For the BH, you can use the command
?p.adjust(..., "BH")
For reading the files in and doing the analysis, here is an example:
http://www.r-tutor.com/elementary-statistics/non-parametric-methods/wilcoxon-signed-rank-test
You'd then need to extract all the P values like this:
x <- wilcox.test(... )
Hi Liliana,
Try
> table(factor(x, levels = 0:10))
0 1 2 3 4 5 6 7 8 9 10
0 2 0 6 0 2 2 1 0 1 6
Also, see ?factor.
HTH,
Jorge
On Tue, Nov 23, 2010 at 10:29 PM, Liliana Pacheco <> wrote:
> Hi R users
> I want to make a table of frequencies to show how many times the numbe
Hi R users
I want to make a table of frequencies to show how many times the numbers
form 0 to 10 appear in a vector.
For example, if the vector is:
> x
[1] 5 10 3 5 10 10 3 6 10 1 10 3 7 1 10 3 6 9 3 3
In this case I want a table that will show that the zero has frequency 0,
the o
Tena koe John
I got a similar message with a different package, and solved it by loading the
latest version of the package (which was built under 2.12.0). So you may need
to get around your institutional firewall!
HTH
Peter Alspach
> -Original Message-
> From: r-help-boun...@r-p
Hello, I wonder if I could get advice on the above problem
I have just installed V 2.12.0 (I chose only the 32-bit version) into a new
directory (C:/R) on a 64bit Windows 7 machine
> sessionInfo()
R version 2.12.0 (2010-10-15)
Platform: i386-pc-mingw32/i386 (32-bit)
locale:
[1] LC_COL
On Nov 23, 2010, at 6:25 PM, jt306 wrote:
Is it possible to create a contour plot with the isolines labeled.
I know
you can do this with Matlab. Argh!
It is fairly straightforward with lattice::levelplot. Provide some
sample data.
I tried creating a filled contour plot, then using
Hi there,
I'm a total newbie to R. I'd like to use a Wilcoxon Rank Sum test to compare
two populations of values. Further, I'd like to do this simultaneously for
114 sets of values. The two populations are C and N. The different sets of
values have arbitrary names (I'll call them a, b, c etc). The
Is it possible to create a contour plot with the isolines labeled. I know
you can do this with Matlab. Argh!
I tried creating a filled contour plot, then using par(new=T), followed by
overlaying the contour plot on top. However, the placement of the filled
contour plot and the contour plot do
Thanks Gabor! Work perfectly! This is the code i used.
z <- read.zoo(lines, header = TRUE, index = list(1, 2), FUN = function(d,t)
as.POSIXct(paste(date,time), format = "%m/%d/%Y %H:%M"))
On Tue, Nov 23, 2010 at 8:07 PM, Gabor Grothendieck wrote:
> On Tue, Nov 23, 2010 at 7:56 PM, Nikos Rachman
On Tue, Nov 23, 2010 at 7:56 PM, Nikos Rachmanis
wrote:
> Hi Gabor,
> Thanks very much for the suggestion. It worked well until the point that i
> tried to plot the xts object. here is my code and the error message.
> lines<-data.frame(date,time,open,high, low,close) #create a dataframe with
> all
Hi Gabor,
Thanks very much for the suggestion. It worked well until the point that i
tried to plot the xts object. here is my code and the error message.
lines<-data.frame(date,time,open,high, low,close) #create a dataframe with
all the variables
z <- read.zoo(lines, header = TRUE, index = list(1
Martin -
Type
help(scan)
or
?scan
to see scan's help file, which includes:
quiet: logical: if ‘FALSE’ (default), scan() will print a line,
saying how many items have been read.
So adding quiet=TRUE to your call to scan will suppress the message.
Try this:
scan(..., quiet = TRUE)
On Tue, Nov 23, 2010 at 9:01 PM, Martin Knapp wrote:
> Hi,
> I'm using R2.8.1 in windows vista and the scan function to load in my
> data. That works all fine except that I get a message after the scan
> function finishes. It tells me how many items the function
Hi,
I'm using R2.8.1 in windows vista and the scan function to load in my
data. That works all fine except that I get a message after the scan
function finishes. It tells me how many items the function has read.
For example "Read 202035 items". That would not be a problem but this
is printed in the
I have friendship data which is strong skewed. So it doesn't make sense to
use maximum likelihood methods for fa and cfa.
But I couldn't find any function for asymptotically distributed data for doing
a factor analysis. Only: apca() but there is no possibility to allow for factor
correlations.
This wasn't a completed answer. (I hit the send button by accident.)
The code only fills in a small poertion of the matrix. )
--
David.
On Nov 23, 2010, at 4:48 PM, David Winsemius wrote:
On Nov 23, 2010, at 3:26 PM, Barroso, Judit wrote:
Dear all of you,
I would like to get some help fr
On Tue, 23 Nov 2010, Liang Peng wrote:
Appreciate any suggestions regarding how to fit an unbalanced panel data to
a Tobit model using R functions. I am trying to analyze how real estate
capital expenditures (CapEx) are affected by market conditions using a panel
Tobit model. The CapEx is either
Hi
On 23/11/2010 5:41 p.m., Steve Bellan wrote:
Hi all,
I'm trying to visualize animal movement data characteristics
spatiotemporally by overlaying many different histograms on a map. I
want the histograms to be plotted at coordinates in the map that
matches a region they describe. If I was j
Appreciate any suggestions regarding how to fit an unbalanced panel data to
a Tobit model using R functions. I am trying to analyze how real estate
capital expenditures (CapEx) are affected by market conditions using a panel
Tobit model. The CapEx is either positive or 0, so it is censored. The dat
On Tue, Nov 23, 2010 at 5:03 PM, Manta wrote:
>
>
> Gabor Grothendieck wrote:
>>
>> How to convert a monthly series to a daily series has already been
>> illustrated in multiple ways in this thread.
>>
>
> Fair enough. However, my last question was different. I simply want to know
> if there is a
At this point, I'm giving up on this issue. The stage of our project
requiring daily use of Google Docs ended, thankfully, about a week before
this problem arose. I was able to just export the spreadsheet for further
analysis, rather than relying on the Google Spreadsheet. If this problem had
arise
Gabor Grothendieck wrote:
>
> How to convert a monthly series to a daily series has already been
> illustrated in multiple ways in this thread.
>
Fair enough. However, my last question was different. I simply want to know
if there is a simple neat way to import a monthly series as a zoo object
On Tue, Nov 23, 2010 at 4:41 PM, Manta wrote:
>
>
> Gabor Grothendieck wrote:
>>
>>> 2. How to import using as index 'Date' a monthly series (code below). The
>>> series in the US CPI from November 1979 to October 2010.
>>>
>>> z <- zoo(cpius$Value, as.Date("1979-11-30")+0:372)
>>
>> Assuming the
On Nov 23, 2010, at 3:26 PM, Barroso, Judit wrote:
Dear all of you,
I would like to get some help from you.
Here I attach you a model, that I would like to be stochastic so I
would need each time the value of x changed, the values of
parameters (ss, emrg, gf, spp, sr) did too according to
Thank you all the reply. The use of sys.call() as suggested by Ducan works
pretty well. This is good as the outputs from my function are large. The use
of apply family functions as suggested by Erik is good for small amounts of
outputs.
Edwin Sun
--
View this message in context:
http://r.78969
Gabor Grothendieck wrote:
>
>> 2. How to import using as index 'Date' a monthly series (code below). The
>> series in the US CPI from November 1979 to October 2010.
>>
>> z <- zoo(cpius$Value, as.Date("1979-11-30")+0:372)
>
> Assuming the question is how to turn this daily series into a monthly
Edwin -
I think the usual way to do this would be to use a function
like lapply or mapply to call your function multiple times
with varying arguments.
For example, with one varying argument:
lapply(list(3,5), test, z = 4)
With multiple varying arguments:
mapply(test, y = list(3,5), z = list(4,
On 23/11/2010 4:21 PM, Edwin Sun wrote:
Hello all,
I wrote a function with many arguments. Then I need to call it many times
with changes on some arguments only. Is there any way to write a function or
have a method to "update" it, like the relationship between lm() and
update()?
Many thanks,
Seems to me you're going about it in a bit of a difficult way. Assuming
you know you want to run 20 samples (x = 1 to 20), just generate sets of
20 random variables for each of your inputs.
for example,
ss <-rnorm(20,mean=.3, sd=.08)
Then inside the loop, replace "ss" with "ss[x]"
By the
Hello all,
I wrote a function with many arguments. Then I need to call it many times
with changes on some arguments only. Is there any way to write a function or
have a method to "update" it, like the relationship between lm() and
update()?
Many thanks,
Edwin Sun
This is the sampl
On Tue, Nov 23, 2010 at 3:58 PM, Manta wrote:
>
> Ok this helps definitely! But I still would like to know
>
> 1. How to change from one index to another within a 'zoo' object
Using
z <- zooreg(1:10, start = as.yearmon("2000-01"), freq = 12)
Try
aggregate(z, as.Date, identity)
or
ti
Hi:
Here are some possibilities using ggplot2:
library(ggplot2)
# Create ordered factors
bar$area2 <- ordered(bar$area)
bar$year2 <- ordered(bar$year)
# Side-by-side, aka dodged, bar charts
ggplot(bar, aes(x = area2, y = disc, fill = year2)) +
geom_bar(aes(group = year2), position = 'dodge
I often use code like Curt's encapsulated in the
following isFirstInRun function:
isFirstInRun <- function(x,...) {
lengthX <- length(x)
if (lengthX == 0) return(logical(0))
retVal <- c(TRUE, x[-1]!=x[-lengthX])
for(arg in list(...)) {
stopifnot(lengthX == lengt
Neeti,
I'm pretty sure that the error is related to the confusionMAtrix call,
which is in the caret package, not e1071.
The error message is pretty clear: you need to pas in two factor
objects that have the same levels. You can check by running the
commands:
str(pred_true1)
str(species_tes
Ok this helps definitely! But I still would like to know
1. How to change from one index to another within a 'zoo' object
2. How to import using as index 'Date' a monthly series (code below). The
series in the US CPI from November 1979 to October 2010.
z <- zoo(cpius$Value, as.Date("1979-11-30")
Generating new data from a kernel density estimate is equivalent to choosing a
point from your data at random, then generating a point from your kernel
centered at the chosen point.
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.s...@imail.org
801.408.8111
What Frank was trying to tell you is that the p-values don't have much meaning
if you do stepwise regression (sometimes they are worse than useless). The
p-values are computed based on certain assumptions, once you remove a variable
because it is "Not Significant", then recompute, those assumpt
Dear all of you,
I would like to get some help from you.
Here I attach you a model, that I would like to be stochastic so I would need
each time the value of x changed, the values of parameters (ss, emrg, gf, spp,
sr) did too according to a normal distribution, with mean its value and
standard
You can fit this model with AD Model Builder's random effects module.
there is an example fitting a Poisson and negative binomial to the
venerable
polio data set with ar(1) random effects at
http://admb-project.org/examples/count-data/negative-binomial-serially-correlated-counts
A big
One possibility for what is happening is that when you load the object it does
not automatically load the package as well, so you get an error when working
with the object. Try loading lme4 package in a new session, then load the
saved object and see if things work for you.
--
Gregory (Greg)
2010/11/23 Ignacio Alborés
> I want to obtain a barplot with error bars and a mean plot with error bars
> with other scale on the left in the same plot. I need help to get the
> same
> x axis (centered) when overlay two plots (barplot2 and plotCI (or
> lineplot.CI) with errorbars), with difere
I would suggest reading the discussion (all the posts) starting at:
http://tolstoy.newcastle.edu.au/R/e2/help/07/08/22858.html
This gives reasons for not doing what you are asking along with some
alternatives.
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
gre
Look at the subplot function in the TeachingDemos package.
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.s...@imail.org
801.408.8111
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-
> project.org] On Behalf Of St
On Tue, Nov 23, 2010 at 2:49 PM, Art Burke
wrote:
> Given the data structure below, how can I create a bar plot for the values of
> disc for each area grouped by year?
>
> bar <-structure(list(year = c(2003, 2003, 2003, 2003, 2003, 2003, 2003,
> 2007, 2007, 2007, 2007, 2007, 2007, 2007), area = s
Given the data structure below, how can I create a bar plot for the values of
disc for each area grouped by year?
bar <-structure(list(year = c(2003, 2003, 2003, 2003, 2003, 2003, 2003,
2007, 2007, 2007, 2007, 2007, 2007, 2007), area = structure(c(6L,
4L, 1L, 2L, 3L, 5L, 7L, 6L, 4L, 1L, 2L, 3L,
Aha: capture.output(str(x))
Sorry for the two previous eMails, but sometimes posting to the list helps
me thinking somehow ;-)
Greetz,
Janko
> -Ursprüngliche Nachricht-
> Von: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
> Im Auftrag von Janko Thyson
> Gesendet: Die
On 23/11/2010 2:35 PM, Janko Thyson wrote:
A short annotation:
I already considered deparsing the object via deparse(x), but that's not
exactly the info I'd like. I'm only interested in a short summary like str()
would give me, not in the deparsed object.
I think you need to give more details.
See R.utils package and sourceDirectory(), e.g.
sourceDirectory("R/");
You can also specify that you only want to reload files that have been
modified, e.g.
sourceDirectory("R/", modifiedOnly=TRUE);
/Henrik
On Tue, Nov 23, 2010 at 7:04 AM, Santosh Srinivas
wrote:
> Hello R-Helpers,
>
> I have
A short annotation:
I already considered deparsing the object via deparse(x), but that's not
exactly the info I'd like. I'm only interested in a short summary like str()
would give me, not in the deparsed object.
Thx,
Janko
## SYSTEM INFO ##
Windows XP SP3
R 2.12.0 (patched as of
> > ...
> > # I've found I need to lag a column to mimic SAS' first.
> > # operator, thusly, though perhaps someone else knows
> > # differently. Note this does not work on unordered
> > # dataframes!
> > lag.k1 <- c(NA, tt$k1[1:(nrow(tt) - 1)])
> > tt$r.first.k1 <- ifelse(is.na(lag.k1), 1, tt$k1
Dear list,
I'm looking for a suitable way to sort of "one-line-pretty-print" an
arbitrary R object in some of my log outputs.
Consider this:
cat(paste("The object/value is: ", x, ".", sep=""), sep="\n")
No problem if x is of class:
- character (length=1)
- numeric (length=1)
- logical
Colour affects the lines and points in ggplot2, whilst fill is for
areas (polygons, etc.)
By the way, I can't type either -- I meant last_plot().
baptiste
On 23 November 2010 20:18, John Kane wrote:
> Obviously I cannot type or spell.
>
> Your solution was great but do you have any idea why
>
Obviously I cannot type or spell.
Your solution was great but do you have any idea why
scale_colour_manual(value=mycolours)
does not work. More typos on my part or am I misunderstanding what
scale_colour_manual(value=mycolours) is supposed to do?
Thanks
--- On Tue, 11/23/10, baptiste augu
On Tue, 23 Nov 2010, seeliger.c...@epamail.epa.gov wrote:
Is there any similar function in R to the first. in SAS?
?duplicated
a$d <- ifelse( duplicated( a$a ), 0 , 1 )
a$d.2 <- as.numeric( !duplicated( a$a ) )
Actually, duplicated does not duplicate SAS' first. operator, though
fixing the various typos in your code, this works,
lat_plot() +
scale_fill_manual(value=mycolours)
HTH,
baptiste
On 23 November 2010 19:47, John Kane wrote:
> Someone was asking how to do a 16 category piechart in OpenOffice Calc and it
> appears that it can not be done (which we, probably, s
I want to obtain a barplot with error bars and a mean plot with error bars
with other scale on the left in the same plot. I need help to get the same
x axis (centered) when overlay two plots (barplot2 and plotCI (or
lineplot.CI) with errorbars), with diferent y axis. I use par (new=T) but
the X
Someone was asking how to do a 16 category piechart in OpenOffice Calc and it
appears that it can not be done (which we, probably, should be happy about) but
I thought that I'd try it in ggplot2.
It works but I then thought I'd like to make the colours more distinctive but
fro some reason I do
> > Is there any similar function in R to the first. in SAS?
> ?duplicated
>
> a$d <- ifelse( duplicated( a$a ), 0 , 1 )
>
> a$d.2 <- as.numeric( !duplicated( a$a ) )
Actually, duplicated does not duplicate SAS' first. operator, though it
may suffice for the OP's needs.
To illustra
On Tue, Nov 23, 2010 at 1:11 PM, Manta wrote:
>
> Thanks as always Gabor.
>
> What I was looking for was to plot the daily series on the same graph of the
> monthly using the command 'lines'. That is why I wanted to change the index
> of one of the two, as it seems the reason why at the moment it
Thanks as always Gabor.
What I was looking for was to plot the daily series on the same graph of the
monthly using the command 'lines'. That is why I wanted to change the index
of one of the two, as it seems the reason why at the moment it is not
working!
--
View this message in context:
http:/
On Tue, Nov 23, 2010 at 12:53 PM, Manta wrote:
>
> Dear R community, I have the following two zoo objects:
>
> MONTHLY CPI
>
>> plot(z)
>> par("usr")
> [1] 1977.76333 2011.15333 70.39856 227.03744
>> z=zooreg(cpius$Value,as.yearmon("1979-11"),frequency=12)
>> str(z)
> ‘zooreg’ series from Nov 1
Dear R community, I have the following two zoo objects:
MONTHLY CPI
> plot(z)
> par("usr")
[1] 1977.76333 2011.15333 70.39856 227.03744
> z=zooreg(cpius$Value,as.yearmon("1979-11"),frequency=12)
> str(z)
‘zooreg’ series from Nov 1979 to Oct 2010
Data: num [1:372] 76.2 77 77.8 78.5 79.5 80.3
Hi James,
Here is another option: all.equal() is more general, the sapply()
option I wrote assumes both data frames are basically the same, but
that maybe a few elements are different (e.g., you have two data sets
and are not sure which is more recent).
y <- x <- iris # x & y copies of iris
## c
I'm wondering if R has a package for generating count data subject to
extra-multinomial variation (or extra-binomial variation?)?? In other words,
the effect that cluster sampling has on count data.
For example: Have a multinomial (p,n) distribution with p=(p11,
p12,p21,p22) and n=number of unit
On Tue, 23 Nov 2010, David Winsemius wrote:
On Nov 23, 2010, at 11:04 AM, Charles C. Berry wrote:
On Tue, 23 Nov 2010, Dennis Murphy wrote:
> Interesting. Check this out:
>
> u <- sample(c(TRUE, FALSE), 10, replace = TRUE)
> > u
> [1] FALSE FALSE TRUE FALSE FALSE TRUE FALSE FALSE FALSE FA
Try this:
> input <- textConnection("var1 a
+ var2 b
+ var3 c
+ var4 d
+ var5 e")
> x <- read.table(input, as.is = TRUE)
> close(input)
> # create a list
> xList <- as.list(x$V2)
> names(xList) <- x$V1
> xList
$var1
[1] "a"
$var2
[1] "b"
$var3
[1] "c"
$var4
[1] "d"
$var5
[1] "e"
>
On Tue, N
Try this:
as.list(scan("your_txt_file"))
On Tue, Nov 23, 2010 at 3:05 PM, Guido Leoni wrote:
> Dear List I'm a newbie R user.
> I'm utilizing the list function in order to make a var like this:
> clusters<-list(a=var1,b=var2) My problem is that the total numer of
> variables that I need to
See all.equal, a toy example:
all.equal(iris, Orange)
On Tue, Nov 23, 2010 at 3:17 PM, Jim Edd Jones wrote:
> Is it possible to use R to compare two datasets to look for discrepancies,
> as one would with the SAS procedure PROC COMPARE?
>
> Any help on this would be greatly appreciated.
>
> Ja
Is it possible to use R to compare two datasets to look for discrepancies, as
one would with the SAS procedure PROC COMPARE?
Any help on this would be greatly appreciated.
James E. Jones
135 Salina St.
Lafayette, CO 80026
[[alternative HTML version deleted]]
__
could anyone help me with my last problem. if the question is not clear
please let me know
thank you
Hi everyone
I am trying to do cross validation (10 fold CV) by using e1071:svm method.
I
know that there is an option (cross) for cross validation but still I
wanted to make a funct
@Francial Giscard LIBENGUE please post your query again so that with
different subject
--
View this message in context:
http://r.789695.n4.nabble.com/cross-validation-using-e1071-SVM-tp3055335p3055831.html
Sent from the R help mailing list archive at Nabble.com.
_
Hi Andrea,
Help us help you. Please read this: http://www.R-project.org/posting-guide.html
and then reply to the list with more details on what you have, what
you have tried, and what you expect the results to be. Then we can
get you started (and maybe even finished). To borrow from Sherlock
Ho
> From: tal.gal...@gmail.com
> Date: Tue, 23 Nov 2010 18:14:39 +0200
> To: m.t.h...@ai.rug.nl
> CC: r-help@r-project.org
> Subject: Re: [R] Three-way ANOVA shows me two-way results
>
> It seems your model defines more parameters then your data permits (e.g:
> singular).
>
if you do googl
Hi, I just downloaded R, but I can't create a matrix with a parameter p.
Could someone help me please?
Andrea
--
View this message in context:
http://r.789695.n4.nabble.com/problem-with-parameters-tp3055682p3055682.html
Sent from the R help mailing list archive at Nabble.com.
Dear List I'm a newbie R user.
I'm utilizing the list function in order to make a var like this:
clusters<-list(a=var1,b=var2) My problem is that the total numer of
variables that I need to include in my list is up to 200. I've the text
string with the complete list of my variables but is too l
On Tue, Nov 23, 2010 at 11:49 AM, Santosh Srinivas
wrote:
> Thanks. Sorry about the naming (just cobbled the example).
> The rows are duplicate in the actual data. I am looking for a way to find
> the number of duplicate rows for each unique rowindex combination which is
> based on columns 1:6.
>
Thanks. Sorry about the naming (just cobbled the example).
The rows are duplicate in the actual data. I am looking for a way to find
the number of duplicate rows for each unique rowindex combination which is
based on columns 1:6.
-Original Message-
From: Gabor Grothendieck [mailto:ggrothe
Hi Stephen,
bold is presentation type idea, you can do it in reports (e.g., via
Sweave in LaTeX with \textbf{} or \mathbf{}), or graphs from R (e.g.,
make your title bold: expression(bold("Your Title"))), but it is not
something you do in R itself. Some front ends to R or text editors
may provide
dhacademic gmail.com gmail.com> writes:
> Hi,
>
> I have struggled on this "bound optimization with equality constraint" by
> using optim function for two days, but still fail to prepare a good input.
> Can anyone help to prepare the input for my specific case? Many thanks.
>
> Best,
> Hao
Yo
Hi folks,
I ran following commands to create a data file:
Year <- c(1983, 1984, 1985, 1986, 1987, 1988, 1989, 1990)
Yield <- c(60, 50, 70, 70, 80, 50, 60, 40)
Concentration% <- c(30, 40, 50, 60, 70, 80, 90, 100)
Temperature <- c(56, 47, 53, 53, 56, 47, 44, 44)
datafile <- data.frame(Year,Yield,C
On Nov 23, 2010, at 11:04 AM, Charles C. Berry wrote:
On Tue, 23 Nov 2010, Dennis Murphy wrote:
Interesting. Check this out:
u <- sample(c(TRUE, FALSE), 10, replace = TRUE)
u
[1] FALSE FALSE TRUE FALSE FALSE TRUE FALSE FALSE FALSE FALSE
class(u)
[1] "logical"
u + 0
[1] 0 0 1 0 0 1 0 0
On Tue, Nov 23, 2010 at 11:04 AM, Santosh Srinivas
wrote:
> Hello Group,
>
> I have a huge time series dataset with sample below. I am basically trying
> to read it into a zoo object with columns 1:6 to index. Zoo issues a warning
> that some of the rows have duplicated index.
>
> dput(z)
> struct
On Tue, 23 Nov 2010, Joel wrote:
Hi
When I try to update an number in a large data.frame by its pos It's really
slow it take almost a sec to do this and I wonder why and if where is any
faster way to update a number in a data.frame
ive tried
DF$col[POS]<-number
DF[xPOS,yPOS]<-number
See
It all has to do with the precedence of the ! operator.
Compare !duplicated(v) + 0 with (!duplicated(v)) + 0
Dave
From:
"Charles C. Berry"
To:
Dennis Murphy
Cc:
r-help@r-project.org
Date:
11/23/2010 10:08 AM
Subject:
Re: [R] the first. from SAS in R
Sent by:
r-help-boun...@r-project.org
On
It seems your model defines more parameters then your data permits (e.g:
singular).
What you can do is try something like this:
summary(aov(formula('FP ~ (lum + obj + man)^2 - Error(vp/(lum + obj +
man)^2)),data=dataf))
(notice the use of + and ^)
This is something your data might permit (assumi
On Tue, 23 Nov 2010, Dennis Murphy wrote:
Interesting. Check this out:
u <- sample(c(TRUE, FALSE), 10, replace = TRUE)
u
[1] FALSE FALSE TRUE FALSE FALSE TRUE FALSE FALSE FALSE FALSE
class(u)
[1] "logical"
u + 0
[1] 0 0 1 0 0 1 0 0 0 0
0 + u
[1] 0 0 1 0 0 1 0 0 0 0
v <- rpois(10, 3)
Hello Group,
I have a huge time series dataset with sample below. I am basically trying
to read it into a zoo object with columns 1:6 to index. Zoo issues a warning
that some of the rows have duplicated index.
dput(z)
structure(list(TrdTimestamp = structure(list(sec = c(19, 19,
18, 10, 12, 43, 4
Hi
When I try to update an number in a large data.frame by its pos It's really
slow it take almost a sec to do this and I wonder why and if where is any
faster way to update a number in a data.frame
ive tried
DF$col[POS]<-number
DF[xPOS,yPOS]<-number
Thx
//Joel
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On Tue, Nov 23, 2010 at 10:04 AM, Santosh Srinivas
wrote:
> Hello R-Helpers,
>
> I have a directory with some ".R" files that I execute every day.
> I want to write a script that executes each one of time sequentially.
>
> Is there a statement for this?
>
> Thank you.
>
>
On Tue, 23 Nov 2010, Joel wrote:
Is there any similar function in R to the first. in SAS?
What it dose is:
Lets say we have this table:
a b c
1 1 5
1 0 2
2 0 2
2 0 NA
2 9 2
3 1 3
and then I want do to do one thing the first time the number 1 appers in a
and something else the
Interesting. Check this out:
u <- sample(c(TRUE, FALSE), 10, replace = TRUE)
> u
[1] FALSE FALSE TRUE FALSE FALSE TRUE FALSE FALSE FALSE FALSE
> class(u)
[1] "logical"
> u + 0
[1] 0 0 1 0 0 1 0 0 0 0
> 0 + u
[1] 0 0 1 0 0 1 0 0 0 0
v <- rpois(10, 3)
> !duplicated(v)
[1] TRUE FALSE TRUE T
Hi folks,
> help.search("item_name")
displays a list of dataset containing the item_name. If I need to search
containing more than one item_name.
What will be the command
TIA
B.R.
Stephen L
__
R-help@r-project.org mailing list
https://stat.ethz.
Francial Giscard LIBENGUE gmail.com> writes:
>
> S'il vous plait, j'ai un problème avec la fonction gamma je n'arrive pas à
> bienécrire l'exposant j'ai utilisé ^ et ça ne marche pas.
> cordialemont
>
ici on parle anglais (je ne sais pas s'il y a une autre liste ou on pourrait
poser des ques
Also check out the following post:
http://permalink.gmane.org/gmane.comp.lang.r.general/210897
On Sat, Nov 20, 2010 at 4:32 PM, casperyc wrote:
>
> Hi,
>
> I have count data
>
> x2=rep(c(0:3),c(13,80,60,27))
>
> x2
> 0 1 2 3
> 13 80 60 27
>
> I want to graph to be ploted as
>
> barplot(table
Hello R-Helpers,
I have a directory with some ".R" files that I execute every day.
I want to write a script that executes each one of time sequentially.
Is there a statement for this?
Thank you.
__
R-help@r-project.org mailing list
https://stat.ethz.c
Kendric,
I've seen these too and traceback() usually goes back to ksvm(). This
doesn't mean that the error is there, but the results fo traceback()
from you would be helpful.
thanks,
Max
On Mon, Nov 22, 2010 at 6:18 PM, Kendric Wang
wrote:
> Hi. I am trying to construct a svmLinear model using
La langue de cette liste ci est desolais anglais.
S'il vous plait, nous devon un example. J'ai trouve:
> 2^3
[1] 8
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Hope this helps.
Spencer
On
Hi all,
I'm doing a 3-way ANOVA like this:
summary(aov(formula('FP ~ (lum * obj * man)3 - Error(vp/(lum * obj *
man)3)'),data=dataf))
But in the output I only get 1- and 2-way effects, like this one:
Error: vp:obj:man
Df Sum Sq Mean Sq F value Pr(>F)
obj:man 1 1.5291e-34 1.5291e-34 5.7011 0.05
On Tue, Nov 23, 2010 at 6:05 AM, fbielejec wrote:
> Dear,
>
> I'm doing analysis where I need to work on relatively large (50-60 MB)
> text files, though I'm really interested only in parts with binary
> variables (named indicators1, indicators2, ... etc.)
>
> Every text file contains other numeri
You need to spend some time and learn the basics of R. There are several
excellent tutorials available on the CRAN R-project website. This will be a
very good investment of your time. After you have spent a couple of days
learning the basics, try using my package to solve your problem.
Best,
R
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