Hi
> -Original Message-
> From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of Shivi82
> Sent: Friday, June 12, 2015 8:28 AM
> To: r-help@r-project.org
> Subject: Re: [R] Missing Values in Table Statement
>
> Hi Petr,
>
> Probably i did not explain my scenario clearly.
> table(
Hi Petr,
Probably i did not explain my scenario clearly.
table(test$ORIGIN_NAME,is.na(test$SCH_TIME)) is the syntax with which i am
trying to find per destination wise how many instances are there where
system failed to enter the scheduled delivery time & there are multiple
cases of these. I am e
Hi
> -Original Message-
> From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of Shivi82
> Sent: Friday, June 12, 2015 7:41 AM
> To: r-help@r-project.org
> Subject: [R] Missing Values in Table Statement
>
> HI All,
>
> I need help on 2 issues as highlighted below"
>
> A)I have 2 v
HI All,
I need help on 2 issues as highlighted below"
A)I have 2 variables:- Sch_Time & Origin Name.
Now there are multiple instances where Scheduled time i.e. Sch_Time is
missing from each location hence i need to count how many instances do i
have split on location.
the code i have is :
table(
Cross-posting to SO and Rhelp is deprecated.
On Jun 11, 2015, at 8:40 PM, Bogdan Tanasa wrote:
> Dear all,
>
> please could you please with a simple question : I do have an array of 32
> elements, where each element is indexed by a name : eg :
>
> list_triplet_wells <-c("A1:A2:A3", "A4:A5:A6 ",
Dear all,
please could you please with a simple question : I do have an array of 32
elements, where each element is indexed by a name : eg :
list_triplet_wells <-c("A1:A2:A3", "A4:A5:A6 ", "A7:A8:A9", "A10:A11:A12
")
xxx <-array(0, dim=4)
dimnames(xxx) = list(list_triplet_wells)
>From another
Hi Kevin,
I don't even pretend to try to read it. If David is having a problem I am
going for tea.
I think we need some sample data (see ?dput for the best way to supply it) and
an succinct description in English of what the problem is and what you need to
get out of the data.
I suspect bec
Oh, Swami, gazing into the crystal ball one can see ...
;-}
Cheers,
Bert
Bert Gunter
"Data is not information. Information is not knowledge. And knowledge is
certainly not wisdom."
-- Clifford Stoll
On Thu, Jun 11, 2015 at 4:48 PM, David Winsemius
wrote:
>
> On Jun 11, 2015, at 12:25 PM,
Thanks Don,
I suspected there was a Boxplot() out there by was too lazy to look. I still
don't see how the original code would work if label only had onevalue but I
must admit what Boxplot() is actually doing is still confusing me.
John Kane
Kingston ON Canada
> -Original Message-
>
On Jun 11, 2015, at 12:25 PM, Kevin Kowitski wrote:
> Hey,
>
> I am having an issue with a for loop that is intended to read index values
> by row and column so that it can pull out the valuable information. My issue
> is that I am using a data.frame(which(df==1, arr.ind=TRUE))
That would
Hello,
I am familiar with the basics of statistical regression models, including
GAMs, but I am stumped on a particular implementation issue.
I am constructing a GAMM to fit to data that is autocorrelated. On of my
covariates (covariate "4") is a sin function the represents time of day. It
is sim
Hey,
I am having an issue with a for loop that is intended to read index values by row and
column so that it can pull out the valuable information. My issue is that I am using a
data.frame(which(df==1, arr.ind=TRUE)) to find the index of the values in my data frame
that are equal to 1. Th
I'm trying, and so far failing, to extract data from a very large
SharePoint list.
I have a URL for getting XML from sharepoint:
http://
/_vti_bin/owssvr.dll?Cmd=Display&List={GUID}&Query=*&XMLDATA=TRUE
When I do something like the following:
require(XML)
require(RCurl)
URL <- "http://
/_vti_b
Hi John,
That does help, thanks!
Brittany
> On Jun 11, 2015, at 4:02 PM, John Fox wrote:
>
> Dear Brittany,
>
> There is an essentially perfect linear dependency among the variables in your
> data (note the last eigenvalue, which is 0 within rounding error):
>
>> eigen(cor(problem.data.box
On Jun 11, 2015, at 2:44 PM, MacQueen, Don wrote:
> In addition to the other answers, I would suggest that the next time you
> get the "could not find function" message, try like this:
>
> help.search('Boxplot')
Spencer Graves uses RSiteSearch() as the underlying function for sos::findFn
--
Dear Brittany,
There is an essentially perfect linear dependency among the variables in your
data (note the last eigenvalue, which is 0 within rounding error):
> eigen(cor(problem.data.boxcox[,-1]), only.values=TRUE)
$values
[1] 3.644257e+00 1.821582e+00 1.712152e+00 1.205091e+00 1.007231e
On Jun 11, 2015, at 2:19 PM, John Kane wrote:
> Well it might have worked for your supervisor but I don't see how.
>
> As was mentioned it is boxplot , not Boxplot and the rest of the syntax looks
> dodgy to say the least.
There is a "Boxplot" function in package 'car' and it has a labels ar
In addition to the other answers, I would suggest that the next time you
get the "could not find function" message, try like this:
help.search('Boxplot')
Among the output from that you should see
graphics::boxplot Box Plots
which should lead you to "boxplot" instead o
Well it might have worked for your supervisor but I don't see how.
As was mentioned it is boxplot , not Boxplot and the rest of the syntax looks
dodgy to say the least.
Try
boxplot(Acc_S$Subj ~ Acc_S$Acc)
I don't see how label = will work , ?boxplot says it should be names = and as
the code
Do you already have an R package that will do a Spearman
Correlation Volcano plot ?
What do the data look like?
Have a look at
http://stackoverflow.com/questions/5963269/how-to-make-a-great-r-reproducible-example
and http://adv-r.had.co.nz/Reproducibility.html
for some suggestions on how to as
Start by going through an R tutorial or two? You need to do some minimal
homework BEFORE posting here.
Cheers,
Bert
Bert Gunter
"Data is not information. Information is not knowledge. And knowledge is
certainly not wisdom."
-- Clifford Stoll
On Thu, Jun 11, 2015 at 8:05 AM, Pierlot Gabin
wr
R is case sensitive.
try "boxplot" not "Boxplot"
On Thu, Jun 11, 2015 at 7:20 AM, Kris Singh
wrote:
> Dear Sirm/Madam,
>
> Just wondering if someone could help me. I've tried running a code on R
> and the code includes the following:
>
>> Boxplot(~Acc_S$Acc, label=Acc_S$Subj)
>
> But I receive
Dear all,
I am using ksvm in R for SVM classification. I trained the SVM classifier
with prob.model=TRUE. I have the two following outputs when I predicted.
1) label<-predict(model, TestData)
I have classification accuracy about 93%, where as in another mode
2) Probvalues<-predict(model
Hi all,
I have a data frame composed by 25 numerical variables. I want to do a Spearman
Correlation Volcano plot (i. e. x = correlation coefficient and y = -log10(p
value))
I'm a begginer in R, so how can I do this ?
PS : Sorry for my English, this is not my mother tongue.
Thank you !
[
This is the script i am working from.
library(poLCA) f <-
cbind(bq70,bq72_1,bq72_2,bq72_3,bq72_4,bq72_5,bq72_6,bq72_7,bq73a_1,bq73a_2,bq73a_3,bq73a_4)~
zq88+zq89+dm_zq101_2+dm_zq101_3+dm_zq101_4+dm_zq101_5+dm_zq101_6+dm_zq101_7+dm_zq101_8+dm_zq101_9
for(i in 2:14){max_II<--100 min_bic<-10
The idea is to move from regional dummies interacted with time dummies (model
1) to a smooth spline (defined on longitudes and latitudes) interacted with
time dummies (model 2), i.e.,
Model 1: Log p ~ X\beta + REGION*YEAR
Model 2: Log p ~ X\beta + f(long, lat)*YEAR
Estimating the f's in a loop
Rosa Oliveira wrote:
> Dear Jim,
>
> when I run your code (even the one you send me, not in my data), I get:
>
> Don't know how to automatically pick scale for object of type function.
> Defaulting to continuous
> Error in data.frame(x = c(0.1, 0.2, 0.1, 0.2, 0.1, 0.2, 0.1, 0.2, 0.1, :
> argume
Dear Sirm/Madam,
Just wondering if someone could help me. I've tried running a code on R
and the code includes the following:
> Boxplot(~Acc_S$Acc, label=Acc_S$Subj)
But I receive the following error message:
*Error: could not find function "Boxplot"*
I have tried installing all the packages
Frank,
I'm not sure what is going on. The following test function works for me in both 3.1.1
and 3.2, i.e, the second model matrix has fewer columns. As I indicated to you earlier,
the coxph code removes the strata() columns after creating X because I found it easier to
correctly create the
The main effect trend seems rather dangerous, why not just estimate the f’s in
a loop?
url:www.econ.uiuc.edu/~rogerRoger Koenker
emailrkoen...@uiuc.eduDepartment of Economics
vox: 217-333-4558University of Illinois
fax: 217-244-6678
Thank you John for spending time on this query and helping out.
It really helped me and finally i am able to achieve the desired results.
Thanks a ton to all others as well to spending time and furbishing solution.
Regards, Shivi
--
View this message in context:
http://r.789695.n4.nabble.com/
Thanks a lot Rainer
On Thu, Jun 11, 2015 at 4:24 PM, Rainer M Krug wrote:
>
> Again: please keep this on r-help!
>
> AROONALOK PYNE writes:
>
> > Hi,
> >
> > My system has 4GB memory. On running htop command on linux terminal, for
> > c(999), a() rug on single cpu core showed stats for
Again: please keep this on r-help!
AROONALOK PYNE writes:
> Hi,
>
> My system has 4GB memory. On running htop command on linux terminal, for
> c(999), a() rug on single cpu core showed stats for 100% cpu
> utilization. On reaching b(), the entire system hanged (no response and the
> cp
-- Forwarded message --
From: AROONALOK PYNE
Date: Thu, Jun 11, 2015 at 3:49 PM
Subject: Re: Issue with mcapply
To: Rainer M Krug
Hi,
My system has 4GB memory. On running htop command on linux terminal, for
c(999), a() rug on single cpu core showed stats for 100% cpu
u
-- Forwarded message --
From: AROONALOK PYNE
Date: Thu, Jun 11, 2015 at 2:25 PM
Subject: Re: Issue with mcapply
To: Rainer M Krug
R version 3.0.2 (2013-09-25)
Large value : 999
I rerun the code as c(1000) for which your machine works fine but
my code still hangs on r
Please keep this on the r-help mailing list.
AROONALOK PYNE writes:
> R version 3.0.2 (2013-09-25)
>
>
> Large value : 999
>
>
> I rerun the code as c(1000) for which your machine works fine but
> my code still hangs on reaching b(). I am running it from Linux
> Terminal.
Might be memor
AROONALOK PYNE writes:
> Please check this code :
>
> library(parallel)
> workerFunc <- function(n) { return(n^2) }
> a <- function(){
> CurrentTime <- Sys.time()
> res <- lapply(values, workerFunc)
> TimeTaken <- Sys.time() - CurrentTime
> print(TimeTaken)
> }
> b <- function(){
> Curr
Hi
I (wrongly) understood that Shivi82 wanted to summarise on month values.
Therefore
format(test$CR_DT,"%m")
shall give you month number and list is required by aggregate.
All the problem was in
test$CHG_WT
which seems to be a factor (for whatever reason)
Cheers
Petr
> -Original Mess
Please check this code :
library(parallel)
workerFunc <- function(n) { return(n^2) }
a <- function(){
CurrentTime <- Sys.time()
res <- lapply(values, workerFunc)
TimeTaken <- Sys.time() - CurrentTime
print(TimeTaken)
}
b <- function(){
CurrentTime <- Sys.time()
numWorkers <- detectCore
Thanks John! My eyes aren't good enough to see that. I actually checked (I
thought). This was the default window on Mac console, for others who might care.
Sent from my iPad
> On Jun 10, 2015, at 6:17 PM, John Kane wrote:
>
> You have curly quotes rather than plain ones here :
> col=4,type=“
Dear all,
I would like to estimate a quantile regression model including a bivariate
nonparametric term which should be interacted with a dummy variable, i.e.,
log p ~ year + f(a,b):year.
I tried to use Roger Koenker's quantreg package and the functions rqss and qss
but it turns out that interac
Oh I see, I'm sorry I just plopped it in GitHub for ease of help, I didn't
notice I put it under coursera work. This task is not related to coursera, I
will separate it out.
-Kevin
Sent from my iPhone
> On Jun 10, 2015, at 3:21 PM, David Winsemius wrote:
>
>
>> On Jun 10, 2015, at 9:41 AM,
I did not realize that there is a coursers assignment similar to this. I am
running this for data analysis at work, not for coursers. However I will look
through the link you provided and see if it is applicable.
Thanks,
Kevin
Sent from my iPhone
> On Jun 10, 2015, at 3:21 PM, David Winsemius
Dear Don, thank you very much.
I really wasn’t being able to figure the problem.
You were a big (huge) help.
Seeing the graphs, I think I’ll try to put the 3 settings (sample size) in
different graphs.
I’ll try to use trellis graphs :) using sample size as the “factor”
Thank you very much ;)
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