Dear All,
I would like to to group the Ticker by Industry and create file names from
the
Industry Factor and export to a txt file.
I have tried the folowing
ind=finvizAllexETF$Industry
ind is then Aluminum Business Services Regional Airlines
ind2=gsub( ,,ind)
ind3
[1] Aluminum
Please, do not write Help, Help but give a meaningful header. Most people
here a specialized and check the posts they understand best.
Madhavi Bhave wrote:
(I have already written the required R code which is giving me correct
results for a given single set of data. I just wish to wish
Peter Rote wrote:
I would like to to group the Ticker by Industry and create file names from
the
Industry Factor and export to a txt file.
I have tried the folowing
ind=finvizAllexETF$Industry
ind is then Aluminum Business Services Regional Airlines
ind2=gsub(
Etienne,
I don't see the point in avoiding some 'special' packages. If you are
willing to change your mind in this regard, try something like
library()
-Original Message-
From: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org] On Behalf Of Etienne Stockhausen
Sorry, wrong button. Below a hopefully more helpful solution...
Etienne,
I don't see the point in avoiding some 'special' packages. If you are
willing to change your mind in this regard, try one of the following
solutions that work for me:
library(combinat)
apply(mat, 2, function(x)
On Tue, 2010-01-19 at 16:27 +, Gavin Simpson wrote:
Dear List,
A student in the Department where I work would like to produce a graphic
similar to this one:
http://image.guardian.co.uk/sys-files/Guardian/documents/2009/09/16/Public_spending_160909.pdf
Does anyone know if the figure
Hi r-users,
I have the following code to solve 4 simultaneous eqns with 4 unknowns using
newton iteration method. But I got the error message:
pars - c(1.15, 40, 50, 0.78)
newton.input2 - function(pars)
{ ## parameters to estimate
alp - pars[1]
b1 - pars[2]
b2 - pars[3]
Hello!
I have a question on uniform distribution.
I want to plot n, say 20, points, uniformly distributed, in a circle, with
radius=0.1 and center,say, (0.4, 0.8)
I do not know how~
Thank you for your time.
Yours
Wolfgang Amadeus
[[alternative HTML version deleted]]
Hello
On Wed, Jan 20, 2010 at 7:00 AM, Madhavi Bhave madhavi_bh...@yahoo.com wrote:
Sir, I am not asking for the modification of existing code as it is running
fine with a single set of data (and I have checked that the output tallies
with other methods). I just want to use this code for
I've a table containing two columns
seperated by space, as shown below.
S:C 2.011085038928
S:A 21.496800549900762
S:J 0.183181039138149
P:E 9.641984304606304
I'm reading this table inside a loop
but unable to access the first column
Hi all I have installed quantmod package but when I try to obtain GOOG data
appers this message: Can anyone inform why itappears?
I type getSymbols(GOOG,src=google)
Thanks and Best Regards for all
Error en download.file(paste(google.URL, q=, Symbols.name, startdate=,
:
no fue posible abrir la
Hi,
You didnt put the parameter sep=' ' in the read.table.
If you try to see str(file), i think you are going to see only one column.
read.table(data.txt,header=FALSE,sep=' ')
Atenciosamente,
Leandro Lins Marino
Centro de Avaliação
Fundação CESGRANRIO
Rua Santa Alexandrina, 1011 - 2º andar
Rio
R does vectorized arithmetic. Your loop was superfluous. Observe:
x - read.table(textConnection(
+ S:C 2.011085038928
+ S:A 21.496800549900762
+ S:J 0.183181039138149
+ P:E 9.641984304606304))
x
V1 V2
1 S:C 2.0110850
2 S:A 21.4968005
3 S:J 0.1831810
4 P:E 9.6419843
transform(x,
ang-runif(20,0,2*pi)
x-cos(ang) +0.4
y-sin(ang) +0.8
plot(x,y,ylim=c(-3,3),xlim=c(-3,3))
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Wolfgang Amadeus
Sent: Wednesday, January 20, 2010 5:26 AM
To: r-help
Subject: [R] question on
Hi Leonard,
Thanks for your quick response.
i tired using the sep= in read.table() function, but still can't access the
first column.
checked , str(file), showing the $V1 variable...but can't access the first
column inside the loop.
regards
Gaurav Kumar
www.gauravkumar.org
PhD Student,
Oops forgot to include the radius
ang-runif(200,0,2*pi)
x-cos(ang)*.1 +0.4
y-sin(ang)*.1 +0.8
plot(x,y)
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Wolfgang Amadeus
Sent: Wednesday, January 20, 2010 5:26 AM
To: r-help
Subject:
Thank Dennis!!!.it works
regards
Gaurav Kumar
www.gauravkumar.org
PhD Student, Chemistry and Biomolecular Sciences, Macquarie , Sydney, Australia.
MS (Computational Biology), NCBS-TIFR, Bangalore, India.
--- On Wed, 20/1/10, Dennis Murphy djmu...@gmail.com wrote:
From: Dennis Murphy
Hi,
I have finally managed to set up a basic diagram for my needs and would now
like to make that a bit nicer but dont find the appropiate answers through
wiki or the manual.
First of all i have a plot for the X and Y axis - and would like to show the
ticks in 2 steps, not in 5 steps, but dont
Alfredo's solution will provide n (=200 in his case) points
uniformly distributed along the *circumference* of the circle.
Wolfgang_Amadeus(!), you wanted them *in* a circle.
If what you mean is uniformly distributed over the area within
the circle, then you also need to generate the radii at
Hi,
I've this dataframes:
data1
12 3 4 56
50.4963017 0 0 0.2481509 1.9852069 0.4963017
10 0.000 0 0 0.000 0.6317266 0.000
15 0.000 0 0 0.000 0.000 0.000
20 0.000 0 0 3.3955301 0.000
Try this:
m - merge(data1, data2, by = 0, all = TRUE, sort = FALSE)
sapply(unique(union(names(data1), names(data2))), function(n)Reduce('+',
m[grep(n, names(m))]))
On Wed, Jan 20, 2010 at 10:59 AM, Alfredo Alessandrini
alfreal...@gmail.com wrote:
Hi,
I've this dataframes:
data1
The user wrote in their first post :
I have a lot of observations in my dataset
Heres one way to do it with a data.table :
a=data.table(a)
ans = a[ , list(dt=dt[dt-min(dt)7]) , by=var1,var2,var3]
class(ans$dt) = Date
Timings are below comparing the 3 methods. In this
Note that in the documentaton ?[.data.table where I say that 'by' is slow,
I mean relative to how fast it could be. Its seems, in this specific
example anyway, and with the code posted so far, to be significantly faster
than sqldf and plyr.
Of course the best of both worlds would be to use
On Jan 19, 2010, at 10:04 PM, Eric Ma wrote:
I am having the exact same error, and the suggested work-around does not
help.
My env is:
64-bit SUSE Linux v 10.1 on Xeon processors. gcc and gfortran version is
4.1.2
64-bit R 2.10.1 compiled from soure using CC=gcc -m64 etc.
64-bit
Can you ping the host 'finance.google.com' ??
The error msg is right there, Unable to Resolve
-c
I just tried it now:
getSymbols(GOOG, src=google)
[1] GOOG
head(GOOG)
GOOG.Open GOOG.High GOOG.Low GOOG.Close GOOG.Volume
2007-01-03466.00476.66 461.11 467.59 7527500
When I look at this line:
f4 - pars[1]*pars[2]*pars[3](pars[1]+pars[4])
It appears to my wet-brain interpreter that you are trying to apply a
function par[3] to the argument (pars[1]+pars[4])
When I insert a *, I then get output but I don't know if it is
correct.
newton.input2(pars)
Hello all
My computer is MacBook and I want to draw a plot in R, for example for
x - c(1,3,6,9,12)
y - c(1.5,2,7,8,15)
I use this command plot(x,y) to do but I see this massage:
Erreur dans plot(Di, g) :
argument(s) inutilisé(s) (c(1, 2, 3), c(12, 23, 34))
Could you please help me?
Thank
Sounds like a good idea. Would it be possible to give an example of how to
combine plyr with data.table, and why that is better than a data.table only
solution ?
hadley wickham h.wick...@gmail.com wrote in message
news:f8e6ff051001200624r2175e38xf558dc8fa3fb6...@mail.gmail.com...
Note that in
You have redefined the plot function somewhere. Check your saved
environment.
mario
On 20-Jan-10 15:36, khaz...@ceremade.dauphine.fr wrote:
Hello all
My computer is MacBook and I want to draw a plot in R, for example for
x- c(1,3,6,9,12)
y- c(1.5,2,7,8,15)
I use
Hello,
I am trying to manually choose the width (i.e. number of character
allowed) of columns containing text when creating dbf. files using write.dbf
(library foreign).
In particular, I want to define this width when the column have to contain
text but is empty.
By default, write.dbf give a
Hello,
I was able to bootstrap 1000 times, 20 times (20 columns of 1000 values). Is
there a way to get mean for each column?
Thanks!
Aaron
Date: Sat, 16 Jan 2010 00:04:59 +0100
From: stephan.kola...@gmx.de
To: aaron.fo...@students.tamuk.edu
CC: r-help@r-project.org
Roslina Zakaria wrote:
I have the following code to solve 4 simultaneous eqns with 4 unknowns
using newton iteration method. But I got the error message:
pars - c(1.15, 40, 50, 0.78)
newton.input2 - function(pars)
{ ## parameters to estimate
...
f1 - pars[1]*pars[2]
f2 -
I got the following warnings when I install R.oo. Are these warnings
normal? Should I reinstall the package as mentioned in the warnings?
How to reinstall? The sessionInfo() is at the end.
install.packages(R.oo, dependencies=T)
Warning in install.packages(R.oo, dependencies = T) :
argument
Hi,
I am stuck in one problem when doing nonmetric multidimensional scaling. I use
the function 'metaMDS' in the package 'vegan' to work on the presence/absence
community data. The problem is when two samples are identical (dissimilarity =
0), metaMDS cannot work with zero dissimilarity.
I
Apparently, Bengtsson.pdf is not current any more. For example, the
installation method is outdated. R.oo.pdf is a reference manual rather
than a tutorial. I'm wondering if there is a better and more current
R.oo tutorial so that I can quickly get started on R.oo.
On Jan 20, 2010, at 10:23 AM, aaron.fo...@students.tamuk.edu aaron.fo...@students.tamuk.edu
wrote:
Hello,
I was able to bootstrap 1000 times, 20 times (20 columns of 1000
values). Is there a way to get mean for each column?
?colMeans
Thanks!
Aaron
Date: Sat, 16 Jan
Hi,
Is there a way to stack bars in a barchart as well as beside bars for the
same treatment? eg
I have one barchart like this:
bio-matrix(c(10,23,9,25),nrow=2,byrow=T)
ntreat-c(n0,n96)
colnames(bio)-ntreat
barplot(bio,beside=T)
now i want a similar barchart but with stacked bars:
Dear list,
I have somewhat a small problem with a plot in R. I want to produce a plot
with the two axis intersection being the exact values of ( 5,6).
The code i am using is
Dear all,
Lets say I have several data frames as follows:
set.seed(42)
dates - as.Date(c(2010-01-19, 2010-01-20))
times - c(09:30:00, 11:30:00, 13:30:00, 15:30:00)
shows - c(Red Dwarf, Being Human, Doctor Who)
df1 - data.frame(Date = dates[1], Time = times[1], Show = shows, Score = 1:3)
try this:
plot(BN, xlim=c(5,10),ylim=c(6,14), xaxs = i, yaxs = i)
I hope it helps.
Best,
Dimitris
jpardila wrote:
Dear list,
I have somewhat a small problem with a plot in R. I want to produce a plot
with the two axis intersection being the exact values of ( 5,6).
The code i am using is
On Jan 20, 2010, at 6:42 AM, jpardila wrote:
Dear list,
I have somewhat a small problem with a plot in R. I want to produce
a plot
with the two axis intersection being the exact values of ( 5,6).
The code i am using is
BN-
c
Thanks to all who responded.
I've learn a lot from your comments. The best solution I can see is to use both
'missing()' and
'match.arg()' in my function.
Héctor
Hi,
I'm sure this one is very easy
I am trying to write a function where one of its arguments has two posible
(strings)
On Wed, Jan 20, 2010 at 8:43 AM, Matthew Dowle mdo...@mdowle.plus.com wrote:
Sounds like a good idea. Would it be possible to give an example of how to
combine plyr with data.table, and why that is better than a data.table only
solution ?
Well, ideally, you'd do:
adt - data.table(a)
ans2 -
Try with tapply:
with(do.call(rbind, df.list), tapply(Score, list(Date, Time, Show), length))
On Wed, Jan 20, 2010 at 10:20 AM, Tony B tony.bre...@googlemail.com wrote:
Dear all,
Lets say I have several data frames as follows:
set.seed(42)
dates - as.Date(c(2010-01-19, 2010-01-20))
times
...original data and design were posted at r-forge vegan forum today.
greetings,
kay
--
View this message in context:
http://n4.nabble.com/restricted-permutations-in-permtest-tp1017422p1018531.html
Sent from the R help mailing list archive at Nabble.com.
Dear R users,
I'm making multiple plots within the same pdf page (par(mfcol = c(5,1)), and
want a legend for this at the bottom of all the plots. From previous mails
it has been suggested to use par(xpd=TRUE), increase the margin at the last
plot, and then draw the legend. However, when I do this,
Folks,
I've got a matrix x as follows:
x - matrix(c(1,2,3,5,3,4,3,2,1), ncol = 3, byrow = TRUE)
x
[,1] [,2] [,3]
[1,]123
[2,]534
[3,]321
In each row of x, I want to replace the minimum value by -1, the maximum
value by +1 and all other values by 0.
I see now, thanks for explaining that. Would it be for you to add data.table
methods to ddply then, for this to happen? Or does a ddply method need to
be added to data.table?
hadley wickham h.wick...@gmail.com wrote in message
Hi,
could someone help me with dilemma on the simulation of logistic
regressiondata with multicollinearity effect and high leverage point..
Thank you
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
Hi
I'm hoping someone can help me I am a relative newbie to R.
I have data that is in a similar format to this...
Experiment Score1 Score2
X -0.85 -0.02
X -1.21 -0.02
X 1.05 0.09
Y -1.12 -0.07
Y -0.27 -0.07
Y -0.93 -0.08
Z 1.1 -0.03
Z 2.4 0.09
Z -1.0 0.09
Now I can easily have a look at the
Thanks so much for your assistance. In fact I did not read the documentation
carefully.
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Sent from the R help mailing list archive at Nabble.com.
Hi,
See my example below.
a-array(1:12,c(2,3,2))
a
, , 1
[,1] [,2] [,3]
[1,]135
[2,]246
, , 2
[,1] [,2] [,3]
[1,]79 11
[2,]8 10 12
I want to get a result something like
dim1 dim2 dim3 elements
111
121
1
Thank you for taking the time to reply Henrique. Although your
solution does take away the zeroes and replaces them with NA's (which
i prefer), it unfortunately seems to reduce all of the other scores to
just '1':
x - with(do.call(rbind, df.list), tapply(Score, list(Date, Show, Time),
length))
You can use this instead:
with(do.call(rbind, df.list), tapply(Score, list(Date, Show, Time), invisible))
On Wed, Jan 20, 2010 at 3:02 PM, Tony B tony.bre...@googlemail.com wrote:
Thank you for taking the time to reply Henrique. Although your
solution does take away the zeroes and replaces
On Jan 20, 2010, at 11:07 AM, BioStudent wrote:
Hi
I'm hoping someone can help me I am a relative newbie to R.
I have data that is in a similar format to this...
Experiment Score1 Score2
X -0.85 -0.02
X -1.21 -0.02
X 1.05 0.09
Y -1.12 -0.07
Y -0.27 -0.07
Y -0.93 -0.08
Z 1.1 -0.03
Z 2.4
On Tue, 19 Jan 2010, jshort wrote:
Hi
How does one tell R that one is using an augmented matrix as appose to an
non-augmented matrix?
Explicitly creating an augmented matrix is unnecessary and
awkward if the object is to solve a system of linear equations.
See:
?solve
HTH,
Try this:
sapply(as.data.frame.table(a), as.numeric)
On Wed, Jan 20, 2010 at 3:18 PM, rusers.sh rusers...@gmail.com wrote:
Hi,
See my example below.
a-array(1:12,c(2,3,2))
a
, , 1
[,1] [,2] [,3]
[1,] 1 3 5
[2,] 2 4 6
, , 2
[,1] [,2] [,3]
[1,] 7 9
Hello,
I would like to juxtapose two lattice graphs with common X axes such that the X
axes line up. I am using plot right now but the edges are not neat and it
would be nice if I could just draw 1 X axis and not both of them.
Here is my code:
Hi,
try c.trellis() from the latticeExtra package.
HTH,
baptiste
2010/1/20 George Chen glc...@stanford.edu:
Hello,
I would like to juxtapose two lattice graphs with common X axes such that the
X axes line up. I am using plot right now but the edges are not neat and it
would be nice if
Try googling latticeExtra x.same for some examples. Here's one:
http://www.mail-archive.com/r-help@r-project.org/msg39048.html
On Wed, Jan 20, 2010 at 9:44 AM, George Chen glc...@stanford.edu wrote:
Hello,
I would like to juxtapose two lattice graphs with common X axes such that
the X axes
You'll probably want to look at the 'by' function
d=data.frame(sex=rep(1:2,50),x=rnorm(100))
d$y=d$x+rnorm(100)
head(d)
cor(d)
by(d[,-1],d['sex'],function(df)cor(df))
You might also want to look at the doBy package
--
View this message in context:
I am trying to generate a line graph with quarterly time buckets (with
nice labels) on the x-axis. The first block of code below will
generate the graph with nicely formatted x-axis labels, but the
type= and col= options are not recognized when factors are used
for the x-axis.
The second block,
Try the zoo package:
plot(as.yearqtr(time.val), inc, col = 'red', type = 'l')
On Wed, Jan 20, 2010 at 3:41 PM, mah harwood...@gmail.com wrote:
I am trying to generate a line graph with quarterly time buckets (with
nice labels) on the x-axis. The first block of code below will
generate the
Meyners,Michael,LAUSANNE,AppliedMathematics schrieb:
Sorry, wrong button. Below a hopefully more helpful solution...
Etienne,
I don't see the point in avoiding some 'special' packages. If you are
willing to change your mind in this regard, try one of the following
solutions that work for me:
On Wed, 2010-01-20 at 23:28 +0800, Bingzhang Chen wrote:
Hi,
I am stuck in one problem when doing nonmetric multidimensional
scaling. I use the function 'metaMDS' in the package 'vegan' to work
on the presence/absence community data. The problem is when two
samples are identical
I have an instance where I need to include Greek letters on a plot title that
is multiple lines.
I've searched the forums for an approach to do this, but most of the previous
posts and replies seem to just address instances of single line examples and
problems:, e.g.
Ted, Alfredo, et al: Please stop doing this expletive deleted
jerk's homework for him!!!
cheers,
Rolf Turner
##
Attention:\ This e-mail message is privileged and confid...{{dropped:9}}
Got it.
Thanks
CC: r-help@r-project.org
From: dwinsem...@comcast.net
To: aaron.fo...@students.tamuk.edu
Subject: Re: [R] bootstrapping
Date: Wed, 20 Jan 2010 10:34:20 -0500
On Jan 20, 2010, at 10:23 AM, aaron.fo...@students.tamuk.edu
aaron.fo...@students.tamuk.edu
wrote:
On 21/01/2010, at 4:41 AM, omar kairan wrote:
Hi,
could someone help me with dilemma on the simulation of logistic
regressiondata with multicollinearity effect and high leverage point..
If that is the clearest way in which you can phrase your question then
I doubt that *anyone* can help
I don't understand what you are looking for.
The line
barplot(t(ld))
is placing the four treatments side by side, and is also stacking the
three values of typ.
Please indicate what you would like the figure to look like.
Rich
__
Use outer margins around the whole thing, see argument oma in ?par.
Uwe Ligges
On 20.01.2010 17:53, Lars Skjærven wrote:
Dear R users,
I'm making multiple plots within the same pdf page (par(mfcol = c(5,1)), and
want a legend for this at the bottom of all the plots. From previous mails
it has
Hello, I need to retrieve datas from bloomberg.
I want to retrieve those datas in the fastest way as possible. I have two
options:
writing the datas from bbg to excel and reading from r the excel sheet or
directly
read the datas from from r with a Rbbg connection. Which connection is
faster?
Six times a year labs submit individual Excel spreadsheets to me. There are
usually around 60 labs and the spreadsheets have 12 columns with 20 rows.
Chapter 8 of the R Data Import/Export manual recommends converting .xls
files, to a text file. I have been manually converting the individual
Try this:
t(apply(x, 1, function(x) (x == max(x)) - (x == min(x
[,1] [,2] [,3]
[1,] -101
[2,]1 -10
[3,]10 -1
You can avoid the transpose using plyr:
library(plyr)
aaply(x, 1, function(x) (x == max(x)) - (x == min(x)))
Var1 1 2 3
1 -1 0 1
2
One way to plot subsets of data identified by a grouping variable is to use
lapply() on a list of subsets. The approach is worth mentioning because
similar tactics are useful for many problems.
#List of unique values for grouping variable
#that is not necessarily a factor
names -
1. read.xls in gdata has the capability of
- reading an xls file starting from the first row with a given pattern
- converting an xls file to a csv file
- converting an xls file to a data frame
Would those capabilities be sufficient for the input side? See
examples in ?read.xls
2. If you do a
Thanks Marc for the quick reply. I confirmed the R binary I built is indeed
64-bit.
sqlplus works fine, so is the odbcConnect() call.
Any idea the error is thrown by RODBC or R?
Eric
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Is there a package for growth mixture modeling in R?
Rhoderick
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PLEASE do read the posting guide
Hi,
I'm looking to combine two data frames. Several of the columns are in
common while the others need to be summed up. The apply functions and
the merge functions don't seem to be working. I've included a basic
example of what I'm trying to do below. Thanks!
Sean
Thank you - zoo did exactly what I needed...
On Jan 20, 12:20 pm, Henrique Dallazuanna www...@gmail.com wrote:
Try the zoo package:
plot(as.yearqtr(time.val), inc, col = 'red', type = 'l')
On Wed, Jan 20, 2010 at 3:41 PM, mah harwood...@gmail.com wrote:
I am trying to generate a line
Hi Gabor,
Thanks for your insights and suggestions. There was a post on the Wiki you
mentioned that makes me think this will work.
Unfortunately, the spreadsheet was designed to make it easy for lab staff to
enter their results, but not so easy for another program to read in the
data. That is
Try this:
U -
http://www.mda.state.mn.us/en/sitecore/content/Global/MDADocs/licensing/map/mapreportform.aspx;
library(gdata)
DF - read.xls(U, pattern = SAMPLE, as.is = TRUE)
and now write an R program to create the desired data frame from DF.
On Wed, Jan 20, 2010 at 3:56 PM, Jerry Floren
Tena koe Sean
I suspect the apply() and merge() functions are working, but they may
not be doing what you expect :-) You could try rbind() and aggregate():
data.frame1$HAD - as.numeric(NA)
data.both - rbind(data.frame1, data.frame2)
aggregate(data.both[,-(1:3)], data.both[,1:3], sum,
Hi All,
I am new to python, R and rpy2. I am trying to print an Rvector say v and I
want to print 1st element so the statement will be
*print (v[0])* and this gives me *output [1] 876* but the real value I
should get is 876.
Can any one can help how to get rid of [1] while printing.
I am using
Jason Rupert wrote:
I have an instance where I need to include Greek letters on a plot title that is multiple lines.
I've searched the forums for an approach to do this, but most of the previous posts and replies seem to just address instances of single line examples and problems:, e.g.
I'm attempting to generate matrices where the entries are randomly generated
numbers of specified distribution.
The following code was an attempt to create a 3 by 3 matrix, where my
entries where randomly generated from a uniform (0,1) distribution.
x = matrix(0,ncol = 3, byrow = T)
for(i in
On Jan 20, 2010, at 4:55 PM, jshort wrote:
I'm attempting to generate matrices where the entries are randomly
generated
numbers of specified distribution.
The following code was an attempt to create a 3 by 3 matrix, where my
entries where randomly generated from a uniform (0,1)
Thanks a million Gabor. I was able to quickly import 21 test files. This will
save me hours, and should eliminate some errors.
Thanks,
Jerry Floren
Minnesota Department of Agriculture
--
View this message in context:
Hello!
I have a data frame with a factor and a numeric variable:
x-data.frame(factor=c(b,b,d,d,e,e),values=c(1,2,10,20,100,200))
For each level of factor - I would like to divide each value of
values by the mean of values that corresponds to the level of
factor
In other words, I would like to
R-lang.pdf has the following description in Section 3.1.1.
Any number typed directly at the prompt is a constant and is evaluated.
1
[1] 1
Perhaps unexpectedly, the number returned from the expression 1 is a
numeric. In most
cases, the difference between an integer and a numeric value will be
Jerry Floren wrote:
Hi Gabor,
Thanks for your insights and suggestions. There was a post on the Wiki you
mentioned that makes me think this will work.
Unfortunately, the spreadsheet was designed to make it easy for lab staff to
enter their results, but not so easy for another program to read
On Jan 20, 2010, at 1:36 PM, Eric Ma wrote:
Thanks Marc for the quick reply. I confirmed the R binary I built is indeed
64-bit.
sqlplus works fine, so is the odbcConnect() call.
Any idea the error is thrown by RODBC or R?
Eric
If you are getting the same error as in the original
According to R-lang.pdf (Section 2):
Function mode gives information about the mode of an object in the
sense of Becker,
Chambers Wilks (1988), and is more compatible with other
implementations of the S
language. Finally, the function storage.mode returns the storage mode
of its argument
in the
Hi Peng,
On Wed, Jan 20, 2010 at 5:37 PM, Peng Yu pengyu...@gmail.com wrote:
R-lang.pdf has the following description in Section 3.1.1.
Any number typed directly at the prompt is a constant and is evaluated.
1
[1] 1
Perhaps unexpectedly, the number returned from the expression 1 is a
While I agree that this is someone's homework, it occur to me that we
can mess him up a little more.
He asked for 200 points *uniformly* distributed on (or in) a circle.
Well, he did NOT say uniform random distribution.
So in fact the plot, on a circle would be
theta -seq(0,2*pi, by =
On 1/20/2010 5:37 PM, Dimitri Liakhovitski wrote:
Hello!
I have a data frame with a factor and a numeric variable:
x-data.frame(factor=c(b,b,d,d,e,e),values=c(1,2,10,20,100,200))
For each level of factor - I would like to divide each value of
values by the mean of values that
Sorry Gabor, but I am not quite there yet. Thanks David for your suggestion
on xlsReadWrite. Your message was posted while I was composing this one.
The Excel worksheet I want to read in is named Paste Special
I started with this code and thought it worked:
### Start of Code ###
I don't find a tutorial on S3. Bengtsson.pdf cites MASS (1999
edition). However, I don't think that MASS (2002 edition) clearly
explain what S3 is and help a user who knew very little about S3 to
quickly understand it. Could somebody let me know if there are some
better learning materials to help
On Wed, Jan 20, 2010 at 5:04 PM, Peng Yu pengyu...@gmail.com wrote:
I don't find a tutorial on S3. Bengtsson.pdf cites MASS (1999
edition). However, I don't think that MASS (2002 edition) clearly
explain what S3 is and help a user who knew very little about S3 to
quickly understand it. Could
Hello, I am using te blpGetData() function to retrieve closing prices from
bloomberg on r. This is the code that I wrote:
library(RBloomberg)
conn=blpConnect
blpGetData(conn,ANF UN Equity,PX_LAST,2009/09/01,2009/09/10)
and I get the following error:
Error in substring(paste(0, v$day, sep = ),
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