On a Unix machine I ran caret::rfe using the multicore package, and I
saved the resulting object using save(lm2, file = lm2.RData).
[Reproducible example below.]
When I try to load(lm2.RData) on my Windows laptop, I get
Error in loadNamespace(name) : there is no package called 'multicore'
I
On 08/22/2011 10:04 AM, Jim Silverton wrote:
I have data that ranges from 0.3 to 2 and I want to change the scale to be
from 0 to 1.
Can this be done in R?
Hi Jim,
library(prettyR)
rescale(x,c(0,1))
will linearly transform x into the specified range.
Jim
Hi
Your code is rather baroque and it is difficult to see what is exactly
going on without having appropriate data. I does not consider your process
of reading data from Excel problematic.
Maybe the difference is in that
d- rnorm(whatever)
produces vector while
d- read.delim(clipboard,
On 22.08.2011 03:34, Brad Patrick Schneid wrote:
these guys wont help you with your homework. But have you ever heard of
Google?... if so, R has plenty of online manuals and cheat sheets.
1. This went to R-help (and hence not necessarily to the original poster).
2. You forgot to cite the
Hello,
For the R Wiki, not very realistic to think about fusing it with the
other tools, due to the nature of a wiki on one hand, and the necessity
to share the CRAN site across different repositories on the other hand.
Also, I think that packages development is in the hand of their
On 22.08.2011 08:52, Allan Engelhardt wrote:
On a Unix machine I ran caret::rfe using the multicore package, and I
saved the resulting object using save(lm2, file = lm2.RData).
[Reproducible example below.]
When I try to load(lm2.RData) on my Windows laptop, I get
Error in
please look at the latex() function in package Hmisc.
Sent from my iPhone
On Aug 22, 2011, at 0:55, Alex Ruiz Euler rruizeu...@ucsd.edu wrote:
Dear community,
I had been looking for an easy way to produce latex tables from R
output. xtable() and the package apsrtable produce good
Hi, Emilie.
For your second question. You may check Gleser and Olkin (2009). They gave
several formulas to estimate the sampling covariance for dependent effect
sizes. One of them can be applied in your case.
Gleser, L. J., Olkin, I. (2009). Stochastically dependent effect sizes. In
H. Cooper,
Hi, thanks for the help.
Class says as follows :
class(DF)
[1] SpatialPointsDataFrame
attr(,package)
[1] sp
class(grd)
[1] SpatialPixels
attr(,package)
[1] sp
Anyway, the problem was, as Rmh suggested, with the zerodist().
Tnx, m
-Original Message-
From:
Sorry wrong thread ;)
m
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Matevž Pavlič
Sent: Monday, August 22, 2011 12:14 PM
To: Rmh; Alex Ruiz Euler
Cc: r-help@r-project.org
Subject: Re: [R] Multiple R linear models into one
Hi Daniele,
_ in dati_england is treated as a special character in LaTeX math
mode and causes your LaTeX-compiler trying to switch to math mode (you
might have noticed a warning abaout missing `$' inserted). To produce
a plain _ in TeX you have to mask it as \_. Package Hmisc has some
sanitization
On 08/21/2011 11:52 PM, Allan Engelhardt wrote:
On a Unix machine I ran caret::rfe using the multicore package, and I
saved the resulting object using save(lm2, file = lm2.RData).
[Reproducible example below.]
When I try to load(lm2.RData) on my Windows laptop, I get
Error in
Hello,
I've been dealing with a set of values that contain time stamps and part
of my summary needs to look at just weekend data. In trying to limit the
data I've found a large difference in performance in the way I index a
data frame. I've constructed a minimal example here to try to explain
Hi,
I have two lists (c and h - see below) containing matrices with similar
cases but different values. I want to split these matrices into multiple
matrices based on the values in h. So, I did the following:
years-c(1997:1999)
for (t in 1:length(years))
{
I'm tring the functions to check the cointegration of a matrix.
I'm using **Phillips Ouliaris Cointegration Test**
The function in *tseries* package is **po.test** and **ca.po** in *urca*
The results with **URCA** are:
ca.po(prices, demean='none')
Update: I have recreated an artificial distribution using uniform random
numbers
n - c(runif(Car[1],0,2), runif(Car[2],2,5),runif(Car[3],5,10),
runif(Car[4],10,20),
runif(Car[5],20,30), runif(Car[6],30,40), runif(Car[7],40,60),
runif(Car[8],60,200) )
The resulting density
Any commons is great appreciated.
--
View this message in context:
http://r.789695.n4.nabble.com/How-to-generate-piecewise-cubic-spline-with-many-knots-tp3755419p3759893.html
Sent from the R help mailing list archive at Nabble.com.
__
Hi, I'm a new user of R - Is this how you construct a hat matrix?
H - x %*% solve(t(x) %*% x) %*% t(x)
H
colnames(H) = rep('', 11)
round(H,2)
If so can you make them for more than 2 matrices? Why do you have to use
the 2nd piece of code to round and stuff? Shouldn't it be correct from the
[R] Increase the size of the boxes but not the text in a legend
Jürgen Biedermann
to:
r-help
08/21/2011 06:02 PM
HI there,
I want to add a legend to a plot using the density and angle argument,
so patterns with lines in different angles are used in the plot and
should be referred
[R] Selecting cases from matrices stored in lists
mdvaan
to:
r-help
08/22/2011 07:24 AM
Hi,
I have two lists (c and h - see below) containing matrices with similar
cases but different values. I want to split these matrices into multiple
matrices based on the values in h. So, I did
The problem is that the way you are using *apply, there are
individual calls to the function for each item. In the direct
indexing, you are only making a single call with a vector of values;
Here is a illustration that shows the number of calls:
# count the calls
f.test - function(x) callCnt -
I got it to works. The problem was mainly how data were passed to my
panel and panel.groups function. For reference, here is what I ended
with:
require(lattice)
mybwplot - function(x,y,data,group){
if (missing(group)||is.null(group)) {
group - 'NULL'
ngroups - 1
} else {
Hi,
Much of the tagging/sorting/commenting stuff is already implemented as
http://crantastic.org. Unfortunately few people have taken the time to
contribute reviews. I propose that those of us who would like to bring
more order to the R package universe should start by contributing
reviews, tags,
On Aug 22, 2011, at 9:44 AM, Andrew Campomizzi wrote:
David,
It's fair to question my intentions. I'm running power analyses using
simulations (based on Bolker's Ecological Models and Data in R) and
need to
provide decision-makers with options. So, I'm attempting to make it
clear
that if
Hi Eik,
On Mon, Aug 22, 2011 at 4:14 AM, Eik Vettorazzi
e.vettora...@uke.uni-hamburg.de wrote:
Hi Daniele,
_ in dati_england is treated as a special character in LaTeX math
mode and causes your LaTeX-compiler trying to switch to math mode (you
might have noticed a warning abaout missing `$'
Hi
I used neuralnet for predciton new covarites. Ir give me the predictions as
matrix 1*, . I want to convert the predictions to grid map.
Please help me
Thank you so much
[[alternative HTML version deleted]]
__
R-help@r-project.org
David,
It's fair to question my intentions. I'm running power analyses using
simulations (based on Bolker's Ecological Models and Data in R) and need to
provide decision-makers with options. So, I'm attempting to make it clear
that if the research hypothesis (e.g., response variable declines
Hi all,
Using the exponential distribution to exemplify: The dexp function is
the PDF (1) and pexp is the CDF (2), that is obtained integrating the
PDF. How can I get the qexp and the rexp? Considering that I have the
PDF, how this two are mathematically related to the PDF?
(1) ke^{-kx}
(2)
Hi,
I am interested in ggplot2 and I found this lattice code very interesting
(http://addictedtor.free.fr/graphiques/graphcode.php?graph=48).
Code:
library(lattice)
lattice.options(default.theme = canonical.theme(color = FALSE))
tmp -
expand.grid(geology = c(Sand,Clay,Silt,Rock),
Dear all,
I would like to draw three forest plots to represent results at years 1, 2
and 3. I have the data as point estimates and 95% confidence intervals.
Using the following code I can get three basic forest plots - the first
which has the table of results. I have to plot each separately as
In my orgainzation, the people resonsible for the network are not that keen on
setting up new software, so when I asked them to set up emacs or some other
common R editor, I was told to have a look at ConText. This editor is avaliable
in the network, and there is some R support. Special
Hi,
Let's say that I have a set of column names that begin with the string
Xyz. How do I extract these specific columns? I tried to do the
following:
dataframe1[,grep(Xyz,colnames(dataframe1))]
But it does not work. What is wrong with my expression?
Jean V Adams wrote:
[R] Selecting cases from matrices stored in lists
mdvaan
to:
r-help
08/22/2011 07:24 AM
Hi,
I have two lists (c and h - see below) containing matrices with similar
cases but different values. I want to split these matrices into multiple
matrices based on the
It's not that it's known to be false, rather it's not of interest in this
case. If animal density (response) decreases with increasing year
(predictor), then a change in land management practices might be needed.
Whereas, if animal density is increasing, then the status quo should
suffice.
Thank you both for the replies. While neither produced the exact desired
results, they spurred some new thinking about how to approach the problem. I
came up with a way to get the output desired, but it is probably pretty
clunky. I will post it here anyway, in case someone is interested in the
Hi Josh,
you are absolutly right. Thanks for pointing that out. It is the
Scode- environment which causes the error in TeX.
@Daniele:
have a look at the Sweave user manual (page 7ff) and try
Sweave('example.Rtex',syntax=SweaveSyntaxLatex)
Your Scode block should not be asterisked. I don't know
Read the documentation by typing ?qexp (or whatever other function) at the
command line.
But, since you asked and it won't take long to answer, the general pattern
is:
rDIST gives random numbers sampled from the distribution
dDIST gives the PDF
pDIST gives the CDF
qDIST gives the quantile
Can you say a little more about what you mean it does not work? I'd guess
you have a regular expression mistake and are probably getting more columns
than desired, but without an example, it's hard to be certain.
Use dput() and head() to give a small cut-and-paste-able example.
Michael
On Mon,
On Mon, Aug 22, 2011 at 9:07 AM, Jean V Adams jvad...@usgs.gov wrote:
[R] Selecting cases from matrices stored in lists
mdvaan
to:
r-help
08/22/2011 07:24 AM
Hi,
I have two lists (c and h - see below) containing matrices with similar
cases but different values. I want to split these
Hi Ashz,
On Mon, Aug 22, 2011 at 8:42 AM, ashz a...@walla.co.il wrote:
Hi,
I am interested in ggplot2 and I found this lattice code very interesting
(http://addictedtor.free.fr/graphiques/graphcode.php?graph=48).
Code:
library(lattice)
lattice.options(default.theme = canonical.theme(color
Thanks!
Sebastien
2011/8/20 Uwe Ligges lig...@statistik.tu-dortmund.de
On 19.08.2011 22:27, Sébastien Vigneau wrote:
Hi,
I would like to draw horizontal lines above a bar graph, in order to
display
the p-values of a Fisher test. Here is an
In STATA,
multiple observations correspond to the same individual, the cluster( )
option can be employed to
request that the analysis be clustered by individual.
Any suggestion with aftreg?
Thanks,
J
--
View this message in context:
Re: [R] Selecting cases from matrices stored in lists
mdvaan
to:
r-help
08/22/2011 09:46 AM
Jean V Adams wrote:
[R] Selecting cases from matrices stored in lists
mdvaan
to:
r-help
08/22/2011 07:24 AM
Hi,
I have two lists (c and h - see below) containing matrices
Sorry to anyone who tried but failed to download the data - seems not to be
there.
Here's a new link to it please take a look.
http://ubuntuone.com/p/1C6U/
--
View this message in context:
http://r.789695.n4.nabble.com/Histogram-from-frequency-data-in-pre-made-bins-tp3758198p3760458.html
Sent
Hello,
I have a statistical problem that I am using R for, but I am not making
sense of the results. I am trying to use multiple regression to explore
which variables (weather conditions) have the greater effect on a local
atmospheric variable. The data is taken from a database that has
Hi,
I have the following problem:
I have two vectors:
i - c('a','c','g','h','b','d','f','k','l','e','i')
j - c('a', 'b', 'c')
now I would like to generate a vector with the length of i that
has zeros where i[x] != any element of j
and 1 where i[x] == any element of j.
So for the
%in%
Here,
i %in% j
Hope this helps,
Michael
On Mon, Aug 22, 2011 at 11:51 AM, Martin Batholdy
batho...@googlemail.comwrote:
Hi,
I have the following problem:
I have two vectors:
i - c('a','c','g','h','b','d','f','k','l','e','i')
j - c('a', 'b', 'c')
now I would like to
On 22/08/11 12:26, Martin Morgan wrote:
On 08/21/2011 11:52 PM, Allan Engelhardt wrote:
[...]
Obligatory reproducible example: On the Unix machine do
library(multicore)
a - list(data = 1:10, fun = mclapply)
save(a, file = a.RData)
and then try to load the a.RData file on Windows. The
Hi JC,
You have interactions in your model, which means that your models
specifies that the coefficients for hum, wind, and rain should vary
depending on the value of the other two (and depending on their own
value actually, since you also have quadratic effects for each of
these variables in
Dear all,
I have 100 files which are used as input.and I have to input the name of my
files again and again.the name of the files are 1.out, 2.out..100.out.
I want to know if there is anything like perl so that i can use something like
this-
for($f = 1; $f = 100; $f++) {
$file =
Sorry, my mistake. The thing is that the command return no results at
all. However, when I just tried a simpler version of this (I had no
capital letters or no spaces in the string), it worked fine. I cant
figure it out, I think it all boils down to the fact that I'm no
expert at regexp's...
Hi Vikas,
please do make an effort to search for the answer before posting. A
google search for R read multiple files will give you everything you
need.
Best,
Ista
On Mon, Aug 22, 2011 at 12:08 PM, Bansal, Vikas vikas.ban...@kcl.ac.uk wrote:
Dear all,
I have 100 files which are used as
On 22-Aug-11 15:37:40, JC Matthews wrote:
Hello,
I have a statistical problem that I am using R for, but I am
not making sense of the results. I am trying to use multiple
regression to explore which variables (weather conditions)
have the greater effect on a local atmospheric variable.
The
Dear R community,
I have a data set like the following:
probe_name chr_id position array1 array2 array3 array4 array5 array6
array7
1C-3 10 16566949 10 10 10 10 10 10
10
2C-3AAAB 17 33478940 10 10 10 10 10 10
10
3
Try this:
i %in% j * 1
On Mon, Aug 22, 2011 at 12:51 PM, Martin Batholdy
batho...@googlemail.com wrote:
Hi,
I have the following problem:
I have two vectors:
i - c('a','c','g','h','b','d','f','k','l','e','i')
j - c('a', 'b', 'c')
now I would like to generate a vector with the
Dear Ista
I have searched about the problem and came to know that we can make a list of
our file names.But the thing is I am using this code-
define- function()
{ repeat{
hojy=readline(Enter the name of your file: )
if(file.exists(hojy)==T)
{return(hojy)
break}
else
print(File does not
Hi
I want to create a dependence diagram of a subset of the packages on CRAN
and would therefore like to read the DEACRIPTION files into a list. The list
should be as follow for each package:
- package name: list
- Package: character
- Version: character
- Date: character
- ...
-
I want to select the array columns that are not equal to 10. is ambiguous
to me.
Just to clarify, do you want to simply drop the column named array10 or do
you want to check each column for having one/all 10's as values and drop
based on that test?
Michael
On Mon, Aug 22, 2011 at 12:35 PM,
Hello,
I'd like to send an email from R using Windows Outlook.
The sendmailR package doesn't allow for authentication (usernames and
passwords).
Is there any other way to do this? From the Windows command line?
Right now I am using a .bat file to send an email via a program called Blat.
I'd
HI, Michael,
What I want to do is remove all the rows, for which array1, array2,
..array15 are all equal to 10.
I want to keep all the rows at least one of the array variables are not
equal to 10.
sorry for the confusion.
On Mon, Aug 22, 2011 at 9:52 AM, R. Michael Weylandt
Here is my solution to produce a date value if your data set only has week
values associated with data (ie no date).
It gives the first monday of the week.
Michael Folkes
s - seq(as.Date(2010-01-01), as.Date(2010-12-31), by = day) #produce all
days of the year
On 22.08.2011 18:43, Rainer M Krug wrote:
Hi
I want to create a dependence diagram of a subset of the packages on CRAN
and would therefore like to read the DEACRIPTION files into a list. The list
should be as follow for each package:
- package name: list
- Package: character
- Version:
Why don't you just use list.files() and iterate over the result in
your for loop?
fileNames - list.files(pattern = \\.out)
myFiles - list()
for(i in fileNames) {
myFiles[[i]] - read.table(i, fill=T,colClasses = character)
}
?
Best,
Ista
On Mon, Aug 22, 2011 at 12:34 PM, Bansal, Vikas
This isn't the most beautiful code, but I think it should work for you:
# Some sample data
M =
cbind(matrix(rnorm(10),ncol=2),matrix(sample(c(10,1),15,replace=T),ncol=3))
colnames(M) = c(Thing1,Thing2,paste(array,1:3,sep=))
colsToCheck = grepl(array,colnames(M)) # Isolate the array columns
Ben qant ccquant at gmail.com writes:
Hello,
I'd like to send an email from R using Windows Outlook.
The sendmailR package doesn't allow for authentication (usernames and
passwords).
I don't know about Outlook, but you can try
Campomizzi, Andrew J acampomizzi at neo.tamu.edu writes:
On 20/08/11 10:20, Andrew Campomizzi wrote:
Hello,
I'm having trouble figuring out how to calculate a p-value for a 1-tailed
test of beta_1 in a linear model fit using command lm. My model has only 1
continuous, predictor
Because i have to use the value of i in for loop also.example-
for(i in 1:100){
df=read.table($i.out,fill=T,colClasses = character)
if(i=50){
df$V6 - sapply(df$V6, function(a)
paste(as.integer(charToRaw(a)), collapse = ' '))}
else{
df$V5 - sapply(df$V6, function(a)
THANKS SO MUCH, Michael!
Appreciated!
On Mon, Aug 22, 2011 at 10:16 AM, R. Michael Weylandt
michael.weyla...@gmail.com wrote:
This isn't the most beautiful code, but I think it should work for you:
# Some sample data
M =
On Mon, Aug 22, 2011 at 1:25 PM, Bansal, Vikas vikas.ban...@kcl.ac.uk wrote:
Because i have to use the value of i in for loop also.example-
for(i in 1:100){
df=read.table($i.out,fill=T,colClasses = character)
if(i=50){
df$V6 - sapply(df$V6, function(a)
paste(as.integer(charToRaw(a)),
For H_0: beta = 0, then the correct p-value is
pt(tvalue,df)
regardless of the sign of tvalue. Negative tvalues of large magnitude
will yield small p-values.
albyn
On Mon, Aug 22, 2011 at 05:22:06PM +, Ben Bolker wrote:
Campomizzi, Andrew J acampomizzi at neo.tamu.edu writes:
HI, Michael,
Sorry for my numb, I have one more question.
When you use function(x){any (x != 10), here x is a vector, x!=10 will give
a vector of logical value, right?
If it is, how can vector be compared to a scale, 10 in this case?
Thanks!
On Mon, Aug 22, 2011 at 10:16 AM, R. Michael
Hi:
You need a leading ^ in your grep string. Here's a reproducible
example to illustrate:
df - data.frame(Xyz1 = rnorm(5), Xyz2 = rnorm(5), Xyz3 = rnorm(5),
Abc1 = rnorm(5), Abc2 = rnorm(5))
df[, grep('^Xyz', names(df))]
df[, grep('^Abc', names(df))]
HTH,
Dennis
On Mon, Aug
The different lengths work because R recycles values whenever you try to do
a binary operation on things of different lengths: in essence, R copies 10
however many times needed to make something that has the right length for an
elementwise comparison with x.**
If you did something like
x !=
Hi,
I would like to draw a stacked bar chart with four bars (say a, b, c,
d) . Two bars belong to group A and the two others to group B. Therefore,
I would like to have, on the x-axis, a label for each bar and an additional
label for each group, positioned underneath. To give an idea, the x-axis
Thanks, Michael!
You have an heart of gold! Appreciated!
On Mon, Aug 22, 2011 at 10:53 AM, R. Michael Weylandt
michael.weyla...@gmail.com wrote:
The different lengths work because R recycles values whenever you try to do
a binary operation on things of different lengths: in essence, R
Hi Sébastien,
Not sure about an elegant, general way but here is something quick and dirty:
p - barplot(matrix(1:8, 2))
axis(1, at = p, labels = letters[1:4])
axis(1, at = c(mean(p[1:2]), mean(p[3:4])), labels = paste(\n,
LETTERS[1:2]), padj = 1)
Cheers,
Josh
On Mon, Aug 22, 2011 at 10:14
On Aug 22, 2011, at 1:45 PM, Dennis Murphy wrote:
Hi:
You need a leading ^ in your grep string. Here's a reproducible
example to illustrate:
df - data.frame(Xyz1 = rnorm(5), Xyz2 = rnorm(5), Xyz3 = rnorm(5),
Abc1 = rnorm(5), Abc2 = rnorm(5))
df[, grep('^Xyz', names(df))]
On Aug 22, 2011, at 12:14 PM, Sébastien Vigneau wrote:
Hi,
I would like to draw a stacked bar chart with four bars (say a, b, c,
d) . Two bars belong to group A and the two others to group B. Therefore,
I would like to have, on the x-axis, a label for each bar and an additional
label for
Hello all,
I have two columns of numbers. I would like to do the following:
(1) Plot both cdfs, F1 and F2 on the same graph.
(2) Find smoothed approximations of F1 and F2 lets call them F1hat and F2hat
(3) Find values for F1hat when we substitue a value of x in it.
(4) Find the corresponding
Number 1 can be done as follows:
x = rnorm(50); y = rnorm(50)
xCDF = ecdf(x); yCDF = ecdf(y)
plot(xCDF)
lines(yCDF,col=2)
For the other ones, you are going to have to be a little more specific as to
how you want to do the approximation...but ?density might be a place to
start for #4, assuming
Dear all,
I am having trouble creating LaTex tables using the xtable command. I am
using the bayesm package to analyse data. However, I am unable to generate
LaTex tables converting the output from summary(out$deltadraws.) I have made
several attempts using xtable but have been unsuccessful and
R -
I have 3 variables with data below. Variable Rev is a vector that changes
from 1 to 2, 2 to 3, etc Variable FF is a binary variable with 1's
and 0's. Variable bin is a different binary variable with 1's and 0's.
I want to calculate the number of elements:
1. Starting with the first
On Aug 22, 2011, at 3:18 PM, ea819 wrote:
Dear all,
I am having trouble creating LaTex tables using the xtable command.
I am
using the bayesm package to analyse data. However, I am unable to
generate
LaTex tables converting the output from summary(out$deltadraws.) I
have made
several
Hello all,
I posted a questions on how to terminate a function call that does not
return after a certain time ( I can not modify the function code) some
time ago.
Since I didn't find a solution I just came up with the idea to run the
functions call in a separate thread who I could terminate a
When you read Excel data from the Windows clipboard, the delimiter is a tab,
not a comma.
--
David L Carlson
Associate Professor of Anthropology
Texas AM University
College Station, TX 77843-4352
-Original Message-
From:
Yes. The xCDF/yCDF objects that are returned by the ecdf function can be
called like functions.
For example:
x = rnrom(50); xCDF = ecdf(x); xCDF(0.3)
# This value tells you what fraction of x is less than 0.3
You can also assign this behavior to a function:
F - function(z) { xCDF(z) }
F does
Thanks!
Sebastien
On Mon, Aug 22, 2011 at 2:11 PM, Marc Schwartz marc_schwa...@me.com wrote:
On Aug 22, 2011, at 12:14 PM, Sébastien Vigneau wrote:
Hi,
I would like to draw a stacked bar chart with four bars (say a, b,
c,
d) . Two bars belong to group A and the two others to group B.
On Mon, Aug 22, 2011 at 12:39 PM, Immanuel mane.d...@googlemail.com wrote:
Hello all,
I posted a questions on how to terminate a function call that does not
return after a certain time ( I can not modify the function code) some
time ago.
Since I didn't find a solution I just came up with the
On Aug 22, 2011, at 3:50 PM, R. Michael Weylandt wrote:
Yes. The xCDF/yCDF objects that are returned by the ecdf function
can be
called like functions.
Because they _are_ functions.
function %in% class(xCDF)
[1] TRUE
is.function(xCDF)
[1] TRUE
For example:
x = rnrom(50); xCDF =
Dear list,
I have a spacialPolygonDataFrame where variables were unnecessarily imported as
factors. So I am trying to unfactor variables from spatialPolygonDataFrame@data
with a loop
for (i in (1:length(names( spatialPolygonDataFrame{
Thanks Jean, changing c[[t]] to c[[year]] solved the issue.
Math
Jean V Adams wrote:
Re: [R] Selecting cases from matrices stored in lists
mdvaan
to:
r-help
08/22/2011 09:46 AM
Jean V Adams wrote:
[R] Selecting cases from matrices stored in lists
mdvaan
to:
r-help
Juta,
On Mon, Aug 22, 2011 at 4:29 PM, Juta Kawalerowicz
juta.kawalerow...@stx.ox.ac.uk wrote:
Dear list,
I have a spacialPolygonDataFrame where variables were unnecessarily imported
as factors. So I am trying to unfactor variables from
spatialPolygonDataFrame@data with a loop
for (i in
[R] Counting Elements Conditionally
Edward Patzelt
to:
r-help
08/22/2011 02:33 PM
R -
I have 3 variables with data below. Variable Rev is a vector that
changes
from 1 to 2, 2 to 3, etc Variable FF is a binary variable with
1's
and 0's. Variable bin is a different binary
Re: [R] Counting Elements Conditionally
Jean V Adams
to:
Edward Patzelt
08/22/2011 03:53 PM
[R] Counting Elements Conditionally
Edward Patzelt
to:
r-help
08/22/2011 02:33 PM
R -
I have 3 variables with data below. Variable Rev is a vector that
changes
from 1 to 2,
Awesome, this is close, couple changes. Below is full data set for 1
person. I want the code to look at the first time it sees a 0 in FF after
the transition in Rev. I then want it to test whether bin is also a 0. If
and only if this is the first 0 in FF after the transition, and bin = 0,
then
after it sees the first occurrence of 0 in FF following a transition, I want
it to ignore all further elements until the next transition.
On Mon, Aug 22, 2011 at 3:58 PM, Edward Patzelt patze...@umn.edu wrote:
Awesome, this is close, couple changes. Below is full data set for 1
person. I
On Aug 22, 2011, at 4:34 PM, David Winsemius wrote:
On Aug 22, 2011, at 3:50 PM, R. Michael Weylandt wrote:
Yes. The xCDF/yCDF objects that are returned by the ecdf function
can be
called like functions.
Because they _are_ functions.
function %in% class(xCDF)
[1] TRUE
So, using the full data set, what should the result look like?
c(NA, NA, NA, 3, NA,NA, NA, 2) ?
Jean
Edward Patzelt patze...@umn.edu wrote on 08/22/2011 03:58:38 PM:
[image removed]
Re: [R] Counting Elements Conditionally
Edward Patzelt
to:
Jean V Adams
08/22/2011
that is exactly correct, assuming we did not start at the beginning, but
started at the first transition (this is the correct way to think about it)
On Mon, Aug 22, 2011 at 4:08 PM, Jean V Adams jvad...@usgs.gov wrote:
So, using the full data set, what should the result look like?
c(NA, NA,
Hello,
thanks for the input. Below is a small example, simpler then expected :)
I'm just curious why I can't see any output from print(i).
--
library(multicore)
f_long - function() {
for (i in 1:1){ a=i}
print(i)
return(finished)
}
p_long -
1 - 100 of 134 matches
Mail list logo
