Hi all,
I'm interested in doing a dot plot where *both* the size and color (more
specifically, shade of grey) change with the associated value.
I've found examples online for ggplot2 where you can scale the size of the
dot with a value:
I can actually answer this!! I was trying to figure out how to use sapply
for a function I wrote with multiple arguments.
Suppose the function is called
FUN(a,b), where a is a number and b is a number
You can use mapply(FUN, a = VECTOR, b = VECTOR) where each vector is your
input arguments. It
Try running this and see if it does what you want. It just uses plain
old plot with the cex arg for size and the col arg for colour...
greyDots - function() {
# make up some data
x - runif(50, 0, 10)
y - runif(50, 0, 10)
valueMax - 100
value - sample(valueMax, 50)
# edit these to
This following works for me but I still favor the quick and dirty method
suggested originally by David.
options(scipen = 10)
x - seq(0,2, by = .01)
f - expression(5*cos(2*x)-2*x*sin(2*x))
D(f, 'x')
f.prime - function(x){
-(5 * (sin(2 * x) * 2) + (2 * sin(2 * x) + 2 * x * (cos(2 * x) *
Hi
without toy example it is rather complicated to check your function. So
only few remarks:
Instead of generating 1 random number inside a loop generate whole vector
of random numbers outside a loop and choose a number
Do not mix ifelse with if. ifelse is intended to work with whole vector.
Is there an equivalent package for mixed linear effects models developed using
the package nlme as there is for linear models?
Tschüß
Tony Meissner
Principal Scientist (Monitoring/Statistics)
Resource Monitoring
Science, Monitoring and Information Division
Department for Water
Imagine ©
*(ph)
Hi folks,
OS - Ubuntu 10.04
On R I create a datafile named data. I can evoke it on R with;
data
On R Commander
Data - Active data set - Select active data set - (data) OK
only one data set there data
- View data set
I can read it
- Edit data set
showing 25 rows of data. Clicking the box
To illustrate the second option I proposed,
library(ggplot2)
library(gridExtra)
category - paste(Geographical Category, 1:10)
grp1 - rnorm(10, mean=10, sd=10)
grp2 - rnorm(10, mean=20, sd=10)
grp3 - rnorm(10, mean=15, sd=10)
grp4 - rnorm(10, mean=12, sd=10)
mydat -
By default everything in an R workspace (data objects, functions etc)
will be stored in a file called .RData in the working directory.
Please see http://cran.r-project.org/doc/contrib/Paradis-rdebuts_en.pdf
Michael
__
R-help@r-project.org mailing list
On R I create a datafile named data. I can evoke it on R with;
data
On R Commander
Data - Active data set - Select active data set - (data) OK
only one data set there data
- View data set
I can read it
- Edit data set
showing 25 rows of data. Clicking the box shows a thick
- Original Message
From: Michael Bedward michael.bedw...@gmail.com
To: Stephen Liu sati...@yahoo.com
Cc: r-help@r-project.org
Sent: Thu, August 12, 2010 3:50:12 PM
Subject: Re: [R] Where the data file is stored?
By default everything in an R workspace (data objects, functions etc)
- Original Message
From: Philipp Pagel p.pa...@wzw.tum.de
To: r-help@r-project.org
Sent: Thu, August 12, 2010 3:54:53 PM
Subject: Re: [R] Where the data file is stored?
You dont't tell us what you did to create a datafile - to me it
sounds like you created an object (probably a data
Hi,
You can find your current working directory with the getwd() function.
Alain
On 12-Aug-10 11:22, Stephen Liu wrote:
- Original Message
From: Philipp Pagelp.pa...@wzw.tum.de
To: r-help@r-project.org
Sent: Thu, August 12, 2010 3:54:53 PM
Subject: Re: [R] Where the data file is
Dear all,
I have a few points that I am unsure about using scan. I know that it is
covered in the intro to R, and also has been discussed here:
http://www.mail-archive.com/r-help@r-project.org/msg04869.html
but nevertheless, I cannot get it to work.
I have a potentially very large matrix that
Hi all,
I have used this library to create a (360 day)calendar for my rainfall
data (which is divided over 9 gridcells):
## CODE##
library(udunits)
utInit()
calendar - att.get.nc(nc,'time','calendar')
T - var.get.nc(nc,time)
times.list - utCalendar(T,days since
Dear all,
I have a time series and I like to fit S(t) = T(t) + P(t)
T(t) = Linear Trend = a + bt
P(t) = Periodic Signal = sum(Ai*cos(wt) + Bi*sin(wt) ) for i=1,...,3.
Thanks in advance,
Best regards,
__
R-help@r-project.org mailing list
Hello all,
I would like to know what the difference is between chisq.test and
fisher.test when using the Monte Carlo method with simulate.p.value=TRUE?
Thank you
--
View this message in context:
On Aug 12, 2010, at 11:30 AM, Martin Tomko wrote:
c-scan(file=f,what=list(c(,(rep(integer(0),cols, skip=1)
m-matrix(c, nrow = rows, ncol=cols,byrow=TRUE);
for some reason I end up with a character matrix, which I don't want. Is this
the proper way to skip the first column (this is
Hello all,
I apply Repeated measures ANOVA for analyzing ERPs data. To do this I
use lme() function and glht() for the post hoc analysis.
I followed the very useful suggestions found in this help list but I
have a problem. I do not know how to test the interaction between
different levels of
Hi Peter,
thank you for your reply. I still cannot get it to work.
I have modified your code as follows:
rows-length(R)
cols - max(unlist(lapply(R,function(x) length(unlist(gregexpr(
,x,fixed=TRUE,useBytes=TRUE))
c-scan(file=f,what=rep(c(list(NULL),rep(list(0L),cols-1),rows-1)), skip=1)
- Original Message
From: Alain Guillet alain.guil...@uclouvain.be
To: Stephen Liu sati...@yahoo.com
Cc: r-help@r-project.org
Sent: Thu, August 12, 2010 5:28:32 PM
Subject: Re: [R] Where the data file is stored?
You can find your current working directory with the getwd() function.
Hi
On Thu, Aug 12, 2010 at 12:37 PM, Stephen Liu sati...@yahoo.com wrote:
- Original Message
From: Alain Guillet alain.guil...@uclouvain.be
To: Stephen Liu sati...@yahoo.com
Cc: r-help@r-project.org
Sent: Thu, August 12, 2010 5:28:32 PM
Subject: Re: [R] Where the data file is stored?
Not quite what I was trying to say. The process generates a random uniform
number between 0 and 1 and compares to a specific conditional probability.
It is looking for this in particular:
random number Pr( rain(station=i,day=d)=1 | rain(station=i,day=d-1)=0
rain(station=j,day=d)=0
Hi!
I want to use the DRC package in order to calculate the IC50 value of an
enzyme inhibition assay.
The problem is that the estimated ED50, is always out of the fitted curve.
In the example below, I had a ED50 value of 2.2896,
But when I predict the response level for this concentration I
Dear all,
I am using plot.circular(x, stack=TRUE) to plot a histogram from a list of
angle. I would also like to draw a line from the origin at the angle of the
mean (mean.circular), preferably with the resultant's length (rho.circular) as
length.
How do I achieve this on the circular plot,
We're using Rserve as a back-end in our application. A problem that we
experienced is that the Rserve connection gets stuck when R is asking for
some kind of user interaction. For example, when executing an
'install.packages' command for the first time without specifying the repos
argument, R
Hi,
I don't know if this can be useful to you, but I recently wrote a small
function to read a large datafile like yours in a number of steps, with the
possibility to save each intermediate block as .Rdata. This is based on
read.table --- not as efficient as lower-level scan() but it might be
You're not seeing the .Rdata file containing the data objects. Try:
list.files(getwd(),full.name=TRUE, all.files=TRUE)
Stephen Liu sati...@yahoo.com wrote in message
news:961426.85478...@web113203.mail.gq1.yahoo.com...
- Original Message
From: Alain Guillet alain.guil...@uclouvain.be
Hi baptiste,
thanks a lot. Could you please comment on that code, I cannto figure out
what it does. Appart from the file name, what parameters does it need?
Seems to me like you need to know the size of the table a priori. Is
that right? Do you have to set up the block size depending on that
Hello Nikhil, hope you are well today.
I am sorry to be a pain but I have one follow up question, I am trying to
express my results in a grid, which would look like a 6 by 6 matrix and would
have just Yes or NO in each grid..
So what I am thinking is a way to store every result I get on
On Aug 12, 2010, at 1:34 PM, Martin Tomko wrote:
Hi Peter,
thank you for your reply. I still cannot get it to work.
I have modified your code as follows:
rows-length(R)
cols - max(unlist(lapply(R,function(x) length(unlist(gregexpr(
,x,fixed=TRUE,useBytes=TRUE))
Notice that the above
The adfresd function that you have, prints the outcome of the test rather
than `return' ing a value. If you would modify that function to return a
value (True or false/ or p-value) you should automatically get vector that
you desire. Then is a simple task of naming the resultant vector with.
This command
cdmoutcome- glm(log(value)~factor(year)
+log(gdppcpppconst)+log(gdppcpppconstAII)
+log(co2eemisspc)+log(co2eemisspcAII)
+log(dist)
+fdiboth
+odapartnertohost
+corrupt
Hi useRs,
A reminder. The IEEE ICDM Contest: Road Traffic Prediction, will end in
24 days. Hurry up if you want to play with traffic data and solve the
problem of jams prediction. Prizes of $5,000 in total will be awarded
for the best solutions.
http://tunedit.org/challenge/IEEE-ICDM-2010
Hi Peter,
apologies, too fast copying and pasting.
So, here is the explanation:
f-C:/test/mytab.txt;
R-readLines(con=f);
where mytab.txt is a table formatted as noted in previous post (space
delimited, with header, rownames, containing integers).
Now, my understandign of scan was that I have
There appears to be a problem in both regressions, as a singularity is
also reported in the second regression analysis as well. It appears that
the litrate variable is considered a factor in the first analysis and
continuous in the second. There also appears to be collinearity between
the
Duncan and David, thank you so much.
You are right. We can use
z1 - outer(x, y, function(x,y) x^2+3*y^2)
rather than
xy - meshgrid(x,y)
z2 - xy$x^2+ 3*xy$y^2
to get right answer. I run these codes on my computer and found that z2 is
the transpose of z1.
So I guess in order to obtain
baptiste,
I have two more questions. How can I get the category labels right
justified? It seems that I need to change the size of the text too. Since in
my real data the text size are big and they appears to be even bigger than
my plot area. So, my second question is how can I change the text
Thanks for the help,
I tried to apply this to a vector with two columns, well I suppose it is not
a vector but for instance like this:
[,1] [,2]
[1,]1 2
[2,]2 3
[3,]1 2
[4,]1 2
[5,]3 4
and return a vector :
1,2,1,1,3, so that it recognises
I've tried lm, but something is wrong.
I've made a test dataset of 599 data points, my original equation is
zz = 1 +0.5*xx -3.2*xx*xx -1*yy +4.2*yy*yy
but the R gives this result:
---
mp - read.csv(file=sample.csv,sep=;,header=TRUE)
lm(zz ~
Hi guys,
I have a code in R and it was work well but when I decrease the epsilon
value (indicated in the code) , then I am getting this error
Error: evaluation nested too deeply: infinite recursion /
options(expressions=)?
any help please
y = 6.8;
w = 7.4;
z = 5.7;
muy = 7;
muw = 7;
muz =
right. How does it come that if I devide the result vector with
10*interception, I get a much better result?
zz2 - 25.86 -2239.86*mp$xx -595.01*mp$xx*mp$xx + 2875.54*mp$yy +
776.84*mp$yy*mp$yy
mp$zz2 - zz2
library(lattice)
cloud(zz2/258.6 + zz ~ xx * yy, data=mp)
looks quite pretty.
I have a simple dataset of a numerical dependent Y, a numerical independent X
and a categorial variable Z with three levels. I want to do linear
regression Y~X for each level of Z. How can I do this in a single command
that is without using lm() applied three isolated times?
--
View this message
Hi,
I'm newbie with R and don't really know how to add a median line to each
of the groups that is not all the plot long.
Here is a small working code that i have adapted for my purpose. If
somebody could tell me how to draw median lines on each group and not
all plot long.
ctl -
On Aug 12, 2010, at 10:35 AM, asdir wrote:
This command
cdmoutcome- glm(log(value)~factor(year)
+log(gdppcpppconst)+log(gdppcpppconstAII)
+log(co2eemisspc)+log(co2eemisspcAII)
+log(dist)
+fdiboth
+odapartnertohost
You're not seeing the .Rdata file containing the data objects. Try:
list.files(getwd(),full.name=TRUE, all.files=TRUE)
Hi Keith,
Thanks for your advice
On R console running
list.files(getwd(),full.name=TRUE,all.files=TRUE)
The output is similar to running following command on Ubuntu
I think I understand your question and the following would produce the result
you've posted.
(x - matrix(c(1, 2, 2, 3, 1, 2, 1, 2, 3, 4), nrow=5, byrow=TRUE))
On Aug 12, 2010, at 5:41 AM, clips10 wrote:
Thanks for the help,
I tried to apply this to a vector with two columns, well I suppose
?RsiteSearch
or consult package sos.
Learn how to use R's search resources!
--
Bert Gunter
Genentech Nonclinical Statistics
On Wed, Aug 11, 2010 at 8:21 PM, Geoffrey Smith g...@asu.edu wrote:
Hello, does anyone know how to compute the following two normality tests
using R:
(1) the
From: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org] On Behalf Of David martin
Sent: Thursday, August 12, 2010 7:42 AM
To: r-h...@stat.math.ethz.ch
Subject: [R] Median abline how-to ?
Hi,
I'm newbie with R and don't really know how to add a median
line to each
of
Hi Barry,
Following 2 commands are useful to me;
row.names(subset(file.info(list.files(getwd(),full.name=TRUE)),isdir))
showing directories.
row.names(subset(file.info(list.files(getwd(),full.name=TRUE)),!isdir))
showing files
What is ! for? TIA
B.R.
Stephen L
- Original Message
I would like to build an XML file from scratch using the XML package.
I would like to save the following vector in it:
1:10
Could someone help me by briefly outlining how I go about it ?
And maybe provide a few lines of code?
Thanks!
Mark
Mark Heckmann
thanks !!!
On 12/08/10 17:49, William Dunlap wrote:
segments(x0=ix-w, x1=ix+w, y0=mediansByGroup, col=ix)
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R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
On Aug 12, 2010, at 11:51 AM, Stephen Liu wrote:
Hi Barry,
Following 2 commands are useful to me;
row
.names(subset(file.info(list.files(getwd(),full.name=TRUE)),isdir))
showing directories.
row.names(subset(file.info(list.files(getwd(),full.name=TRUE)),!
isdir))
showing files
What is
On Thu, Aug 12, 2010 at 4:51 PM, Stephen Liu sati...@yahoo.com wrote:
Hi Barry,
Following 2 commands are useful to me;
row.names(subset(file.info(list.files(getwd(),full.name=TRUE)),isdir))
showing directories.
row.names(subset(file.info(list.files(getwd(),full.name=TRUE)),!isdir))
@JLucke:
As for the africa variable: I took it out of the model, so that we can
exclude this variable itself and collinearity between the africa and the
litrate variable as causes for the litrate-problem. This also removed the
singularity remark at the top. However, the problem with
Hello,
when I ran R CMD INSTALL circular_0.4.tar.gz on a machine with Debian stable,
the command also created the documentation in various fomrats (latex, html,
online).
The same command on Debian testing only provides the online documentation.
How can I persuade R to create the different
On 12/08/2010 10:35 AM, szisziszilvi wrote:
I've tried lm, but something is wrong.
I've made a test dataset of 599 data points, my original equation is
zz = 1 +0.5*xx -3.2*xx*xx -1*yy +4.2*yy*yy
but the R gives this result:
---
mp -
On 12/08/2010 12:10 PM, Tim Gruene wrote:
Hello,
when I ran R CMD INSTALL circular_0.4.tar.gz on a machine with Debian stable,
the command also created the documentation in various fomrats (latex, html,
online).
The same command on Debian testing only provides the online documentation.
How
Martin Tomko wrote:
Hi Peter,
apologies, too fast copying and pasting.
So, here is the explanation:
f-C:/test/mytab.txt;
R-readLines(con=f);
where mytab.txt is a table formatted as noted in previous post (space
delimited, with header, rownames, containing integers).
Now, my
Thanks everyone for your help and advice. For the R-help archives, here is
what I ended up doing.
First creating a separate function to handle one day at a time -
byrow.gen2 - function(genmat,rownum,use1,use2,num,ortho_obs_used){
prev = rownum-1
ran = runif(length(rownum),0,1)
I'm running r 2. on a mac running 10.6.4 and a dual-core macbook pro. I'm
having a funny time with multicore. When I run it with 2 cores, mclapply, R
borks with the following error.
The process has forked and you cannot use this CoreFoundation functionality
safely. You MUST exec().
Break on
Hi,
I know how to read fixed width format data with read.fwf, but suddenly I need
to read in a large number of old fwf files with 2 types of lines. Lines that
begin with 3 in first column carry one set of variables, and lines that begin
with 4 carry another set, like this:
…
On Thu, 12 Aug 2010, Jarrett Byrnes wrote:
I'm running r 2. on a mac running 10.6.4 and a dual-core macbook pro. I'm
having a funny time with multicore. When I run it with 2 cores, mclapply, R
borks with the following error.
The process has forked and you cannot use this CoreFoundation
Hi,
I am a brand new user and may be my question is too simple. I have R on
our (not Unix) server. I am trying to build a decision tree and the error
message says couldn't find function rpart. Does it mean I have to ask our
server guy to install an additional package?
Thank you,
Olga
I think your code will work but only for the two columns I gave. I used those
as an example but my actual data is 200 in length with two columns and I
need code that will give a label to each unique pair but still have the
original length for instance, one that will turn something such as
You can use following scriptI think
#create a vector of random numbers on which to test script
v-sample(1:3,size=90,replace=TRUE)
#creates two matrixes out of vector v which can be assigned to M to test
script
M2-matrix(v,ncol=2)
M3-matrix(v,ncol=3)
M- #Assign you're matrix or a
Ah. Indeed, this is from the glmulti. I had not realized there would be
problems using Java. Is there a way around this to still use a multicore
approach? For what other packages that use multiple cores will this not be a
problem?
-Jarrett
On Aug 12, 2010, at 11:02 AM, Thomas Lumley
Olga Shaganova wrote:
Hi,
I am a brand new user and may be my question is too simple. I have R on
our (not Unix) server. I am trying to build a decision tree and the error
message says couldn't find function rpart. Does it mean I have to ask our
server guy to install an additional package?
On Wed, Aug 11, 2010 at 10:14 PM, Brian Tsai btsa...@gmail.com wrote:
Hi all,
I'm interested in doing a dot plot where *both* the size and color (more
specifically, shade of grey) change with the associated value.
I've found examples online for ggplot2 where you can scale the size of the
Hi
I am trying to calculate the row sums of a matrix i have created
The matrix ( FeaturePresenceMatrix) has been created by
1) Read csv
2) Removing unnecesarry data using [-1:4,] command
3) replacing all the NA values with as.numeric(0) and all others with
as.numeric
(1)
When I carry out the
Amit Patel wrote:
Hi
I am trying to calculate the row sums of a matrix i have created
The matrix ( FeaturePresenceMatrix) has been created by
1) Read csv
2) Removing unnecesarry data using [-1:4,] command
3) replacing all the NA values with as.numeric(0) and all others with as.numeric
(1)
Yes, please do as Erik said in the future but here's one way to do it.
(A - matrix(data = rnorm(n = 9, mean = 0, sd = 1), nrow = 3, ncol = 3, byrow =
FALSE, dimnames = NULL))
matrix(rowSums(A))
On Aug 12, 2010, at 11:28 AM, Amit Patel wrote:
Hi
I am trying to calculate the row sums of a
Another suggestion: compare
-1:4
with
-(1:4)
-Peter Ehlers
On 2010-08-12 12:28, Amit Patel wrote:
Hi
I am trying to calculate the row sums of a matrix i have created
The matrix ( FeaturePresenceMatrix) has been created by
1) Read csv
2) Removing unnecesarry data using [-1:4,] command
hi Peter,
There's no single function for ICC with variable number of judges. To
estimate the variances in that case you need hierarchical linear modeling. I
posted code for this at Stackoverflow in answer to your question there:
http://stackoverflow.com/questions/3205176/
Stats questions get
Dear helpeRs,
I have a dataframe (14947 x 27) containing measurements collected
every 5 seconds at several different sampling locations. If one
measurement at a given location is less than zero on a given day, I
would like to delete all measurements from that location on that day.
Here is a
Hi how can print x-axis labels in 45 degree in boxplot() (or plot in general)?
I
can use las=2 to print in 90 degree, but it looks ugly. Is there a simple
option
to do 45 degree easily?
Thanks
John
__
R-help@r-project.org mailing list
On Aug 12, 2010, at 2:14 PM, array chip wrote:
Hi how can print x-axis labels in 45 degree in boxplot() (or plot in
general)? I
can use las=2 to print in 90 degree, but it looks ugly. Is there a simple
option
to do 45 degree easily?
Thanks
John
John,
See R FAQ 7.27 How can I
On Aug 12, 2010, at 3:11 PM, Toby Gass wrote:
Dear helpeRs,
I have a dataframe (14947 x 27) containing measurements collected
every 5 seconds at several different sampling locations. If one
measurement at a given location is less than zero on a given day, I
would like to delete all
Try this:
subset(toy, !rowSums(mapply(is.element, toy[c('CH', 'DAY')], subset(toy,
SLOPE 0, CH:DAY))) 1 | SLOPE 0)
On Thu, Aug 12, 2010 at 4:11 PM, Toby Gass tobyg...@warnercnr.colostate.edu
wrote:
Dear helpeRs,
I have a dataframe (14947 x 27) containing measurements collected
every 5
On Aug 12, 2010, at 2:11 PM, Toby Gass wrote:
Dear helpeRs,
I have a dataframe (14947 x 27) containing measurements collected
every 5 seconds at several different sampling locations. If one
measurement at a given location is less than zero on a given day, I
would like to delete all
On Aug 12, 2010, at 2:24 PM, Marc Schwartz wrote:
On Aug 12, 2010, at 2:11 PM, Toby Gass wrote:
Dear helpeRs,
I have a dataframe (14947 x 27) containing measurements collected
every 5 seconds at several different sampling locations. If one
measurement at a given location is less than
On Aug 12, 2010, at 3:17 PM, Marc Schwartz wrote:
On Aug 12, 2010, at 2:14 PM, array chip wrote:
Hi how can print x-axis labels in 45 degree in boxplot() (or plot
in general)? I
can use las=2 to print in 90 degree, but it looks ugly. Is there a
simple option
to do 45 degree easily?
Hi Olga,
not directly related to your question. We have also a server
installation and subsequently our IT department determines which version
and packages I can use on R.
A few days ago I have switched to R-portable. Works without any problems
from my USB stick on any locked-for-installation
I'd like to plot a point at the intersection of these two curves. Thanks
x - seq(.2, .3, by = .01)
f - function(x){
x*cos(x)-2*x**2+3*x-1
}
plot(x,f(x), type = l)
abline(h = 0)
__
R-help@r-project.org mailing list
Than you Marc.
John
- Original Message
From: Marc Schwartz marc_schwa...@me.com
To: array chip arrayprof...@yahoo.com
Cc: r-help@r-project.org
Sent: Thu, August 12, 2010 12:17:12 PM
Subject: Re: [R] x-axis label print in 45 degree
On Aug 12, 2010, at 2:14 PM, array chip wrote:
Hi
I searched with print x-axis label in 45 degree which didn't return useful
links. Apparently I used poor search keywords.
- Original Message
From: David Winsemius dwinsem...@comcast.net
To: Marc Schwartz marc_schwa...@me.com
Cc: array chip arrayprof...@yahoo.com;
Yes, I'm playing around with other things but the points() function is what I
was looking for. Thanks
On Aug 12, 2010, at 12:47 PM, David Winsemius wrote:
On Aug 12, 2010, at 3:43 PM, TGS wrote:
I'd like to plot a point at the intersection of these two curves. Thanks
x - seq(.2, .3, by =
Actually I spoke too soon David.
I'm looking for a function that will either tell me which point is the
intersection so that I'd be able to plot a point there.
Or, if I have to solve for the roots in the ways which were demonstrated
yesterday, then would I be able to specify what the
I did. Did not work. Did you try your code? The matrix did not result into
integer numbers as expected. MY approach resulted in a correct scan
result, at least.
M.
Martin Tomko wrote:
Hi Peter,
apologies, too fast copying and pasting.
So, here is the explanation:
f-C:/test/mytab.txt;
On Aug 12, 2010, at 3:54 PM, TGS wrote:
Actually I spoke too soon David.
I'm looking for a function that will either tell me which point is
the intersection so that I'd be able to plot a point there.
Or, if I have to solve for the roots in the ways which were
demonstrated yesterday,
I don't know if it's elegant enough for you, but you could split the file into
two files with 'grep ^3 file file_3' and 'grep ^4 file file_4'
and then read them in separately.
Tim
On Thu, Aug 12, 2010 at 01:57:19PM -0400, Denis Chabot wrote:
Hi,
I know how to read fixed width format data
I was meaning something like the following:
x - seq(.2, .3, by = .01)
f - function(x){
x*cos(x)-2*x**2+3*x-1
}
plot(x,f(x), type = l)
abline(h = -.1)
But I'm guessing uniroot will do this?---I haven't looked far into the
uniroot function to see if it will solve this.
On Aug 12, 2010,
Thank you all for the quick responses. So far as I've checked,
Marc's solution works perfectly and is quite speedy. I'm still
trying to figure out what it is doing. :)
Henrique's solution seems to need some columns somewhere. David's
solution does not find all the other measurements,
What did you try?
R CMD build (build all lower case) does work for me
Uwe Ligges
On 09.08.2010 22:30, Hintzen, Niels wrote:
Dear all,
As I couldn't find any thread on the internet I hope the help-list might help
me out.
I've tried to update Rtools from R210 used in combination with
Hi listers...
I am working o a scatterplot where I would like to plot the variables
according with another frequency variable.
Another friend here proposed this code...
x - rnorm(10)
y - rnorm(10)
ind - c(1,0,1,0,1,0,1,1,0,0)
plot(x, y, col = ind + 1, pch = 16) # 1 is black, 2 is red
But in my
Hi,
I want to eliminate an element of a list:
list - seq(1,5,1)
s - sample(list,1)
lets say s=3
Now I want to remove 3 from the list: list2 = {1,2,4,5}
Can someone give me a tip?
Thanks,
André
[[alternative HTML version deleted]]
__
Try this:
setdiff(list, s)
On Thu, Aug 12, 2010 at 5:32 PM, André de Boer rnie...@gmail.com wrote:
Hi,
I want to eliminate an element of a list:
list - seq(1,5,1)
s - sample(list,1)
lets say s=3
Now I want to remove 3 from the list: list2 = {1,2,4,5}
Can someone give me a tip?
On Aug 12, 2010, at 4:06 PM, Toby Gass wrote:
Thank you all for the quick responses. So far as I've checked,
Marc's solution works perfectly and is quite speedy. I'm still
trying to figure out what it is doing. :)
Henrique's solution seems to need some columns somewhere. David's
solution
Hi Marcio,
Your friend has given the answer.
x - rnorm(10)
y - rnorm(10)
ind - c(3,0,1,0,3,0,2,2,0,0)
plot(x, y, col = grey(0:max(ind)/max(ind))[ind], pch = 16)
Mestat wrote:
I am working o a scatterplot where I would like to plot the variables
according with another frequency
# just to clean it up for my own understanding, the difference approach as
you had suggested would be
x - seq(.2, .3, by = .1)
f1 - function(x){
x*cos(x)-2*x**2+3*x-1
}
plot(x,f1(x), type = l)
abline(h = -.1)
abline(v = x[which.min(abs(diff((f1(x) - (-.1))**2)))], lty = 'dotted')
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