Hi all,
Does anyone know of a way to force utils::bibentry to mimic the BibTex
behaviour of using double { to force a "corporate name" in the author
field to print correctly? For example take this bibentry:
entry <- utils::bibentry(
bibtype = "Manual",
title = "The Thing",
author = "The
ussion has
> already occurred there.
>
> On February 28, 2020 3:35:09 PM PST, Sam Albers
> wrote:
> >Great question Will. If it were my code I would definitely do this.
> >However the problem is manifesting itself for my work with Dirk's
> >great digest pac
atch to deal with problems
> opening or
> reading the file.
>
> Bill Dunlap
> TIBCO Software
> wdunlap tibco.com
>
>
> On Fri, Feb 28, 2020 at 2:54 PM Sam Albers wrote:
>>
>> Thanks Jeff. I am probably not explaining myself very well but my
>> question under wha
ed in both of these functions?
On Fri, Feb 28, 2020 at 2:37 PM Jeff Newmiller wrote:
>
> Dunno. They agree for me. Maybe look closer at all permissions via Windows
> File Manager?
>
> On February 28, 2020 2:06:34 PM PST, Sam Albers
> wrote:
> >Some additional follow-up:
>
e.R
-1
Can anyone think why file.access and file would be contradicting each other?
Sam
On Fri, Feb 28, 2020 at 10:47 AM Sam Albers wrote:
>
> Hi there,
>
> Looking for some help in diagnosing or developing a work around to a
> problem I am having on a Windows machine. I am
Hi there,
Looking for some help in diagnosing or developing a work around to a
problem I am having on a Windows machine. I am running R 3.6.2.
I have two identical files, one stored locally and the other stored on
a network drive.
For access:
> file.access(local_file, 4)
local.R
0
>
Hello all,
I am experience some issues with building a package that we are
hosting on GitHub. The package itself is quite large. It is a data
package with a bunch of spatial files stored as .rds files.
The repo is located here: https://github.com/bcgov/bcmaps.rdata
If we clone that package to
Is it possible for someone to explain what is going on here? I would expect
that `as.POSIXlt` would be able to accept `datestring` and return all the
elements without having to convert it using `as.POSIXct` first. Do
`as.POSIXlt` and `as.POSIXct` do different things with the `tz` arg?
datestring
Hi there,
I am trying to distinguish between getOption() and Sys.getenv(). My
understanding is that these are both used to set values for variables.
getOption is set something like this: option("var" = "A"). This can be
placed in an .Rprofile or at the top of script. They are called like this
Hi folks,
I am pleased to announce that the rsoi is now up on CRAN (v0.2.1
https://CRAN.R-project.org/package=rsoi). rsoi is a minimal but
hopefully useful package to folks that are looking for easy access in
R to Southern Oscillation Index and Oceanic Nino Index data.
rsoi uses data collected
Hello there,
I am wondering if anyone on this list has ever encountered an
implementation of the Split and Merge algorithm for R. This algorithm
is reasonably well known and was first developed in this paper:
https://www.computer.org/csdl/trans/tc/1974/08/01672634-abs.html
>From that paper here
Hello,
I have a problem to which I am certain grep or gsub or that family of
functions are the solution. However, I just can't seem to wrap my mind
around exactly how. I have a dataframe below that has the dimensions
of a net. I am given the data is the "W X H" format. For calculations
I'll like
Hi all,
This is a process question. How do folks efficiently identify column
numbers in a dataframe without manually counting them. For example, if I
want to choose columns from the iris dataframe I know of two options. I can
do this:
str(iris)'data.frame':150 obs. of 5 variables:
$
like
a - data.frame(a=rnorm(10), b=rnorm(10), c=rnorm(10), d=rnorm(10))
toMatch - c(a, d)
grep(paste(toMatch,collapse=|), colnames(a))
#to subset
a[,grep(paste(toMatch,collapse=|), colnames(a))]
On Tue, Aug 25, 2015 at 10:17 AM, Sam Albers tonightstheni...@gmail.com
wrote:
Hi all
Hi all,
I have a question about using R in a way that may not be correct but I
thought I would ask anyway.
I have an instrument that outputs a text file with comma separated data. A
new line is added to the file each time the instrument takes a new reading.
Is there any way to configure R such
Hello,
I am having some trouble figuring out how to deal with data that has some
observations that are detection limits and others that are integers denoted
by greater and less than symbols. Ideally I would like a column that has
the data as numbers then another column with values Measured or
Hello,
I have recently received a dataset from a metal analysis company. The
dataset is filled with less than symbols. What I am looking for is a
efficient way to subset for any whole numbers from the dataset. The column
is automatically formatted as a factor because of the symbols making it
$Cedar.Creek - as.numeric(metals$Cedar.Creek)
R str(metals)
'data.frame':19 obs. of 2 variables:
$ Parameter : Factor w/ 20 levels Antimony,Arsenic,..: 1 2 3 4 6
7 8 9 10 11 ...
$ Cedar.Creek: num 100 100 500 100 10 1000 100 516 550 10 ...
Sarah
On Wed, Jul 9, 2014 at 1:19 PM, Sam
Hello,
I am trying to define a different interval for a year. In hydrology,
a water year is defined as the period between October 1st and
September 30 of the following year. I was wondering how I might do
this in R. Say I have a data.frame like the following and I want to
extract a variable with
Hello all,
Let me first say that this isn't a question about outliers. I am using
the outlier function from the outliers package but I am using it only
because it is a convenient wrapper to determine values that have the
largest difference between itself and the sample mean. Where I am
running
Hello,
I am having a problem making use of some data outputted from an
instrument in a somewhat weird format. The instrument outputs two
columns - one called JulianDay.Hour and one called Minutes.Seconds. I
would like to convert these columns into a single column with a time.
So I was using
Apologies. I was searching using the wrong search terms. This is
clearly a string issue. I've added the solution below.
Sam
On Tue, May 29, 2012 at 11:39 AM, Sam Albers tonightstheni...@gmail.com wrote:
Hello,
I am having a problem making use of some data outputted from an
instrument
Hello all,
I can't seem to figure out how to format a date as a title. I have
something like this:
plot(x=1:10, y=runif(10,1,18), main=paste(as.Date(2011-05-03,
format=%Y-%m-%d)))
## When I would really like this
plot(x=1:10, y=runif(10,1,18), main=paste(May-03-2011))
## I thought to try this
Hello,
I am having trouble figuring out how to convert a Day of Year integer
back into a Date format. For example I have the following:
date -
c('2008-01-01','2008-01-02','2008-01-03','2008-01-04','2008-01-05','2008-01-06','2008-01-07',
Hello,
I am struggling to produce an MDS plot using the randomForest package
with a moderately large data set. My data set has one categorical
response variables, 7 predictor variables and just under 19000
observations. That means my proximity matrix is approximately 133000
by 133000 which is
On Mon, Mar 19, 2012 at 9:11 PM, Gabor Grothendieck
ggrothendi...@gmail.com wrote:
On Mon, Mar 19, 2012 at 8:03 PM, Sam Albers tonightstheni...@gmail.com
wrote:
Hello R-ers,
I just wanted to update this post. I've made some progress on this but
am still not quite where I need to be. I
Hello all,
I need to figure out a way to lag a variable in by a number of days
without using the zoo package. I need to use a remote R connection
that doesn't have the zoo package installed and is unwilling to do so.
So that is, I want a function where I can specify the number of days
to lag a
Hello R-ers,
I just wanted to update this post. I've made some progress on this but
am still not quite where I need to be. I feel like I am close so I
just wanted to share my work so far.
Thanks in advance!
Sam
On Mon, Mar 19, 2012 at 1:10 PM, Sam Albers tonightstheni...@gmail.com wrote
Hello all,
I have become somewhat confused with options available for dealing
with a highly unbalanced data set (1 in one class, 50 in the
other). As a summary I am unsure:
a) if I am perform the two class weighting methods properly,
b) if the data are too unbalanced and that this type of
Hello all,
## I am trying to convert some year and month data into a single
variable that has a date format so I can plot a proper x axis.
## I've made a few tries at this and search around but I haven't found
anything. I am looking for something of the format %Y-%m
## A data.frame
df -
Hello,
I am looking for a way to subset a data frame by choosing the top ten
maximum values from that dataframe. As well this occurs within some
factor levels.
## I've used plyr here but I'm not married to this approach
require(plyr)
## I've created a data.frame with two groups and then a id
Hello all,
This is one of those Is there a better way to do this questions. Say
I have a dataframe (df) with a grouping variable (z). This is my base
data. Now I know that there is a higher order level of grouping that
exist for my group variable. So what I want to do is create a new
column that
for the response!
Sam
On Thu, Jan 19, 2012 at 3:05 PM, Sam Albers tonightstheni...@gmail.com
wrote:
Hello all,
This is one of those Is there a better way to do this questions. Say
I have a dataframe (df) with a grouping variable (z). This is my base
data. Now I know
,
Check the examples in
require(car)
?recode
HTH,
Jorge.-
On Thu, Jan 19, 2012 at 6:05 PM, Sam Albers wrote:
Hello all,
This is one of those Is there a better way to do this questions. Say
I have a dataframe (df) with a grouping variable (z). This is my base
data. Now I know
Thanks for getting me on the right path Gabor! I have one outstanding
issue though.
On Mon, Jan 9, 2012 at 4:21 PM, Gabor Grothendieck
ggrothendi...@gmail.com wrote:
On Mon, Jan 9, 2012 at 6:39 PM, Sam Albers tonightstheni...@gmail.com wrote:
Hello all,
I am trying to determine how
Hello all,
I am trying to determine how to calculate rolling means in R using a
grouping variable. Say I have a dataframe like so:
dat1 - data.frame(x = runif(2190, 0, 125), year=rep(1995:2000,
each=365), jday=1:365, site=here)
dat2 - data.frame(x = runif(2190, 0, 200), year=rep(1995:2000,
Hello all,
I am trying write a fairly simple function that provide a quick way to
calculate several distributions for a data set. I am trying to provide
a function that has a argument that specifies which distribution is
outputted (here norm or cumu). I also have a melt argument but
that seems to
## Hello there,
## I have an issue where I need to use the value of column names to
multiply with the individual values in a column and I have many
columns to do this over. I have data like this where the column names
are numbers:
mydf - data.frame(`2.72`=runif(20, 0, 125),
Thanks for the response David.
On Tue, Aug 16, 2011 at 1:13 PM, David Winsemius dwinsem...@comcast.net wrote:
On Aug 16, 2011, at 3:37 PM, Sam Albers wrote:
## Hello there,
## I have an issue where I need to use the value of column names to
multiply with the individual values in a column
Hello list,
## I have been doing the following process to convert data from one
form to another for a while but it occurs to me that there is probably
an easier way to do this. I am often given data that have column names
which are actually data and I much prefer dealing with data that are
sorted
Hello R-help,
I have some data collected at regular intervals but for a varying
length of time. I would like to standardize the length of time
collected and I can do this by standardizing the number of records I
use for my analysis.
Take for example the data set below:
library(plyr)
x -
Many thanks Dr. Ripley. I was made aware that the problem for me was that I
was sending HTML in my email message. Transgression noted.
On Wed, Jun 29, 2011 at 10:17 PM, Prof Brian Ripley
rip...@stats.ox.ac.ukwrote:
On Wed, 29 Jun 2011, Sam Albers wrote:
I know that this has been asked before
Thanks for the response Dr. Ripley. Much appreciated.
On Wed, Jun 29, 2011 at 10:17 PM, Prof Brian Ripley
rip...@stats.ox.ac.ukwrote:
On Wed, 29 Jun 2011, Sam Albers wrote:
I know that this has been asked before in other variations but I just
can't
seem to figure out my particular
I know that this has been asked before in other variations but I just can't
seem to figure out my particular application from previous posts. My
apologies if I have missed the answer to this question somewhere in the
archives. I have indeed looked.
I am running Ubuntu 11.04, with R 2.12.1 and
Hello all,
I can't seem to figure how to use a greek character in expression() in
plot() labels without adding a space. So for example below when plotting
this out
x-1:10
plot(x,x^2, xlab=expression(Chlorophyll~italic(a)~mu~g~cm^-2))
the axis label read as μ g cm^-2 because I have space there
If I understand what you want (which I may very well not) you could use
something like this:
If this is an example of your type of data:
564589,+
substr(x, 1, 6)
as.numeric(x)
Please try to post something more thorough if you would like a better
answer.
Sam
--
View this message in context:
Hello all,
I have been using an instrument that collects a temperature profile of a
water column. The instrument records the temperature and depth any time it
takes a reading. I was sampling many times at discrete depth rather than a
complete profile of the water column (e.g. interested in 5m,
Hello,
I am having a little trouble finding the right set of criteria to subset a
portion of data. I am using an instrument that does depth profiles of a
water column. The instrument records on the way down as well as the way up.
## So I am left with data like this:
dat - data.frame(var =
Hello,
Apologies for a similar earlier post. I didn't include enough details in
that one.
I am having a little trouble subsetting some data based on a grouping
variable. I am using an instrument that does depth profiles of a water
column. The instrument records on the way down as well as the way
$day.hour, format=%j%H)
data$Time - strptime(data$min.sec, format=%M%S)
data
Using Ubuntu 10.10 and R 2.11.1. Thanks in advance
Sam
--
*
Sam Albers
Geography Program
University of Northern British Columbia
University Way
Prince George
Hello Venerable List,
I am trying to loop (I think) an operation through a list of columns in a
dataframe to remove set of #DIV/0! values. I am trying to do this like so:
#Data.frame
test - read.csv(http://dl.dropbox.com/u/1574243/sample_data.csv;,
header=TRUE, sep=,)
#This removes all the
!')), ]
HTH,
Jorge
On Wed, Mar 9, 2011 at 3:29 PM, Sam Albers wrote:
Hello Venerable List,
I am trying to loop (I think) an operation through a list of columns in a
dataframe to remove set of #DIV/0! values. I am trying to do this like so:
#Data.frame
test - read.csv(http://dl.dropbox.com/u
that I would like to compare to dat.
dat2 - data.frame(y = runif(2, 0, 125), date =
as.Date(c(2009-09-27,2009-10-01), format=%Y-%m-%d))
--
*
Sam Albers
Geography Program
University of Northern British Columbia
University Way
Prince George
-integer?
Sorry for the simple question and thanks in advance.
Sam
--
*
Sam Albers
Geography Program
University of Northern British Columbia
University Way
Prince George, British Columbia
Canada, V2N 4Z9
phone: 250 960-6777
On Wed, Nov 17, 2010 at 3:49 PM, David Winsemius dwinsem...@comcast.netwrote:
On Nov 17, 2010, at 6:27 PM, Sam Albers wrote:
Hello all,
I have a fairly simple data manipulation question. Say I have a dataframe
like this:
dat - as.data.frame(runif(7, 3, 5))
dat$cat - factor(c
Prior to creating a plot I usually just order the factor levels to the order
I want them in. So for your example I would do:
#Create some data
library(lattice)
x - runif(100, 0, 20)
df - data.frame(x)
df$y - (1:10)
df$Month - c(October, September, August, July,June)
#Plot the figure
plt
it as an example.
with(wf, contourplot(speed ~ width*length,
region=TRUE,
contour=FALSE
))
--
*
Sam Albers
Geography Program
University of Northern British Columbia
University Way
Unless you aren't writing scripts why wouldn't you just use something like
this?
x=c(1,2,3)
pdf(RRules.pdf)
plot(x,x)
dev.off()
--
View this message in context:
http://r.789695.n4.nabble.com/Can-I-set-default-parameters-for-the-default-graphics-device-tp2290827p2290836.html
Sent from the
more
than two distributions.
Thanks in advance.
Sam
--
*
Sam Albers
Geography Program
University of Northern British Columbia
University Way
Prince George, British Columbia
Canada, V2N 4Z9
phone: 250 960-6777
Hello,
I am curious if anyone has had any success with finding a R version of a
k-sample Kolmogorov-Smirnov test. Most of the references that I have able to
find on this are fairly old and I am wondering if this type of analysis has
fallen out of favour. If so, how do people tend to compare
Bimal,
in the memisc packages:
?panel.errbars
This might be a good option for you.
HTH,
Sam
--
View this message in context:
http://r.789695.n4.nabble.com/plotting-data-when-all-you-have-is-the-summary-data-tp2173026p2173303.html
Sent from the R help mailing list archive at Nabble.com.
I'm not sure about plotmeans but this is usually the way I plot means with
lattice:
library(lattice)
x - runif(48, 2, 70)
data - data.frame(x)
data$factor1 - factor(c(A, B, C, D))
data$factor2 - factor(c(X, Y, Z))
data.mean - with(data, aggregate(data$x, by=list(factor1=factor1,
; returning Inf
2: In max(x) : no non-missing arguments to max; returning -Inf
3: In min(x) : no non-missing arguments to min; returning Inf
4: In max(x) : no non-missing arguments to max; returning -Inf
--
*
Sam Albers
Geography Program
University
))
--
*
Sam Albers
Geography Program
University of Northern British Columbia
University Way
Prince George, British Columbia
Canada, V2N 4Z9
phone: 250 960-6777
*
[[alternative HTML version deleted
Hi Ivan,
Can you educate me a little bit on the use of barchart?
Unfortunately no... For this post I eventually used the barplot2() in the
gplots packages. I got bogged down trying to do it in lattice so I looked
for an alternative. It was quite straight forward which was nice and I was
able to
0.50866 down BU1 0.34869 up BU2 0.32831 up BU3 0.59877
up BM1 0.52518 mid BM2 0.94387 mid BM3 0.94387 mid BD1 0.46872 down BD2
0.63115 down BD3 0.45239 down
n down down down down down down
--
*
Sam Albers
Geography Program
University
Julia,
I think exporting to excel takes you a step back. Likely it would be easier
to work solely in R and sort the P values like that. I had to do something
similar only with a bunch of regressions a while back. I found this post
extremely helpful as well as the plyr package
as.character(Sample) in place of Sample in the last line.
On Sat, Apr 3, 2010 at 12:18 PM, Sam Albers tonightstheni...@gmail.com
wrote:
Good Morning,
I am trying to create a new column of character strings based on the
first
two letters in a string in another column. I believe that I need
--
*
Sam Albers
Geography Program
University of Northern British Columbia
University Way
Prince George, British Columbia
Canada, V2N 4Z9
phone: 250 960-6777
*
[[alternative HTML version deleted
Thanks!
On Fri, Apr 2, 2010 at 11:35 AM, Erik Iverson er...@ccbr.umn.edu wrote:
Hello,
Sam Albers wrote:
Hello there,
I have a situation where I would like to select the first row of a
particular factor for a data frame (data example below). So that is, I
would
like to select
5699.8 none 0
2 265.6 syto -5434.2
2 329.6 sytolec -5370.2
2 383 sytolec -5316.8
2 968.8 sytolec -4731
3 2466.8 none 0
3 1303 syto -1163.8
3 1290.6 sytolec -1176.2
3 110.2 sytolec -2356.6
3 15086.8 sytolec 12620
--
*
Sam Albers
,
scales=list(x=list(rot=45)),
ly=lower.se,
uy=upper.se,
prepanel=prepanel.ci,
panel=panel.superpose,
panel.groups=panel.ci
))
--
*
Sam Albers
Geography Program
University of Northern British
and Ubuntu 9.04.
Thanks in advance!
Sam
--
*
Sam Albers
Geography Program
University of Northern British Columbia
University Way
Prince George, British Columbia
Canada, V2N 4Z9
phone: 250 960-6777
Downstream,125.04,11/09/09,0.13
--
*
Sam Albers
Geography Program
David Winsemius, MD
Heritage Laboratories
West Hartford, CT
--
*
Sam Albers
Geography Program
University of Northern
/09,0.12
Middle,125.04,11/09/09,0.1
Downstream,125.04,11/09/09,0.11
Downstream,125.04,11/09/09,0.13
Downstream,125.04,11/09/09,0.13
--
*
Sam Albers
Geography Program
University of Northern British Columbia
University Way
Prince George
/2009,80.37
26/10/2009,20.69
26/10/2009,63.37
26/10/2009,70.91
26/10/2009,22.7
26/10/2009,23.89
--
*
Sam Albers
Geography Program
University of Northern British Columbia
University Way
Prince George, British Columbia
Canada, V2N 4Z9
phone: 250
/10/2009,21.38
19/10/2009,80.37
26/10/2009,20.69
26/10/2009,63.37
26/10/2009,70.91
26/10/2009,22.7
26/10/2009,23.89
--
*
Sam Albers
Geography Program
University of Northern British Columbia
University Way
Prince
Hello,
Can anyone recommend a good example of web implementation of R? Can't seem
to find anything on my own.
Thanks in advance!
Sam
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
So I managed to solve this for myself in a very roundabout kind of way. So I
figured that I should share in case anyone else needed something like this.
filled.contour(contour, axes=F, frame.plot=F, color=terrain.colors, ylab=
, key.title = title(main=Velocity\n(m/s)),asp=2, key.axes = axis(4,
Hello there,
I have a fairly simple request (I hope!)
I have produced a filled contour plot like this:
library(grDevices)
library(gplots)
library(plotrix)
filled.contour(contour, axes=F, frame.plot=TRUE, color=terrain.colors, ylab=
Length Along Flume (m), key.title =
Right equidistant was clearly the wrong word. Sorry. I just meant that any
given point should have an equal distance from the four points immediately
surrounding it (x,-x,y-y) aside from those on the edge which will obviously
only have two or three points surrounding.
On Wed, Jul 9, 2008 at 3:12
Basically, I want 181 points equally spaced over a 40 x 40 cm area. I want
to be able to specify the number of points and the area to which they are
plotted on. I think you are right that grid is what I am looking for but I
was the grid to have axes which your code below, although appreciated, did
Ahhh. That worked perfectly. Thank you very much.
On Wed, Jul 9, 2008 at 4:19 PM, Dylan Beaudette [EMAIL PROTECTED]
wrote:
On Wednesday 09 July 2008, hippie dream wrote:
This might not possible in R but I thought I would give it shot. I am
have
to set up a 40 x 40 cm grid of 181 points
I have constructed a Trellis style xyplot.
lengthf - factor(length)
xyplot(SLI$velocity ~ SLI$width | SLI$lengthf, layout = c(2,7), xlab =
Width (cm), ylab = Velocity (m/s^2), col = black)
This produces a lovely little plot. However, the grouping factor(lengthf)
isn't in the right order. My
I have trying to figure this out all day so hopefully the answer isn't too
obvious. I am able to view a graph in the viewer window. However, I need to
export graph outside of the viewer window. Here is the script I am using:
png(Compare.png)
plot(compare$DepthSLI, compare$DischargeSLI,
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