Thanks Chuck and Rolf.
While Rolf’s code also works on the dput that I actually gave you (a smaller
subset of the full dataset), it failed to work on the larger dataset, because
there are further exceptions:
input[[i]]$content[[1]] is sometimes a list, sometimes a character vector, and
Dear Antonello,
You can specify the colours manually with scale_colour_manual(). See
http://docs.ggplot2.org/0.9.3.1/scale_manual.html for some examples. The
last examples uses greys.
Best regards,
ir. Thierry Onkelinx
Instituut voor natuur- en bosonderzoek / Research Institute for Nature and
You want to use a generalized linear model of some sort
glm(count ~ flow + gravity + group, data=mydata, family=poisson)
would be a start, however, the effects of flow rate are nonlinear, so
you might use a natural spline term like ns(flow,5) to allow
nonlinearity, and there also seem to be
Hi, I'm using ggplot2 to make a plot of the regression of a variable x (let
say, levels of depression),
on a variable y (let say, degree of social impairment),
by taking into account a binary factor (having had or not a past admission
to a psychiatric service),
and age of partecipants.
After some
John et al
Thank you for your advice below. It was sloppy of me not to verify my
reproducible code below. I have tried a few of your suggestions and wrapped
the working code into the function below called pl2. The function properly
lands on the right model parameters when I use the optim or
Dear all
I know I am missing something obvious but after few hours of trials I ask for
some help.
I have some sequence of values (days)
x - 1:30
and an indication of event start and end day
mimo-c(5,10, 13,16, 21,27)
or
events - structure(list(start = c(5, 13, 21), end = c(10, 16, 27)),
Hi All,
The most difficult challenge that I face in “learning R” is to do data munging.
I have reviewed Hadley’s advanced R programming guide, familiarized myself with
data structures, subsetting, plyr, dplyr, tidy, the lapply() family of
functions, basic string manipulation and grepping, SQL
Hmm…Chuck’s solution may actually be problematic because there are several
entries which at the deepest level are called “sha”, but that should not be
included, such as:
input[[67]]$content[[1]]$commit$tree$sha
and
input[[67]]$content[[1]]$parents[[1]]$sha
it’s only the “sha”
Hi,
On Fri, Feb 20, 2015 at 9:27 AM, PIKAL Petr petr.pi...@precheza.cz wrote:
Dear all
I know I am missing something obvious but after few hours of trials I ask for
some help.
I have some sequence of values (days)
x - 1:30
and an indication of event start and end day
mimo-c(5,10,
Dear all members
I have the following matrix in R
thd - matrix(testJAGSdata$thd, ncol=6, byrow=TRUE)
head(thd)
[,1] [,2] [,3] [,4] [,5] [,6]
[1,] -200 -2.517 -1.245 -0.444 0.848 200
[2,] -200 -1.447 -0.420 0.119 1.245 200
[3,] -200 -1.671 -0.869 -0.194 0.679 200
[4,] -200
Hi,
On Fri, Feb 20, 2015 at 9:48 AM, thanoon younis
thanoon.youni...@gmail.com wrote:
Dear all members
I have the following matrix in R
thd - matrix(testJAGSdata$thd, ncol=6, byrow=TRUE)
head(thd)
[,1] [,2] [,3] [,4] [,5] [,6]
[1,] -200 -2.517 -1.245 -0.444 0.848 200
[2,]
This is not the proper venue for a discussion of the mathematics of
optimization, no matter that it is interesting. Please take it off
list.
Cheers,
Bert
Bert Gunter
Genentech Nonclinical Biostatistics
(650) 467-7374
Data is not information. Information is not knowledge. And knowledge
is
Hello yet again.
I am trying to create a zip file for a friend who has a Windows machine.
He needs to access this via the local zip file packages option.
When I use R CMD INSTALL --compile-both, it produces an item in the library
tree (as promised).
However, I would like to have an actual .zip
A little shorter version of the SQL solution after consulting with my
SQL expert:
require(sqldf)
timeline - data.frame(time = 1:30)
events - structure(list(start = c(5, 13, 21), end = c(10, 16, 27)), .Names =
c(start,
+ end), row.names = c(NA, -3L), class = data.frame)
# add event number
Hi,
Your data may look like the following and named speed.txt in working
directory. Then the cod follows the data. The graph is attached as
speed.pdf.
Speed
50
52
55
57
58
59
60
61
62
63
64
65
65
65
67
68
68
68
68
69
69
70
71
72
72
72
73
73
73
73
75
76
77
78
79
speed -
Hello,
I am having trouble generating a correlation matrix on multiple cores.
I have a matrix myMat for which I would like to do a Pearson correlation.
Lets say dim(myMat) is 100 200. I want a 200x200 correlation matrix and
corresponding p-value matrix.
I like to use rcorr(myMat) in the
Hi,
Herr is one implementation with function named eventList.
start
[1] 5 13 21
start
[1] 5 13 21
end
[1] 10 16 27
x
[1] 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24
25 26 27 28 29 30
eventList
function(start, end, x)
{
result - character(0)
for (i in
All,
I'm a newbie to R and am interested in seeing a simple example of calling a 3rd
party Visual Studio generated DLL from RStudio. Does anyone have a simple
example which also walks through the preliminary steps of setting up the
INCLUDE path and the library path to either a DLL or LIB
Using lapply() where Jim used sapply() would keep the types
right and be a fair bit faster than a solution based on repeatedly
appending to a list (like your getFirst).
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Fri, Feb 20, 2015 at 1:52 PM, JS Huang js.hu...@protective.com wrote:
Hi,
On 21/02/15 15:02, Jeff Newmiller wrote:
R CMD INSTALL --build packagename
That will create a *.tar.gz file, not a *.zip file. The latter being
what Erin wanted, if I understand correctly.
I have worked around the problem in the past with a shell script
like unto:
#! /bin/csh
set vnum =
Hi,
Jim's answer is neat. There is an issue on the result. All are
characters even though some are numeric or logic. The following
implementation retains the variable type.
x
[[1]]
[1] 2 3 5
[[2]]
[1] aa bb cc
[[3]]
[1] TRUE FALSE TRUE
getFirst
function(aList)
{
result - list()
Hello All,
I have a data frame of two columns for wind. The first column is for wind
speed and the second wind direction. I'm trying to replace the values
in the first column and the 999 values in the second column with NA. I
tried to use the function ltdl.fix.df but it doesn't seem to do
You did not say how you imported the data, but if you used one of the
read.table variants (including read.csv) then you can use the na.strings
argument as documented in the help file for read.table.
Next time please read the posting guide, as there are some useful tips in
there, such as
R CMD INSTALL --build packagename
---
Jeff NewmillerThe . . Go Live...
DCN:jdnew...@dcn.davis.ca.usBasics: ##.#. ##.#. Live Go...
How can you expect a solution if you cannot specify the problem?
-- Bert
Bert Gunter
Genentech Nonclinical Biostatistics
(650) 467-7374
Data is not information. Information is not knowledge. And knowledge
is certainly not wisdom.
Clifford Stoll
On Fri, Feb 20, 2015 at 6:13 AM, Aron Lindberg
And just to muddy the waters more here's another way to do it using the handy
plyr package where the data.frame is dat1
library(plyr)
ddply(dat1, .(Participant.ID), summarize, mean = mean(Score))
John Kane
Kingston ON Canada
-Original Message-
From: js.hu...@protective.com
Sent:
Simon,
You missed a key request from Sven. He asked for data in dput() format. This
is essential for dealing with many problems. Do ?dput for info on the function
but esssentially it let's the reader see your data exactly as you see it,
unfazed buy any special setting the reader may have for
You cannot do that in one step. Do it right after:
names(out) - df$nm
Please don't post using HTML format.. it scrambles code, and since we cannot
see what you saw it doesn't help in any way.
Also note that df is a function in the base stats package... not a good name
to use.
First, please reply to the list, not just me.
On Fri, Feb 20, 2015 at 10:22 AM, thanoon younis
thanoon.youni...@gmail.com wrote:
thank you very much for your help
actually, my data set like this
#Data Set
testJAGSdata = list(N1=200, N2=200, P=18,
R=structure(
Hello all,
I've been trying to figure out how to return a named list from foreach.
Given that the order of the returned list is guaranteed to be in the
order in which the object is passed to foreach, list members can be
named afterwards. However, I'm wondering if there's a better way to do
Hi John everyone else,
My apologies for not providing you all with a good reproducible example. I
will get to work on this and reply in due course.
Thanks John for pointing me in the right direction with this.
Regards,
On 20 February 2015 at 16:06, John Kane jrkrid...@inbox.com wrote:
Here is a solution using the sqldf package:
require(sqldf)
timeline - data.frame(time = 1:30)
events - structure(list(start = c(5, 13, 21), end = c(10, 16, 27)), .Names =
c(start,
+ end), row.names = c(NA, -3L), class = data.frame)
# add event number
events$num - paste0(A,
Hi Sven,
Many thanks for the reply and my apologies for not posting any code. So
far, I have been able to write this (but it's very basic and just getting
me to the 'complicated' stage).
setwd(C:\\Users\\simon.tarr\\Documents\\GIS\\Test Data)
require(raster)
require(rgdal)
Hi,
I was trying to calculate the CCC metric in R with the help of NbClust
source code and the available SAS manual for CCC. All the Pseudo-F and
R-square are matching exactly with the SAS output except for the E_R2 and
hence CCC. I have tried and tested in multiple ways but couldn’t get any
Dear list,
Let's say I have setup the following list:
a = c(2, 3, 5)
b = c(aa, bb, cc)
c = c(TRUE, FALSE, TRUE)
x = list(a, b, c)
I want to access the first second dimension element of each first dimension
element so that the result is something like:
(2, aa, TRUE)
In my real life problem
Dear List,
Consider this example
df - data.frame(matrix(rnorm(9*9), ncol=9))
names(df) - c(c_1, d_1, e_1, a_p, b_p, c_p, 1_o1, 2_o1,
3_o1)
row.names(df) - names(df)
indx - gsub(.*_, , names(df))
I can split the dataframe by the index that is given in the column.names
after the underscore
try this:
a = c(2, 3, 5)
b = c(aa, bb, cc)
c = c(TRUE, FALSE, TRUE)
x = list(a, b, c)
x
[[1]]
[1] 2 3 5
[[2]]
[1] aa bb cc
[[3]]
[1] TRUE FALSE TRUE
sapply(x, '[[', 1)
[1] 2aa TRUE
Jim Holtman
Data Munger Guru
What is the problem that you are trying to solve?
Tell me what you
I'm supposed to return for class data with the ID and the value.
I'm returning just the correct value. Here's the code and the output.
nobs - data.frame()
files_list - list.files(directory, full.names=TRUE)
dat - data.frame()
for (i in id){
dat - (read.csv(files_list[i]))
nobs -
This is off-topic here (read the posting guide). You would probably proceed
most effectively by studying how GCC interacts with VS object code, e.g. [1],
and studying the Writing R Extensions manual.
[1]
On Windows it builds a zip file. If you are on Linux, you might [1] need [2].
[1] https://stat.ethz.ch/pipermail/r-help/2005-January/063596.html
[2] http://cran.r-project.org/doc/contributed/cross-build.pdf
---
Jeff Newmiller
On 21/02/2015 07:31, Jeff Newmiller wrote:
On Windows it builds a zip file. If you are on Linux, you might [1] need [2].
[1] https://stat.ethz.ch/pipermail/r-help/2005-January/063596.html
[2] http://cran.r-project.org/doc/contributed/cross-build.pdf
But the first is from 2005 and the second
Hello everyone!
I've been messing with this .Rd file and am having forest/trees problem by
now.
Here is the section of the .Rd file that is the troublemaker:
\usage{
plot.fore4nodate(y, sim, dates, date.fmt = %Y-%m-%d, gof.leg = FALSE,
gof.digits = 2, legend = , leg.cex = 1, bands.col =
On 20/02/2015 12:58 PM, Erin Hodgess wrote:
Hello everyone!
I've been messing with this .Rd file and am having forest/trees problem by
now.
Here is the section of the .Rd file that is the troublemaker:
\usage{
plot.fore4nodate(y, sim, dates, date.fmt = %Y-%m-%d, gof.leg = FALSE,
Hello,
If the vector of observed frequencies is:
f-c(0,0,0,2,3,6,17,15,21,21,14,10,5,1,5)
and the vector of probability :p11-c(7.577864e-06, 1.999541e-04
,1.833510e-03, 9.059845e-03, 2.886977e-02, 6.546229e-02 ,1.124083e-01,
1.525880e-01, 1.689712e-01, 1.563522e-01, 1.232031e-01,
I think
?tapply
and friends: ?by ?aggregate ?ave
is what you want.
-- Bert
Bert Gunter
Genentech Nonclinical Biostatistics
(650) 467-7374
Data is not information. Information is not knowledge. And knowledge
is certainly not wisdom.
Clifford Stoll
On Fri, Feb 20, 2015 at 9:33 AM, Tim
On Fri, 20 Feb 2015, Aron Lindberg wrote:
Hmm…Chuck’s solution may actually be problematic because there are several
entries which at the deepest level are called “sha”, but that should not be
included, such as:
input[[67]]$content[[1]]$commit$tree$sha
and
On 20-02-2015, at 19:05, pari hesabi statistic...@hotmail.com wrote:
Hello,
If the vector of observed frequencies is:
f-c(0,0,0,2,3,6,17,15,21,21,14,10,5,1,5)
and the vector of probability :p11-c(7.577864e-06, 1.999541e-04
,1.833510e-03, 9.059845e-03, 2.886977e-02, 6.546229e-02
Ah! Perfect. Thanks so much!
Sincerely,
Erin
On Fri, Feb 20, 2015 at 1:03 PM, Duncan Murdoch murdoch.dun...@gmail.com
wrote:
On 20/02/2015 12:58 PM, Erin Hodgess wrote:
Hello everyone!
I've been messing with this .Rd file and am having forest/trees problem
by
now.
Here is the
On Feb 20, 2015, at 10:05 AM, pari hesabi wrote:
Hello,
If the vector of observed frequencies is:
f-c(0,0,0,2,3,6,17,15,21,21,14,10,5,1,5)
and the vector of probability :p11-c(7.577864e-06, 1.999541e-04
,1.833510e-03, 9.059845e-03, 2.886977e-02, 6.546229e-02 ,1.124083e-01,
And probably why chisq.test has the rescale.p= argument. Your second problem
with small expected values can be handled with simulate.p.value=.
chisq.test(f, p=p11)
Error in chisq.test(f, p = p11) : probabilities must sum to 1.
1-sum(p11)
[1] 4.3036e-08
chisq.test(f, p=p11, rescale.p=TRUE)
On Feb 20, 2015, at 6:13 AM, Aron Lindberg wrote:
Hmm…Chuck’s solution may actually be problematic because there are several
entries which at the deepest level are called “sha”, but that should not be
included, such as:
input[[67]]$content[[1]]$commit$tree$sh
and
On Feb 20, 2015, at 9:33 AM, Tim Richter-Heitmann wrote:
Dear List,
Consider this example
df - data.frame(matrix(rnorm(9*9), ncol=9))
names(df) - c(c_1, d_1, e_1, a_p, b_p, c_p, 1_o1, 2_o1,
3_o1)
row.names(df) - names(df)
indx - gsub(.*_, , names(df))
I can split the dataframe
The elNamed(x, name) function can simplify this code a bit. The following
gives the same
result as David W's get_shas() for the sample dataset provided:
get_shas2 - function (input) {
lapply(input, function(el) elNamed(elNamed(el, content)[[1]],
sha)[1])
}
Bill Dunlap
TIBCO
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