On Tue, 23 Aug 2011, Tyler Rinker wrote:
Simple question but searching rseek did not yield the results I wanted.
Question: Is there a way to open a help manual for a package from within R.
For instance I would like to type a function in r for the tm package
and R would open that PDF as
I am pretty new to R. So this may be an easy question for most of you.
I would like to calculate the squared distances of a large set (let's say
2) of vectors (let's say dimension of 5) to a fixed vector.
Say I have a data frame MY_VECTORS with 2 rows and 5 columns, and one 5x1
I think I have found my problem, but I dont know how to correct it. I have
found an old post saying that it might be a problem if the starting values
are evaluated at Inf (see link here
http://r.789695.n4.nabble.com/Help-about-nlminb-function-td3089048.html)
But how can I run nlminb without the
Let's say your fixed vector is x, and y is the list of vectors that you want
to create the squared distance to x with, then:
x-c(1:5)
y-list()
y[[1]]-sample(c(1:5),5)
y[[2]]-sample(c(1:5),5)
y[[3]]-sample(c(1:5),5)
y
distances-lapply(y,function(a,b) crossprod(a-b), b=x)
#lapply goes over the
You could do something like this:
# data
nrows - 2L
ncols - 5L
myVec - array(rnorm(nrows * ncols), dim = c(nrows, ncols))
y - rnorm(ncols)
temp - t(myVec) - y
result - colSums(temp * temp)
# check
all.equal(as.numeric(crossprod(myVec[1L, ] - y)), result[1L])
#...
(And don't use a
On 24/08/11 01:55, Jonathan Greenberg wrote:
R-helpers:
Are there any ways to auto-generate R-friendly (e.g. will pass a
compilation check) .Rd files given a set of .R code? How about GUIs
that help properly format the .Rd files? Thanks! I want a basic set
of .Rd files that I can update as I
Hi Wei Wu,
What about:
x - matrix(rnorm(2*5),ncol=5)
y - rnorm(5)
distances - rowSums((x-y)**2)
Cheers,
Tsjerk
On Wed, Aug 24, 2011 at 8:43 AM, Enrico Schumann
enricoschum...@yahoo.de wrote:
You could do something like this:
# data
nrows - 2L
ncols - 5L
myVec -
Dear R list
I am trying to understand the auto-correlation concept. Auto-correlation is the
self-correlation of random variable X with a certain time lag of say t.
The article
http://www.mit.tut.fi/MIT-3010/luentokalvot/lk10-11/MDA_lecture16_11.pdf;
(Page no. 9 and 10) gives the methodology
Hello,
I have a function with a long list of parameters (of different types,
numeric and string)
myFunc -function(p1, p2, p3, p4, p5...etc)
{
do.something(p1,p2,)
}
I want to loop over this to provide a different set of parameters to the
list every time.
for (ii in 1:N)
{
You could use something like the following:
paramsList - list(p1 = 1:10, p2 = -5:5, p3 = 5, p4 = 8)
params - unlist(as.relistable(paramsList))
myFunc - function (params) {
params - relist(params, skeleton = paramsList)
p1 - params$p1
p2 - params$p2
p3 - params$p3
p4 -
Yes, sorry, so the distance is
colSums((t(x)-y)**2)
(I knew that) :S
Tsjerk
On Wed, Aug 24, 2011 at 9:19 AM, Enrico Schumann
enricoschum...@yahoo.de wrote:
R will subtract the vector columnwise from the matrix (so the vectors need
be the columns).
x - matrix(0, nrow = 10L, ncol = 5L)
y -
I think you have change options(expression = default_value)
you can input the command options()in Rconsole
And check the value of the expression.
and change the value to 100.
Hope you over this problem.
--
View this message in context:
R-helpers:
I'm trying to build a package, but I'm a bit new to the whole S3/S4
methods concept. I'm trying to add a new definition of the zoo
function as.yearmon, but I'm getting the following error when it
gets to this point during a package install:
***
R CMD INSTALL STARStools
* installing
Thanks Kevin! Got started with roxygen tonight! Cheers!
On Tue, Aug 23, 2011 at 6:28 PM, Kevin Wright kw.s...@gmail.com wrote:
I like the roxygen2 package for combining code and documentation. If you
use Emacs + ESS, it will even create much of the roxygen code for you (and
auto-revise it
Hi Gabor,
Thanks. I will try to figure out the solution you suggest. I found out
about melt() from a discussion forum; it seems to me that
melt()$value is similar to c(), and when I modified the script as
below it 'seems' to be running faster. Anyway in the end I only needed
to use a smaller
On 24.08.2011 10:30, Jonathan Greenberg wrote:
R-helpers:
I'm trying to build a package, but I'm a bit new to the whole S3/S4
methods concept. I'm trying to add a new definition of the zoo
function as.yearmon, but I'm getting the following error when it
gets to this point during a package
Your understanding is wrong. For a start, there is no function acf()
in package tseries: it is in stats.
And the autocorrelation at lag one is not the correlation omitting the
first and last values: it uses the mean and variance estimated from
the whole series and divisor n.
Have you
On Wed, 24 Aug 2011, Tsjerk Wassenaar wrote:
Yes, sorry, so the distance is
colSums((t(x)-y)**2)
(I knew that) :S
Did you know that ** is deprecated (and almost undocumented), so your
readers can hardly be expected to understand that? Please use the
documented operator ^ .
Tsjerk
On
Thomas Lumley wrote:
So it looks as though sum(c(...)) is somwhow independent of
the order of its arguments, which implies that the summation
it does is not quite the addition corresponding to +.
I now feel out of my depth, so hope someone else knows/can
find the truth about this!
sum()
Is there anything in R similar to spectogram command in matlab?
As I do not use matlab I can not know what spectogram does.
What is wrong with fft or spectrum functions?
Regards
Petr
On Tue, Aug 23, 2011 at 10:20 AM, Petr PIKAL petr.pi...@precheza.cz
wrote:
Hi
Hi all,
I am
On 24.08.2011 11:09, Jonathan Greenberg wrote:
Hmm, so I moved the function call to the same file as the methods
call, and placed it above the method in the file -- but I'm getting
the same error. Is there something odd about as.yearmon in the zoo
package that it might not be getting defined
Dear all,
I have a dataset of 3 categorical factors, the mean and the standard deviation
of each value. I want to use these values to plot a boxplot, grouped by each of
the 3 categorical factors (24 boxplots in total). I don't have a clue on how to
do the boxplot from mean and SD data already
Hi
Dear all,
I have a dataset of 3 categorical factors, the mean and the standard
deviation of each value. I want to use these values to plot a boxplot,
grouped by each of the 3 categorical factors (24 boxplots in total). I
don't have a clue on how to do the boxplot from mean and SD
Dear Michael
Thank you for your pointers.
On Tue, Aug 23, 2011 at 4:05 PM, Michael Friendly frien...@yorku.ca wrote:
First, you should be using rms::ols, as Design is old.
Good to know. I've always wondered why Design and rms, in many cases,
were providing similar functions and (upon cursory
The attachment seems to have been dropped, so I'm pasting the code
below. Sorry for that
Liviu
On Wed, Aug 24, 2011 at 1:44 PM, Liviu Andronic landronim...@gmail.com wrote:
Second, penalty in ols() is not the same as the ridge constant in lm.ridge,
but rather a penalty on the log likelihood.
Hmm, so I moved the function call to the same file as the methods
call, and placed it above the method in the file -- but I'm getting
the same error. Is there something odd about as.yearmon in the zoo
package that it might not be getting defined as a generic function?
If so, how would I go about
Hello,
I’m working on a simulation using a cellular automata. The scenario is:
Each cell represent a land plot. Each plot can have 3 different states:
-The owner is a private individual (P)
-The owner is a company ( C)
-The land is abandoned/not used (A)
The first year
Hi all,
I'm trying to use gofCopula() function in copula package. But I'm always
(trying different data sets) getting the same p value for Cramer-von Mises
statistic. I'm using:
g-gumbelCopula(1.653, dim = 2)gofCopula(g, x, N = 1000, method =
itau,simulation = pb)
c-claytonCopula(1.306, dim =
Dear list,
to prepare data as input for a stand-alone groundwater model I use the
cat() function to print data.
I want to print a table with print() or cat() but without the row- and
column-names. Is this possible?
Here a small example
my.df - rbind(c(Nr,FraSand,FraSilt,FraClay,pH), c(,
try function write.matrix() from package MASS, e.g.,
library(MASS)
write.matrix(my.df)
I hope it helps.
Best,
Dimitris
On 8/24/2011 1:46 PM, Tim Haering wrote:
Dear list,
to prepare data as input for a stand-alone groundwater model I use the
cat() function to print data.
I want to print a
Dear R community!
I would like to simulate distributions based on a polynomial model. It
consists of 96 values and I want to randomly select new normaly distributed
values around the modeled values. Furthermore, each value should be
correlated with the previous one. Is it correct to therefore use
Hi,
Based on some modification that I did to the R Cookbook Graphs Scatterplots
code, link:http://wiki.stdout.org/rcookbook/Graphs/Scatterplots%20(ggplot2)
I have some questions and I will appreciate a help:
- How do I change the legend title?
- How can I change the for each linear
Hi all,
I hope that i've posted this in the correct place. if not, please accept my
apologies (where should this go?)
I have carried out experimental removal of bivalves at 2 intertidal shores.
Bivalves were removed by raking of surface sediments. I wish compare the
biomass values of for a total
My script is the following:
library(metafor)
yi-c(-0.1, 0.2, 0.3, 0.4)
sei-c(0.4, 0.2, 0.6, 0.1)
vi-sei^2
studi-c(A, B, C, D)
eventi.c-c(10, 5, 7, 6)
n.c-c(11, 34, 25, 20)
eventi.a-c(2, 7, 6, 5)
n.a-c(11, 35, 25, 15)
dfs-rma(yi, vi, method=DL)
dfs
windows(height=6, width=10, pointsize=10)
Hi List,
I'm having this problem when trying to use the PREDICT function.
Here is a way to reproduce the error
library(REEMtree)
data(simpleREEMdata)
REEMresult-REEMtree(Y~D+t+X, data=simpleREEMdata, random=~1|ID/D)
predict(REEMresult, simpleREEMdata, id =
geigercounter120 geigercounter120 at yahoo.co.uk writes:
Hi all,
I hope that i've posted this in the correct place. if not, please accept my
apologies (where should this go?)
I have carried out experimental removal of bivalves at 2 intertidal shores.
Bivalves were removed by raking of
Hi,
Have you looked at the documentation at http://had.co.nz/ggplot2/ ?
You will find the answers to your questions there.
On Wed, Aug 24, 2011 at 7:46 AM, ashz a...@walla.co.il wrote:
Hi,
Based on some modification that I did to the R Cookbook Graphs Scatterplots
code,
On Aug 24, 2011, at 13:18 , Petr PIKAL wrote:
Hi
Dear all,
I have a dataset of 3 categorical factors, the mean and the standard
deviation of each value. I want to use these values to plot a boxplot,
grouped by each of the 3 categorical factors (24 boxplots in total). I
don't have a
Hi,
I am not sure if this is the right list to ask this question (though I did
not find a more appropriate one).
I've started using R a month ago, and small scripts work fine. However, when
I start writing more complex code, it gets messy.
1. Is there any way to debug normally, with breakpoints?
Am 24.08.2011 07:50, schrieb Paola Tellaroli:
My script is the following:
library(metafor)
yi-c(-0.1, 0.2, 0.3, 0.4)
sei-c(0.4, 0.2, 0.6, 0.1)
vi-sei^2
studi-c(A, B, C, D)
eventi.c-c(10, 5, 7, 6)
n.c-c(11, 34, 25, 20)
eventi.a-c(2, 7, 6, 5)
n.a-c(11, 35, 25, 15)
dfs-rma(yi, vi,
In an .Rd example for a package, I want to use data from another
package, but avoid loading the entire
package and avoid errors/warnings if that other package is not available.
If I don't care about loading the other package, I can just do:
if (require(ElemStatLearn, quietly=TRUE)) {
On Wed, Aug 24, 2011 at 4:20 PM, Eran Eidinger e...@taykey.com wrote:
Hi,
I am not sure if this is the right list to ask this question (though I did
not find a more appropriate one).
I've started using R a month ago, and small scripts work fine. However, when
I start writing more complex
Another great tool is debugonce()
wrap your function name in it and then execute your function call.
debugonce(my.function)
out-my.function(df)
And you'll be brought into the same interactive browser. (its Vi if im not
mistaken which can take a little getting used to.)
Justin
On Wed, Aug
Richard,
Thanks for your observation and tip.
My apologies that the 'expr' seemed undefined. That was intentional on my part
as I only wanted to show the form of the non-working code. Let me be clearer
by updating the code with what I actually type at the command line. The code
below does
Wow, thanks Justin and Liviu,
DebugOnce and browser. great!
Eran
On Wed, Aug 24, 2011 at 5:34 PM, Justin Haynes jto...@gmail.com wrote:
Another great tool is debugonce()
wrap your function name in it and then execute your function call.
debugonce(my.function)
out-my.function(df)
And
Also check out the 'debug' package
On Wed, Aug 24, 2011 at 10:59 AM, Eran Eidinger e...@taykey.com wrote:
Wow, thanks Justin and Liviu,
DebugOnce and browser. great!
Eran
On Wed, Aug 24, 2011 at 5:34 PM, Justin Haynes jto...@gmail.com wrote:
Another great tool is debugonce()
wrap your
?switch
If you read the help page, you will see that if the EXPR evaluates to
a character string, then is matches on the names of the elements;
'x[1]' is not a name, it is a value.
You want to probably use 'match'
match(Choice 2, x)
[1] 2
On Wed, Aug 24, 2011 at 10:52 AM, Mauricio Cornejo
Hello,
I'd like to rank rows of a data frame similar to what rank() does for
vectors. However, ties should be broken by columns that I specify. If it
is not possible to break a ties (because the row data is essentially the
same), I'd like to have the same flexibility that rank() offers. Is
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
On Behalf Of Mauricio Cornejo
Sent: Wednesday, August 24, 2011 7:53 AM
To: Richard M. Heiberger
Cc: r-help@r-project.org
Subject: [R] THX-- How to use 'switch' with strings containing
Dear All,
I would like to ask a question on how to find the index of the minimum of
entries of a numeric vector, without using loops or user defined functions.
Suppose we have a vector:
a - c(3,1,2)
then,
min(a) = 1
and its index is 2.
Target: how to get the index of this minimum? How to get
Your code example doesn't work because x[3]='My 3rd choice' is not a valid
named parameter assignment for a function, and that is because x[3] is not a
valid name for a function argument. The _content_ of x[3] might be, but
argument names aren;t parsed in this context (and indeed only would
which.min()
more generally
which(a==min(a))
Michael Weylandt
On Wed, Aug 24, 2011 at 12:03 PM, Chee Chen chee.c...@yahoo.com wrote:
Dear All,
I would like to ask a question on how to find the index of the minimum of
entries of a numeric vector, without using loops or user defined
I didn't see an answer to this and I THINK I sorted it out myself so I
thought I'd post it for anyone who is interested (in using it or correcting
it):
My original question:
Can someone show me how to modify one (R.oo) class's object inside another
(R.oo) class's method? Is that possible with
Try which.min(a)
On Wed, Aug 24, 2011 at 1:03 PM, Chee Chen chee.c...@yahoo.com wrote:
Dear All,
I would like to ask a question on how to find the index of the minimum of
entries of a numeric vector, without using loops or user defined functions.
Suppose we have a vector:
a - c(3,1,2)
.packages(all = TRUE) will give you a list of all available packages
without really loading them like require().
Regards,
Yihui
--
Yihui Xie xieyi...@gmail.com
Phone: 515-294-2465 Web: http://yihui.name
Department of Statistics, Iowa State University
2215 Snedecor Hall, Ames, IA
On Wed, Aug
To avoid a possible confusion, please note it is roxygen2 rather than
roxygen; both are packages on CRAN, but roxygen2 is an improved
version and under maintenance; I guess the development of the original
roxygen package has stopped.
Emacs+ESS helps a whole lot in writing documentation! I really
Thank you, Bernd, for looking into this.
Yes, at the moment, the color of the summary estimate for models without
moderators is hard-coded (as black). I didn't think people may want to change
that. I guess I was wrong =)
A dirty solution for the moment is to add:
addpoly(dfs, efac=6, row=-1,
Actually it is recommended to test for the availability of a valid
package with find.package(), particularly in this case where the name of
the package is already know.
Best,
Uwe
On 24.08.2011 18:29, Yihui Xie wrote:
.packages(all = TRUE) will give you a list of all available packages
On Aug 24, 2011, at 11:09 AM, Sebastian Bauer wrote:
Hello,
I'd like to rank rows of a data frame similar to what rank() does
for vectors. However, ties should be broken by columns that I
specify. If it is not possible to break a ties (because the row data
is essentially the same), I'd
I have a data set with about 6 million rows and 50 columns. It is a
mixture of dates, factors, and numerics.
What I am trying to accomplish can be seen with the following
simplified data, which is given as dput output below.
head(myData)
mydate gender mygroup id
1 2012-03-25 F
Hi All,
I have a fitted model called glm.fit which I used glm and data dat is my data
frame
pred= predict(glm.fit, data = dat, type=response)
to predict how it predicts on my whole data but obviously I have to do
cross-validation to train the model on one part of my data and predict on the
Hi!
I'd like to rank rows of a data frame similar to what rank() does
for vectors. However, ties should be broken by columns that I
specify. If it is not possible to break a ties (because the row data
is essentially the same), I'd like to have the same flexibility that
rank() offers. Is
What you describe is not cross-validation, so I am afraid we do not
know what you mean. And cv.glm does 'prediction for the hold-out
data' for you: you can read the code to see how it does so.
I suspect you mean you want to do validation on a test set, but that
is not what you actually
Hi,
Thanks for the reply. What I meant is that, I would like to partition my dat
data (a data frame) into training and testing data and then evaluate the
performance of the model on test data. So, I thought cross validation is the
natural choice to see how the prediction works on the hold-out
Hi Juliet:
Here's a Q D solution:
# (1) plyr
f - function(d) length(unique(d$mygroup)) - 1
ddply(myData, .(id), f)
id V1
1 1 0
2 2 2
3 3 1
4 4 0
# (2) data.table
myDT - data.table(myData, key = 'id')
myDT[, list(nswitch = length(unique(mygroup)) - 1), by = 'id']
If one can switch
On Aug 24, 2011, at 1:11 PM, Sebastian Bauer wrote:
Hi!
I'd like to rank rows of a data frame similar to what rank() does
for vectors. However, ties should be broken by columns that I
specify. If it is not possible to break a ties (because the row data
is essentially the same), I'd like to
Thanks Dennis! I'll check this out.
Just to clarify, I need the total number of switches/changes
regardless of if that state
had occurred in the past. So A-A-B-A, would have 2 changes: A to B and B to A.
Thanks again.
On Wed, Aug 24, 2011 at 1:28 PM, Dennis Murphy djmu...@gmail.com wrote:
Hi
Hi!
in R? Basically, what I need is a mixture of order() and rank().
While the former allows to specify multiple vectors, it doesn't
provide the flexibility of rank() such that I can specify what
happens if ties can not be broken.
An example of this simple problem would clarify this greatly.
What is your sample size? I've had trouble getting reliable estimates using
simple data splitting when N 20,000.
Note that the following functions in the rms package facilitates
cross-validation and bootstrapping for validating models: ols, validate,
calibrate.
Frank
Andra Isan wrote:
Hi,
On Aug 24, 2011, at 1:37 PM, Sebastian Bauer wrote:
Hi!
in R? Basically, what I need is a mixture of order() and rank().
While the former allows to specify multiple vectors, it doesn't
provide the flexibility of rank() such that I can specify what
happens if ties can not be broken.
An
Hi R users,
I was using read.table to read a file. The data.fame looked alright, but I
found not all rows are read by the read.table. What's wrong with it? It
didn't give me any warning or error messages. Why the data are truncated?
Thanks.
$ wc -l all/isoform_exp.diff
42847 all/isoform_exp.diff
Dear All,
As always, I appreciate all your help.
I would like to know the easiest way to convert each of the homogeneous
elements of a numeric list into a matrix. Each element of this list is also a
list such that when displayed, looks like a 2-by-3 matrix , I would like to
convert each of them
Hi,
On Wed, Aug 24, 2011 at 2:18 PM, zhenjiang xu zhenjiang...@gmail.com wrote:
Hi R users,
I was using read.table to read a file. The data.fame looked alright, but I
found not all rows are read by the read.table. What's wrong with it? It
didn't give me any warning or error messages. Why the
I'm not sure I understand your question: a[[2]] is a matrix.
a - list(matrix(1:6,2),matrix(5:10,2))
is.matrix(a[[2]])
TRUE
x = a[[2]]
is.matrix(x)
TRUE
x+2
[,1] [,2] [,3]
[1,] 7 9 11
[2,] 810 12
a[[2]] + 2
[,1] [,2] [,3]
[1,] 7 9 11
[2,] 810 12
What
Rereading your email, still not sure what the question is -- perhaps you
could give a better code example to illustrate the difference between a[[2]]
and mat1 -- but, since you mentioned briefly lists of lists, have you looked
at unlist(, recursive = F)? If applied to a list of lists, it won't
On 8/24/2011 12:40 PM, Uwe Ligges wrote:
Actually it is recommended to test for the availability of a valid
package with find.package(), particularly in this case where the name
of the package is already know.
Best,
Uwe
Thanks. So I guess the idiom I'm looking for is
Hi,
In order to find the best models I use AIC, more specifically I calculate
Akaike weights then Evidence Ratio (ER) and consider that models with a ER
2 are equally likely.
But the same problem remain each time I do that. I selected the best models
from a set of them, but I don't know if those
Hi all,
I have a vector xm say: xm = c(1,2,3,4,5,5,5,6,6)
I want to return a vector with the corresponding probabilities based on the
amount of times the numbers occurred. For example, I should get the
following vector for xm:
prob.xm = c(1/9, 1/9, 1/9, 1/9, 3/9, 3/9, 3/9, 2/9, 2/9)
Any help
Try this:
prob.xm - (table(xm)/length(xm))[match(xm, sort(unique(xm)))]
Jean
Jim Silverton wrote on 08/24/2011 02:31:05 PM:
Hi all,
I have a vector xm say: xm = c(1,2,3,4,5,5,5,6,6)
I want to return a vector with the corresponding probabilities based on
the
amount of times the numbers
If your numbers are all positive integers, this should work:
(tabulate(xm)[xm])/length(xm)
it can be put into a function for ease of use:
probVec - function(x) {(tabulate(x)[x])/length(x)}
You'll have some trouble if you have non-positive integers or non-integers.
Let me know if you need to
1. As this is not really appropriate for R, I suggest replies be private.
2. You might try posting on various statistical forums, e.g. on
http://stats.stackexchange.com/
-- Cheers, Bert
On Wed, Aug 24, 2011 at 12:15 PM, Arnaud Mosnier a.mosn...@gmail.com wrote:
Hi,
In order to find the best
Apparently my request to view the help pages is not a popular method among R
users for gaining information. for me these pages are very helpful so I will
follow up to completed this thread for future searchers.
First thanks fo Prof. Brian Ripley. Your idea was spot on what I was looking
This should be easy but it does not work
I have 3 vectors*(activeT,inactT, activeR)*,
the idea is that if the last value in inactT is higher than the last in
activeT
this value has to be append in active T
and the last value in another vector call activeR has to be repeated.
(at the bottom you can
Claudio Zanettini wrote on 08/24/2011 03:04:39 PM:
This should be easy but it does not work
I have 3 vectors*(activeT,inactT, activeR)*,
the idea is that if the last value in inactT is higher than the last in
activeT
this value has to be append in active T
When you say this value which one
On Aug 24, 2011, at 3:31 PM, Jim Silverton wrote:
Hi all,
I have a vector xm say: xm = c(1,2,3,4,5,5,5,6,6)
I want to return a vector with the corresponding probabilities based
on the
amount of times the numbers occurred. For example, I should get the
following vector for xm:
prob.xm =
I'm still a little confused about lastV and lastI. The code you provide
uses lastV, but your description seems to refer to lastI. Test out this
code and see if it is doing what you want it to do.
lastI
lastA
activeT
activeR
if(lastI lastA) {
activeT - c(activeT, lastI)
Thank you, this work fine,
and is not contorted like mine:)
In this case lastV=LastI but depending on the data that I obtain
lastV can be = LastA.
Any way it works very good:)
Thank you
very much :)
PS: but I still do not understand what was wrong in the script that I used,
It was not very
Hi there,
I have length data of an organism over the year and I want to make a
boxplot. I get the boxplot just fine but the months are all out of order. In
the data set they are in order from Jan-Dec...how can I fix this problem?
Thanks so much in advance!!
Phoebe
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Dear all,
How can I covert lm data to text in the form of y=ax+b, r2 and how do I
calculate R-squared(r2)?
Thanks.
Code:
x=18:29
y=c(7.1,7,7.7,8.2,8.8,9.7,9.9,7.1,7.2,8.8,8.7,8.5)
res=lm(y~x)
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Hi,
I'm really new to R so I aoplogise if this is a stupid question.
I'm trying to import data from a .txt file into R using the read.table
command, the headers for the data columns are already in the text file so I
add Header = True after the file location. The problem is I keep getting the
Thank you Dan and Ista!
Both of you are correct, I should have used NA rather than NA in my
example. So the correct code should be:
X -as.data.frame(matrix(c(9, 6, 1, 3, 9, NA, NA,NA,NA,NA,
6, 4, 3,NA, NA, NA, 5, 4, 1, 3), ncol=2))
names(X)-c(X1,X2)
Hello everyone,
I was asked to repost this again, sorry for any inconvenience.
I'm looking replacement for ddply function from plyr package.
Function allows to apply function by category stored in any column/columns.
Regular loops or lapplys slow down greatly because my unique combination
count
Many thanks for your response.
unfortunately, it appears that I'm the closest thing in the vicinity to a
local expert (chilling times indeed...), but i will certainly look at the
booklist
in terms of the number of data points, we have:
two shores,
three treatments,
three replicates of each
answered my own question, just use the call shell function in vb
woohoo
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hello
I need to know the dfn and dfd of my Anova. But in the Anova output there is
only Df.
Is this the dfn or the dfd? and how do I get both of it in R?
Thanks for any answers
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Hello
did you happen to figure this out? I am just learning about using R, i have
a whack of fish data in MSAccess...and i want to take whatever functions
access is limited by with stats, and then call R to do them
i know the package RODBC works great to read data from your mdb, but i want
to
Apologies for the elementary nature of the question (yes, I'm another
newbie)...
I'd like to perform a multiple regression on a single data set containing a
representation of energy consumption and temperatures containing account
number, usage (KWh), heating degree days (HDD) and cooling degree
This should be easy but it does not work
I have 3 vectors*(activeT,inactT, activeR)*,
the idea is that if the last value in inactT is higher than the last in
activeT
this value has to be append in active T
and the last value in another vector call activeR has to be repeated.
(at the bottom you can
Hi!
I'm confused by this:
as.numeric(as.POSIXct(518400,origin=2001-01-01))
[1] 978822000
I guess the problem is that as.numeric() assumes a different origin, but cannot
find
any default origin.
How can I get back the seconds from the POSIXct format? In other words, which
the inverse
Hello,
I am a new user of R, and I'd be grateful if someone could help me with the
following:
I would like to compute the mean of variable trust in dataframe foo, but
separately for each level of variable V2. That is, I'd like to compute the
mean of trust at each level of V2.
I have done this:
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