
Warning message:
In if (MonthsBack = CurrentMonth) { :
the condition has length 1 and only the first element will be used
##
DateBack -
Hello,
It works for me:
R MaxUpdated_row - NULL
R MaxUpdated_val - 2013-06-28
R
R
R rbind(MaxUpdated_row, as.Date(MaxUpdated_val, %Y-%m-%d))
[,1]
[1,] 15884
R c(MaxUpdated_row, as.Date(MaxUpdated_val, %Y-%m-%d))
[1] 15884
R c(MaxUpdated_row, MaxUpdated_val)
[1] 2013-06-28
R
Regards,
Hi jcrosbie ,
Re:
I'm trying to create a function to return the date x months in the past.
With the code below I'm getting the warning message:
Warning message:
In if (MonthsBack = CurrentMonth) { :
the condition has length 1 and only the first element will be used
Use ifelse(),
On Wed, 19 Mar 2014, Rich Shepard wrote:
I need to plot and analyze irregular time series. Despite reading the 2005
JSS article on zoo, ?zoo, ?plot.zoo, and the quickref vignette I'm still not
doing this correctly. dput() output of a matrix and the zoo object are at
the end of this message.
On Thu, 20 Mar 2014, Achim Zeileis wrote:
It's sufficient to send such messages once even if others don't reply within
minutes.
Achim,
I sent it but my postfix spam filter would not let it out. There should
have been only a single copy.
Please have a look at read.zoo() and the
On 03/20/2014 09:57 AM, Jason Rupert wrote:
Not sure I follow.
I would like the date formatting information preserved as is implemented
when using as.character:
MaxUpdated_row-NULL
MaxUpdated_val- 2013-06-28
c(MaxUpdated_row, as.character(MaxUpdated_val))
[1] 2013-06-28
However, when I try to
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