On 17/12/2014, 9:35 PM, Andre Mikulec wrote:
Hi,
when I am trying to deug a package, I am getting the error.
library(quantstrat)
setBreakpoint(strategy.R#3,envir=environment(strategy))
No source refs found.
In R studio, I get a strange message, Breakpoints will be activated when an
What you are describing sounds like a very spreadsheet-y thing.
- The information is already IN your dataframe, and easy to get out by
subsetting. Depending on your usecase, that may actually be the best.
- If the number of CaseIDs is large, I would use a hash of lists (if the data
is
Not sure how much help it will be but there is a package on CRAN called
icd9. Although clearly the codes are different in ICD 10 it may give you
some hints. I suppose you could even email the maintainer to see whether
there is an icd10 in the pipeline.
On 17/12/2014 20:14, Robert Strother
Of the tools I know (and things change every day!), only package trust
uses the Hessian explicitly.
It would not be too difficult to include explicit Hessian by modifying
Rvmmin which is all in R -- I'm currently doing some cleanup on that, so
ask offline if you choose that route.
Given that
Hi all,I am looking for a function that would give me all the combinations
between two vectors.Lets take as example the
test-seq(1,3,by=5000)
Browse[2] test
[1] 1 5001 10001 15001 20001 25001
I want all the combinations between two times the test... I think this is
called permutation
Of course, but why? As Brian S says you have not given us enough information to
know exactly what you are after.
Have a look at https://github.com/hadley/devtools/wiki/Reproducibility or
http://stackoverflow.com/questions/5963269/how-to-make-a-great-r-reproducible-example
for some information
I can't quite tell what you want: your example output is either
unclear to me or mangled by posting in HTML (please don't).
Is
expand.grid(test, test)
what you want, or partway to what you want?
Sarah
On Thu, Dec 18, 2014 at 9:56 AM, Alaios via R-help r-help@r-project.org wrote:
Hi all,I am
Depending on what you want, you probably want to start with expand.grid():
# All combinations of test with test
pairs1 - expand.grid(test, test)
nrow(pairs1)
[1] 36
# Exclude cases that differ only in the order of the values
# E.g. (1, 5001), but not (5001, 1), also (1, 1), etc are included
I would like to know if it is allowable to build an R-package made of one
function only, to launch a GUI to take input data and users to select the
models they intend to use, and to start running the processes?
Thanks a lot,
Diogo Alagador
On 18/12/2014 14:56, Alaios via R-help wrote:
Hi all,I am looking for a function that would give me all the combinations
between two vectors.Lets take as example the
test-seq(1,3,by=5000)
Browse[2] test
[1] 1 5001 10001 15001 20001 25001
I want all the combinations between two times
Make a table that looks like... sounds like a use case that would benefit
from some reflection.
Anyway, at least don't put your IDs *in* the table.
# Your data
CaseID - c('1015285',
'1005317',
'1012281',
'1015285',
'1015285',
'1007183',
'1008833',
'1015315',
'1015322',
'1015285')
No guarantees on best... but one way using base R could be:
# Note that CaseID is actually not a valid PViol.Type as you had it
PViol.Type - c( BW.BackWages
, LD.Liquid_Damages
, MW.Minimum_Wage
, OT.Overtime
, RK.Records_FLSA
IANAL and this is not a legal advice forum, but to the best of my
knowledge... Yes it is allowable. Read the license. Your responsibilities
and limitations have more to do with what to do if you decide to
distribute your code. Also, there is no guarantee that CRAN will accept
your
I like the example provided by David L. Carlson using the function
'expand.grid()' as shown in the previous email.
You may try 'outer()' and 'upper.tri()' instead 'expand.grid()'. Please
see the below:
x - outer(test, test, paste, sep=,)
x
[,1] [,2] [,3] [,4]
anybody has any hint on this?
Subject: Add encoded special characters (greek characters) as text to plot
To: r-help@r-project.org
Date: Wednesday, December 17, 2014, 9:25 PM
Dear all,
I read my a character matrix from a text file. Some of them
On 18/12/2014 16:59, heyi xiao via R-help wrote:
anybody has any hint on this?
Yes, ?plotmath does. But you will need to know what encoding this is
(and hence what Unicode points are meant by \246 and \302).
If this really were Greek, common encodings are UTF-8, CP1253 and ISO
8859-7;
Read the posting guide. The solution is likely to depend on your operating
system and graphics devices.
---
Jeff NewmillerThe . . Go Live...
DCN:jdnew...@dcn.davis.ca.us
I do hope this doesn't upset anyone. But it appears rather interesting to
me, despite the fact that I'm not a statistician. So I thought that it
might be of interest to some others here.
Hi all,
I have a data set like this:
Test.cox file:
V1V2 V3Survival Event
ann 13 WTHomo 41
ben 20 *51
tom 40 Variant 61
Hi
I write my masterthesis and don't know how I can count points in a spatial net.
In practical I have a data set for carsharing usage in Berlin. It includes the
Idletime of the cars with Long/lat coordinates for the certain places. So if I
plot those points I have something like a cloud of
I want to achieve a table that looks like a grid of 1's for all cases in a
survey. I'm an R beginner and don't have a clue how to do all the things you
just suggested. I really appreciate the time you took to explain all of those
options, though. -- BNC
-Original Message-
From:
Hi all,
I am trying to analyse bird data to investigate carry-over effect
using structural equation model.
I failed to run properly a big model with several latent variables
with both L - M block and M - L block.
Rather than trying again and again with the huge model, I am now
looking to a subset
I am trying to compute max, min, and mean from Global Circulation Models
(GCM) for the US. The data is in 3-hour blocks for 2026-2045 and 2081-2100.
Sample Data:
tmp1 - structure(list(FIPS = c(1001L, 1003L, 1005L), X2026.01.01.1 =
c(285.5533142,
285.5533142, 286.2481079), X2026.01.01.2 =
Depending how your data are stored, this could be solved with a very basic use
of R, such as the ifelse and aggregate functions. Try reading [1] for
suggestions on clarifying your problem statement and follow up.
[1]
I like the approach presented by Jeff Newmiller as shown in the previous
post (I really like his way). As he suggested, it would be good to
start with 'factor' since you have all values of 'Primary.Viol.Type'.
You may try to use 'split()' function for creating table that you wish
to build.
Why don't you try this approach if you cannot use 'expression()'?
x - c(alpha, beta, gamma, delta)
plot(0, type=n)
for(i in 1:length(x)) text(x=1, y=i/10, labels=parse(text=x[i]))
Please see the output in R. Is this what you are looking for? I hope
this helps. I would also appreciate it
Dear John,
Thank you for your suggestions.
I'll have a look at the trust package - the trust zone may be doing what
I need.
The tanh transformation could be a good alternative too.
Best wishes
Xavier
On 18. 12. 14 15:10, Prof J C Nash (U30A) wrote:
Of the tools I know (and things change every
Please keep the list in the loop.
Take a look at my code again... the factor function can accept a vector of all
levels you want it to include.
---
Jeff NewmillerThe . . Go Live...
Please take a look at my code again. The error message says that object
'Primary.Viol.Type' not found. Have you ever created the object
'Primary.Viol.Type'? It will be working if you replace
'Primary.Viol.Type' by 'PViol.Type.Per.Case.Original$Primary.Viol.Type'
where 'factor()' is used.
Can you show what the first few rows of your table look like so that we
understand the structure?
--
View this message in context:
http://r.789695.n4.nabble.com/Calculating-mean-median-minimum-and-maximum-tp4700862p4700919.html
Sent from the R help mailing list archive at Nabble.com.
I know this has been explained a few times here in different scenarios, but I
am having a hard time digesting this.
The following code works fine as long as it's not inside a function (see below).
df$season - as.character(df$season) temp - model.matrix( ~ season - 1,
data=df) df -
Hi:
I have one million names of city and their population and want to make
a hash so that by giving key = city, the hash function will return its
population.
I use the hash package in R as follows:
h - hash(c(as.vector(df$city)), c(as.vector(df$population)))
But getting the following error:
I do not think that you need regular expressions for your problem.
Please see the below:
d0 - dat_unmatched
tmp - apply(d0, 1, function(x){
+ first - substr(x,1,1)
+ idx - which(c(T, Y) == first)
+ comb - paste(x[idx[1]-1], x[idx], collapse= )
+ unlist(strsplit(comb, ))
+ })
names(tmp) -
Jose Manuel Veiga del Baño chem...@um.es
writes:
Hola a todos,
Agradeceros de antemano vuestro tiempo y paciencia ya que soy un poco
novato y tal vez esto sea un poco trivial.
Lo que quiero hacer es que me represente en eje de las x las fechas
(columna fecha) y los valores de z (columna
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