You forgot the 'at a minimum' information required by the posting
guide.
Most likely this is a limitation of the locale you used (and failed to
tell us about) on the OS you used (...).
On Sat, 16 Jul 2011, Sverre Stausland wrote:
Dear helpers,
I'm trying to replace a character with a
Hello,
I have plotted a ggplot, the image is attached below.
Now I have two questions:
1. How to order the pvalue displayed in the plot.
I want the pvalue in the order : 0 to 1E-12, 1E-12 to 1E-6, 1E-6 to 1E-3,
1E-3 to 0.05 and finally 0.05
2. Add name to the X-axis.
I wanted to add a label to
Hello all,
I am trying to figure out how to construct several new variables without
having to write a code for each one. Specifically, I want to calculate
ratios for several variables. For instance, let's say I have:
df
male1990 male1995 male2000 male2005 female1990 female1995 female2000
Thanks, Joshua. This is exactly what I was looking for.
On Sat, Jul 16, 2011 at 11:17 AM, Joshua Wiley jwiley.ps...@gmail.com wrote:
Hi e,
Here is a simple function to use the attributes from one scaled object
to scale another.
x = data.frame(a=1:10,b=11:20)
y =
Hi,
On Sat, Jul 16, 2011 at 10:19 PM, hrishi hrishi...@gmail.com wrote:
Hello,
I have plotted a ggplot, the image is attached below.
Now I have two questions:
1. How to order the pvalue displayed in the plot.
I want the pvalue in the order : 0 to 1E-12, 1E-12 to 1E-6, 1E-6 to 1E-3,
1E-3
Hi Carlos,
Assuming your actual dataset has as nice a structure as your example
one, it is extremely easy:
## Your data (in convenient form using dput(df))
df - structure(list(male1990 = c(10L, 23L, 32L), male1995 = c(11L,
22L, 33L), male2000 = c(15L, 25L, 37L), male2005 = c(17L, 28L,
36L),
Thanks David and Peter!
I tried to improve my R-script to get closer to my goal.
I guess I have to use nls(), because later I want to work with
Levenberg-Marquardt-Algorithm and when I got it right, LM-Algorithm uses
least squares as well, fitdistr() instead uses Maximum Likelihoods. Anyway,
I
Thanks I didn't know about that. I ended up with something like this.
Is there a more elegant way to do it?
myplots - list();
hasplots - FALSE;
setHook(before.plot.new, function(...) {
if(hasplots == FALSE){
hasplots - TRUE;
} else {
myplots[[length(myplots)+1]] -
Hello R-people!
I have a general statistical question about regressions. I just want
to describe my case:
I have got a dataset of around 150 observations and 1 dependent and 2
independent variables.
The dependent variable is of metric nature (in my case meters in a
range from around
On Sat, Jul 16, 2011 at 9:35 PM, Peter Maclean pmaclean2...@yahoo.com wrote:
I am estimating Value at Risk using PerfomanceAnalytics package.
The variables are stored in a data frame. I formated the data variables using
zoo() and as.xtx() but it is not working. The working example is below.
On Jul 11, 2011, at 22:08 , Greg Snow wrote:
Note the warning on the help page for recordPlot. If your colleague is using
a different version of R than you there could be problems.
Also notice that the process isn't perfect, as people occasionally discover
when they use dev.copy() and its
Hello everyone,
I have the following x variable:
DATE-as.Date(as.character(data12$DATE),%m/%d/%Y)
This works fine, however, I want to plot multiple years of data on separate
plots by month. All years have all 12 months data except 2011; HOWEVER I
need the x axis limits to remain the same. I
Hi:
The default time origin in R is '1970-01-01'.
as.Date('1999-12-02') - as.Date('1970-01-01')
Time difference of 10927 days
Therefore, the difference in days between Matlab's and R's origins is
10927 - 730456
[1] -719529
as.Date(-719529, origin = '1970-01-01')
[1] 000/-12-31
If we try this
I really sorry if I understood your statement correctly :(
You said:
To put a backslash in the replacement expression of sub or gsub
(when fixed=FALSE) use 4 backslashes
I understood it is okay if I want to replace something with 2
backslashes. what if I want to replace that with just 1
On 11-07-17 8:22 AM, Dan Abner wrote:
Hello everyone,
I have the following x variable:
DATE-as.Date(as.character(data12$DATE),%m/%d/%Y)
This works fine, however, I want to plot multiple years of data on separate
plots by month. All years have all 12 months data except 2011; HOWEVER I
need the
On 17.07.2011 15:18, Nipesh Bajaj wrote:
I really sorry if I understood your statement correctly :(
You said:
To put a backslash in the replacement expression of sub or gsub
(when fixed=FALSE) use 4 backslashes
I understood it is okay if I want to replace something with 2
backslashes. what
I'd write to the Amelia maintainer. Maybe it is a bug in the package, or
just an inappropriate error message (where the error should have been
caught earlier).
Uwe Ligges
On 14.07.2011 20:09, Eric Miner wrote:
I am fairly new at using R/programming in general so I apologize if I am
leaving
On 13.07.2011 23:00, Joshua Gubler wrote:
Hi all,
I have used the apsrtable package to generate tables (using LaTeX and Sweave)
for quite some. However, suddenly the notes option in the package appears to
have stopped working (or I am doing something wrong). So, when I try to run
the
Johannes:
R is not a statistical tutorial service, although kind and able
helpeRs sometimes do reply to such queries. You should try such a
service, for example:
http://stackoverflow.com/
FWIW, this is an example of censoring in regression. R has packages
for this, but you need to learn more or
On 15.07.2011 19:01, ScottM wrote:
Hello all,
I'm relatively new to R and programming in general - I had previously used
MatLab, but decided to make the transition to R, as the computational times
are much better!
Anyway, I'm trying to use R to run a gamma distribution model to estimate
mean
You could treat the dependent variable as a nominal variable. And scale the
indepent variables to have a Mean:0 and StDev:1. Stick all these in a
multinomial regression package such as mlogit. Or a non -parametric method
such as randomForest.
--
View this message in context:
The par(new = TRUE)-Trick seems to be fine, if you just want to add one or
two other lines.
Whenever I plot a couple of graphs ontop of each other i find that the nice
vectorgraphics become cluttered.
Is there any way to use plotmeans() and to restrict it to the line only?
Sincerly
Henning
--
On 15.07.2011 13:50, Zd Gibbs wrote:
Hi All, I have been requested to come up with an out-of-sample prediction
einterval and point estimate. I have never done this and I am hoping for help
from you all.
First can R do this?
If so, what are the steps?
What do I need?
Sounds like you should
- Please also reply to the original poster who may not be subscribed to
the list.
- Please cite the original question (and other relevant parts of the
thread) since some readers of this list will delete messages before an
answer arrives.
Uwe Ligges
On 17.07.2011 15:22, saskay wrote:
You
On 17.07.2011 14:36, h_bumann wrote:
The par(new = TRUE)-Trick seems to be fine, if you just want to add one or
two other lines.
Whenever I plot a couple of graphs ontop of each other i find that the nice
vectorgraphics become cluttered.
Is there any way to use plotmeans() and to restrict it
On 14.07.2011 23:03, Anil acharya wrote:
Hi
I am posting in the topic related to the non-numeric argument to binary
operator as I got similar problem while running the netcdf code. I have
attached the file with this post. It is a climate data from NOAA site. The code
follows as:
On 14.07.2011 19:59, donvolencia wrote:
Hi all,
I have just begun to use R and am hoping to receive some advice about the
problem I need to solve. I have a file containing xy points that I need to
find all significant clusters and write each of their xy coordinates to
file(total points ~
Perhaps you want to ask on the SIG mailing list for spatial statistics.
I guess there is something available that makes use of cloud() [from
lattice], plot3d [from rgl] or scatterplot3d() [identically names
package] or others to satisfy your needs of plotting points atop of a
map, for example.
On Jul 17, 2011, at 4:12 AM, Jochen1980 wrote:
Thanks David and Peter!
I tried to improve my R-script to get closer to my goal.
I guess I have to use nls(), because later I want to work with
Levenberg-Marquardt-Algorithm and when I got it right, LM-Algorithm
uses
least squares as well,
Dear R-list members,
I would like to report some problems, in R 2.13.1, with the function
hist(), which constructs histograms.
Sample code:
x - rnorm(1000)
hist(x, breaks=16)
In R 2.13.1, in the histogram so produced, the bottom of the bars
are not exactly aligned in a perfect horizontal,
df is a very large data frame with arrival estimates for many flights
(DF$flightfact) at random times (df$PredTime). The error of the estimate
is df$dt.
My problem is that I want to know the prediction error at each minute
before landing. This code works, but is very slow, and dominates
Hi, I am searching for a script or set of scripts written in R that
perform Quantitative Microbial Risk Assessment (QMRA) computations for
risk of infection from pathogens in recreational waters. My initial
searches have come up empty. If there are other places I should look for
this or email
Sorry for not including those details. Here is a more detailed description:
data.frame(animals=c(dog,wolf,cat))-my.data
gsub(o,\u0254,my.data$animals)-my.data$animals
my.data$animals
[1] dɔg wɔlf cat
sessionInfo()
R version 2.13.1 (2011-07-08)
Platform: i386-pc-mingw32/i386 (32-bit)
Hi there dear R users,
By using the code below that plots two distributions and calculates a bootstrap
hypothesis test of equality, similarity of these two distributions was
calculated. What I'm interested now is to find out a size of the area where
both distributions are overlapping. I know
On 11-07-17 1:17 PM, Paulo Barata wrote:
Dear R-list members,
I would like to report some problems, in R 2.13.1, with the function
hist(), which constructs histograms.
Sample code:
x- rnorm(1000)
hist(x, breaks=16)
In R 2.13.1, in the histogram so produced, the bottom of the bars
are not
It would be nice if you had some sample data included so that we could
see how the code worked. Have you use Rprof on the code to see where
you are spending your time? You might want to use 'matrix' instead of
'data.frames' since there is a big performance impact with dataframes
when indexing.
Dear List,
I used survival package of R and was able to generate KM plots. Now I need
help in finding the top curve in a non-visualization way. I mean to say that
I would like to know the order of curves through any existing methods rather
than looking at the KM plots. I need this information as
Dear R-list members,
I would like to report some problems, in R 2.13.1, with the function
hist(), which constructs histograms.
Sample code:
x - rnorm(1000)
hist(x, breaks=16)
In R 2.13.1, in the histogram so produced, the bottom of the bars
are not exactly aligned in a perfect horizontal,
Hi all,
I have an interval (e.g [-3,3]) and a numeric vector z (-1.4,0.5,4.7). How
can I test whether an element in z lies between between -3,3? I particularly
need a TRUE/FALSE response.
Thanks
Manuel
--
View this message in context:
I have just joined this list (and just started using R), so please
excuse any etiquette breaches as I do not yet have a feel for how the
list operates.
I am in the process of teaching myself statistics using R as my utility
as my ultimate goals cannot be satisfied by Excel or any of the
dear all,
a simple question, I need to write a function in which I retrive an
undetermined number of vector in the function.
I solved the problem thanks to this link
http://stackoverflow.com/questions/2540232/how-to-allow-multiple-inputs-from-user-using-r
so my function will look alike this
I have a data frame (gom) or a matrix of trace metal data and some other
observations from water column samples taken at sea (e.g., 19 samples
(rows), 19 variables)
I can calc. the rank individually from each column of the attached object.
How can I create a matrix that contains the ranked data
try this:
z - c(-1.4,0.5,4.7)
(z = -3) (z = 3)
[1] TRUE TRUE FALSE
On Sun, Jul 17, 2011 at 10:54 AM, Manuel K. b8220...@klzlk.com wrote:
Hi all,
I have an interval (e.g [-3,3]) and a numeric vector z (-1.4,0.5,4.7). How
can I test whether an element in z lies between between -3,3? I
On Jul 17, 2011, at 11:07 AM, David Winsemius wrote:
On Jul 17, 2011, at 4:12 AM, Jochen1980 wrote:
Thanks David and Peter!
I tried to improve my R-script to get closer to my goal.
I guess I have to use nls(), because later I want to work with
Levenberg-Marquardt-Algorithm and when I got
On Jul 17, 2011, at 11:27 AM, Kishor Tappita wrote:
Dear List,
I used survival package of R and was able to generate KM plots. Now
I need
help in finding the top curve in a non-visualization way. I mean to
say that
I would like to know the order of curves through any existing
methods
Hi John,
Look at ?match.call you will need to make sure that expand.dots =
TRUE, but I think it does what you want. The names you want might be:
names(my_list) - as.character(match.call(expand.dots = TRUE))[-1L]
I will point out that I am not sure you will get the desired results
with your
Also check out the 'cut' and 'findIntervals' functions if you need to
check for many intervals at once.
On Sunday, July 17, 2011, jim holtman jholt...@gmail.com wrote:
try this:
z - c(-1.4,0.5,4.7)
(z = -3) (z = 3)
[1] TRUE TRUE FALSE
On Sun, Jul 17, 2011 at 10:54 AM, Manuel K.
On Jul 17, 2011, at 5:29 PM, jim holtman wrote:
try this:
z - c(-1.4,0.5,4.7)
(z = -3) (z = 3)
[1] TRUE TRUE FALSE
Another way:
findInterval(z, c(-3,3)) == 1
[1] TRUE TRUE FALSE
z=c(-50,-1.4,0.5,4.7)
And just to prove to myself that it behaves as I expect with values
below
On Sun, Jul 17, 2011 at 11:18 AM, christian krahforst
ckrahfo...@gmail.com wrote:
I have a data frame (gom) or a matrix of trace metal data and some other
observations from water column samples taken at sea (e.g., 19 samples
(rows), 19 variables)
I can calc. the rank individually from each
On 17-Jul-11 16:27:25, Jack Sofsky wrote:
I have just joined this list (and just started using R), so please
excuse any etiquette breaches as I do not yet have a feel for how the
list operates.
I am in the process of teaching myself statistics using R as my utility
as my ultimate goals
Dear R Beginners, Experts and Users,
I already asked this question on the R-sig-DB mailing list, but sadly
didn't got a reply :-( .
Keeping fingers crossed that I have more luck on the main R mailinglist :) .
I'm trying to connect to a MySQl database via Gnu R using the RODBC
interface. The
Did you register the database with ODBC?
Dennis
On Sun, Jul 17, 2011 at 3:46 PM, Henri Mone henrim...@gmail.com wrote:
Dear R Beginners, Experts and Users,
I already asked this question on the R-sig-DB mailing list, but sadly
didn't got a reply :-( .
Keeping fingers crossed that I have more
Hi Dennis,
Yes the database is registered with ODBC. I posted a copy of my ODBC
setup ~/.odbc.ini.
Still looking for the problem
Thanks,
Henri
On Mon, Jul 18, 2011 at 12:51 AM, Dennis Murphy djmu...@gmail.com wrote:
Did you register the database with ODBC?
Dennis
On Sun, Jul 17, 2011
I thought I had included the data... Here it is again.
What I want to do is to make box and whisker plots with each flight
counted the same number of times in each time bin. Hence the
interpolation to minute time hacks.
On 7/17/2011 4:16 PM, jim holtman wrote:
It would be nice if you had some
The Minitab and TI results are (modulo different levels of rounding)
just what you'd get from
doing the problem ``by hand'' in the good old-fashioned way. :-)
The Excel result appears to be the same with an excessive level of rounding.
The ``by-hand'' procedure uses the plug-in method to
Dear Help-list, How can I do a multiple comparison test (mct) on selected
contrasts from a linear model while using packages lme4 and multcomp? I am
running R 2.13.0 under Windows 7. The following linear model and mct produces
a global mct of 15 paired contrasts of the combined (Site,
Hi David,
my first attempt to work through your code was successful, my predicted line
is pretty close to the ecdf-function. I have no idea why you inverted the
gumbel-function and what the advantage of this strategy is?
I interpret your (1:100/100)-trick like this: you build a sequence of 100
Hello,
@Bert: I didn't expect a full tutorial service but probably a hint of
the Masters of statistics ;)
Anyway I posted my question again on a special statistic
forum. Your hint about the censored regression: I don't think
that this is the case here. As so far as I understand it is there
the
All -
I'm having an issue with trying to plot a model derived from nls()
onto a simple plot. I have included a sample data set and the code
that I've been using.
year month day date location mileage cost gallon cpg
mpg x
2009 1 4 1/4/2009 BZN 124585 19.39
On Jul 17, 2011, at 8:13 PM, Jochen1980 wrote:
Hi David,
my first attempt to work through your code was successful, my
predicted line
is pretty close to the ecdf-function. I have no idea why you
inverted the
gumbel-function and what the advantage of this strategy is?
I was advised by a
Here is what I did; convert the data to a numeric matrix for faster
processing. You can convert back to a dataframe since you have the
indices into the levels for the flights and runways.
# read in data
source('/temp/df/df')
# convert to matrix
df.mat - cbind(pt = as.numeric(df$PredTime)
+
Hi all:
I have a question about linear mixed model.
my linear mixed model with randomized slope and intercept with interaction
of time and group(g1,g2,g3):
model- glmmPQL(log10(CD4) ~ time + factor(group)+ time:factor(group),
random = ~time|id)
What I get is only the main and interaction of time
Hi David,
PFB the details of my query. Request your help in getting this resolved.
# TESTING is my dataset with almost 40K rows. I am importing this dataset
from my local desktop
TESTING - read.table(/Users/madana/Desktop/testing.txt, header=FALSE,
sep=\t, na.strings=, dec=.,
On Jul 17, 2011, at 11:27 PM, Madana_Babu wrote:
Hi David,
PFB the details of my query. Request your help in getting this
resolved.
# TESTING is my dataset with almost 40K rows. I am importing this
dataset
from my local desktop
TESTING - read.table(/Users/madana/Desktop/testing.txt,
On Jul 17, 2011, at 11:47 PM, David Winsemius wrote:
On Jul 17, 2011, at 11:27 PM, Madana_Babu wrote:
Hi David,
PFB
What ever that TLA means
the details of my query. Request your help in getting this resolved.
# TESTING is my dataset with almost 40K rows.
A small dataset.
I am
Dear David,
Many thanks for your reply.
On Fri, Jul 15, 2011 at 5:24 PM, David Winsemius dwinsem...@comcast.netwrote:
On Jul 15, 2011, at 5:20 AM, Ashim Kapoor wrote:
Dear R helpers,
Please have a look at the following : -
Note : My goal is to find and replace all Inf's in a data
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