Your rather sarcastic comment about knowledge given by John's mother seems
inappropriate, given that he told you where his information came from and it is
the first place you should have looked.
The bit about the decimal leading to a shift in the decimal place pointed out
by Bill is a bit obsc
That's great, John. Your mother told you when you were born? How am I
supposed to know? Thank you both.
The following format statement did it!! I just change F5.3 to F5, 5F8.4
to 5F8. I also change 2E15.9 to 2A9, and then use the following
as.numeric to convert the alphanumerical to numerical. T
You have set yourself an impossible goal. Either you can reformulate your
problem as non-iterative and can process your data as arrays, or you have to
use some kind of for loop. The lapply and Vectorize functions are popular
"pretty" ways to do this, but they amount to hidden for loops.
Note t
I want to dynamically populate a vector by iteratively applying a
function to its previous element, without using a 'for' cycle. My
solution, based on a question I posted some times ago for a more
complicated problem (see "updating elements of a list of matrixes
without 'for' cycles") was to define
Dear Rxperts!
Is there a way to compute relative values.. using within().. function?
Any assistance/suggestions are highly welcome!!
Thanks again,
Santosh...
___
A sample dataset and the computation "outside" within() function is sh
It has been a while since I used Fortran formatted input, but the following,
without dots in the format, works:
> txt <- "1950. .614350 .026834 .087227 .006821 .180001 4.56E-2"
> print(read.fortran(textConnection(txt), c("f5", "6f8")), digits=10)
V1 V2 V3 V4 V5 V6
Thanks John. That helped, but I got a mixed of good thing and bad thing.
Good is R does not like the scientific number format "3E15.9" but I was
able to read with alphanumerical format "3A15" (and convert to
numerical). Bad is R does not like the numbers .1234, .2345 without the
zeros before th
On 28/05/16 06:08, Lida Zeighami wrote:
Hi Dears!
Would you please let me know how I can install package pcalg for R version
3.1.3 ?
I've tried different ways but got error!
Thanks inadvance!
An answer the helpfulness of which is commensurate with the clarity of
your question: Do it correc
Greetings David,
I am new to R and neglected to check vigorously for missing values.
Apologize for posting without checking and finding the one NA.
I appreciate your help.
Thanks.
James F. Henson
On Fri, May 27, 2016 at 1:49 PM, David Winsemius wrote:
>
>> On May 27, 2016, at 10:07 AM, James Hen
Use reply-all to keep the list in the loop.
I happen to have this old version of R still around on Windows 7 and it
installs just fine for me.
You may be able to get it to install by choosing a different CRAN mirror, but I
still recommend that you upgrade R and try with that before posting ag
Hi Josh
Thanks a lot - appreciate the answer. Figured it may have worked that way but
investigating the url called by the package was a smart move. Guess I'll need
to get the info from Bloomberg!
Thanks again
David
Sent from my iPhone
> On 27 May 2016, at 9:52 PM, Joshua Ulrich wrote:
>
>
> On May 27, 2016, at 10:07 AM, James Henson wrote:
>
> Greetings Jeff,
> You are correct that the unequal number of levels is not the problem.
> I revised the data frame so that the number of levels was equal and
> the same error message occurred. The code is below, and the
> Eboni2.txt file i
If you read the Posting Guide it warns you against posting in HTML (it doesn't
say why but basically what you think you sent is not necessarily what we saw).
It also mentions that you should update to the latest version of R (yours is
not) if you want help (though we might try anyway). It also m
On Fri, May 27, 2016 at 12:56 PM, Steven Yen wrote:
> Dear fellow R users:
> I am reading a data (ascii) file with fortran fixed format, containing
> multiple records. R does not recognize fortran's record break (a slash).
> I tried to do the following but it does not work. Help appreciated.
>
>
Hi Dears!
Would you please let me know how I can install package pcalg for R version
3.1.3 ?
I've tried different ways but got error!
Thanks inadvance!
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list -- To UNSUBSCRIBE
In bulk processing streamflow data available from an online database, I'm
wanting to trim the beginning and end of the time series so that daily data
associated with incomplete "water years" (defined as extending from Oct 1st
to the following September 30th) is trimmed off the beginning and end of
Dear fellow R users:
I am reading a data (ascii) file with fortran fixed format, containing
multiple records. R does not recognize fortran's record break (a slash).
I tried to do the following but it does not work. Help appreciated.
60 FORMAT(1X,F6.0,5F8.6/1X,5F8.4,F10.6/1X,2F6.0,3E15.9,F8.0,F
Greetings Jeff,
You are correct that the unequal number of levels is not the problem.
I revised the data frame so that the number of levels was equal and
the same error message occurred. The code is below, and the
Eboni2.txt file is attached. This problem baffles me. I appreciate
any help.
Best r
A possible (simple) solution is to use a binomial GLM which guarantees
fitted values (percentiles) in (0,1):
plot(percentile, score)
o<-glm(percentile~sc, family=binomial)
points(fitted(o), sc, col=2)
You can "predict" percentiles given score via predict.glm()
best,
vito
Franco Danilo Roca L
>If you want to drop levels, use droplevels() either on the factor or on
the >subset of your data frame. Example:
>droplevels(f[1]) #One element, only one level
Calling factor() on a factor, as the OP did, also drops any unused levels,
as the examples showed.
> str(factor(factor(letters)[11:13]))
If your data.frame is a mix you can loop over each column - along the lines
of:
library(plyr)
adply(test, 2, function(x){
if(!is.numeric(x[[1]]){
gsub(",", ";", x[[1]])
}else{
x[[1]]
}
})
Ulrik
On Fri, 27 May 2016 at 17:21 Jun Shen wrote:
> Thanks Ulrik and Bob for your reply.
>
> gsub worked
> sapply(test, gsub, pattern=",", replacement=";")
C1C2
"a;b;c;d" "g;h;f"
-
David L Carlson
Department of Anthropology
Texas A&M University
College Station, TX 77840-4352
-Original Message-
From: R-help [mailto:r-help-boun...@r-projec
Thanks Ulrik and Bob for your reply.
gsub worked for one column!
If I want to replace the whole data frame, gsub doesn't seem to work. Any
idea
On Fri, May 27, 2016 at 11:14 AM, Ulrik Stervbo
wrote:
> use gsub()
>
> On Fri, 27 May 2016 at 17:12 Jun Shen wrote:
>
>> Dear list,
>>
>> Say I have
use gsub()
On Fri, 27 May 2016 at 17:12 Jun Shen wrote:
> Dear list,
>
> Say I have a data frame
>
> test <- data.frame(C1=c('a,b,c,d'),C2=c('g,h,f'))
>
> I want to replace the commas with semicolons
>
> sub(',',';',test$C1) -> test$C1 will only replace the first comma of a
> string.
>
> How do
You can use gsub() instead of sub()
On Fri, May 27, 2016 at 11:10 AM, Jun Shen wrote:
> Dear list,
>
> Say I have a data frame
>
> test <- data.frame(C1=c('a,b,c,d'),C2=c('g,h,f'))
>
> I want to replace the commas with semicolons
>
> sub(',',';',test$C1) -> test$C1 will only replace the first com
Dear list,
Say I have a data frame
test <- data.frame(C1=c('a,b,c,d'),C2=c('g,h,f'))
I want to replace the commas with semicolons
sub(',',';',test$C1) -> test$C1 will only replace the first comma of a string.
How do I replace them all in one run? Thanks.
Jun
[[alternative HTML versio
Hi Elahe,
Based on this and your other questions I think you could benefit from
spending time with tutorials on ggplot and on R in general.
Google should give you plenty and I quickly found this:
http://www.ats.ucla.edu/stat/r/ and this:
https://www.nceas.ucsb.edu/files/scicomp/Dloads/RProgrammin
I think you need to read those questions again more carefully... particularly
the second one.
--
Sent from my phone. Please excuse my brevity.
On May 27, 2016 7:41:25 AM PDT, "ch.elahe via R-help"
wrote:
>Thanks Ulrik,
>But in these examples they want to mark the percentage or frequency of
>p
Thanks Ulrik,
But in these examples they want to mark the percentage or frequency of plot
variables, in my case I want to mark the bars with a different variable of my
df. DO you have any idea?
On Friday, May 27, 2016 3:22 PM, Ulrik Stervbo wrote:
Hi Elahe,
maybe this will help:
http://s
Hm
> -Original Message-
> From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of Jim Lemon
> Sent: Friday, May 27, 2016 12:46 PM
> To: Lorenzo Isella
> Cc: r-help@r-project.org
> Subject: Re: [R] Getting Rid of NaN in ts Object
>
> Hi Lorenzo,
> Maybe:
>
> tt<-tt[!is.nan(tt)]
Th
Hello,
I hope you can help me. In class, we were given an Excel worksheet with
specified formulas that take the total score from a survey (or from a specific
section) and convert it to a percentage, according to a table that assigns
scores to a percentile. Since the formulas are too long and co
Greetings-
I am seeking to fit a model using Heckman-style selection; however the
wrinkle is that the selection is into multiple categories, not a binary
in/out. In this case, selection is into the type of higher-education
institution a student attended; the goal is to estimate post-graduation
On Mon, May 23, 2016 at 8:57 PM, David Menezes
wrote:
> Hi
>
> Apologies if this has already been asked and answered or if I've labelled
> the subject incorrectly but I can't find a solution using the search
> function for this group; the vignette documentation for quantmod or general
> google sea
Hi Elahe,
maybe this will help:
http://stackoverflow.com/questions/6455088/how-to-put-labels-over-geom-bar-in-r-with-ggplot2
or this:
http://stackoverflow.com/questions/12018499/how-to-put-labels-over-geom-bar-for-each-bar-in-r-with-ggplot2
Best,
Ulrik
On Fri, 27 May 2016 at 15:12 ch.elahe via R
Hi all ,
I have the following bar chart and I want to add SLC variable values to the
charts but I don't know how to use geom_text:
ggplot(df,(Protocol,fill=quant))+geom_bar()+coord_flip()
Thanks for any help!
Elahe
__
R-help@r-project.org mai
Hi,
I am hosting a small repository (of partly non-public repositories) and was
wondering if it is possible to create the package-index page on CRAN (e.g.
here for drat https://cran.rstudio.com/web/packages/drat/drat.pdf)
automatically somehow out of an existing repo (i.e. create the package from
Perfect!
Exactly what I was looking for.
Thanks
Lorenzo
On Fri, May 27, 2016 at 01:50:03PM +0200, Christian Brandstätter wrote:
Hi Lorenzo,
Try:
tt[is.nan(tt)] <- NA
tt <- na.omit(tt)
Best,
Christian
Am 27.05.2016 um 13:38 schrieb Lorenzo Isella:
On Fri, May 27, 2016 at 08:46:20PM +1000,
Hi Lorenzo,
Try:
tt[is.nan(tt)] <- NA
tt <- na.omit(tt)
Best,
Christian
Am 27.05.2016 um 13:38 schrieb Lorenzo Isella:
On Fri, May 27, 2016 at 08:46:20PM +1000, Jim Lemon wrote:
Hi Lorenzo,
Maybe:
tt<-tt[!is.nan(tt)]
Jim
Not really.
tt<-structure(c(NaN, NaN, NaN, NaN, NaN, NaN, NaN,
On Fri, May 27, 2016 at 08:46:20PM +1000, Jim Lemon wrote:
Hi Lorenzo,
Maybe:
tt<-tt[!is.nan(tt)]
Jim
Not really.
tt<-structure(c(NaN, NaN, NaN, NaN, NaN, NaN, NaN, NaN, NaN, NaN,
+ NaN, NaN, NaN, NaN, NaN, NaN, NaN, NaN, 1133.09, 1155.77, 1179.12,
+ 1182.85, 1133.43, 1103.36, 1081.19, 10
Hi Lorenzo,
Maybe:
tt<-tt[!is.nan(tt)]
Jim
On Fri, May 27, 2016 at 8:14 PM, Lorenzo Isella
wrote:
> Dear All,
> I am sure the answer is a one liner, but I am banging my head against
> the wall and googling here and there has not helped much.
> Consider the following time series
>
> tt<-structu
> You did not change df$quant - you made a new object called 'subdf'
> containing a column called 'quant' that had only one level. Changing subdf
> has
> no effect on df.
Also, subsetting a factor _intentionally_ does not change the number of levels.
Example:
f <- factor(sample(letters[1:3], 30
Thank you for your reply. I am trying to use the over function - but having
trouble. Asked at R-sig-geo. Thanks again!
Sincerely,
Milu
On Fri, May 27, 2016 at 12:49 AM, MacQueen, Don wrote:
> Perhaps the
> over()
> function in the sp package.
>
> (in which case, R-sig-geo might be a better p
Dear All,
I am sure the answer is a one liner, but I am banging my head against
the wall and googling here and there has not helped much.
Consider the following time series
tt<-structure(c(NaN, NaN, NaN, NaN, NaN, NaN, NaN, NaN, NaN, NaN,
NaN, NaN, NaN, NaN, NaN, NaN, NaN, NaN, 1133.09, 1155.77,
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