n, I want to fill up x[2] with value, in our case it's 111.
>
> Mike
>
> On Thu, Nov 29, 2012 at 5:39 PM, Jean V Adams wrote:
> Mike,
>
> Based on this example, what do you want
> samples
> to look like?
>
> It's not clear to me what you&#
Josh,
The code you submitted is not reproducible. I don't have the following
objects:
barchart, DataToPlot..SeCl, or Colors.
However, I think I can answer your questions, with some modified code ...
A rowname can't be a factor. But you can still use a factor to sort the
rows of your
Other readers of this list may have better suggestions for how to read in
data with interspersed header rows, but here's a work-around to do
specifically what you requested ...
# find the rows where "Loan" is in the Date column
sel <- grep("Loan", dat$Date)
# create a new vector with these row
Mike,
Based on this example, what do you want
samples
to look like?
It's not clear to me what you're trying to do with
i-1
Jean
C W wrote on 11/29/2012 03:55:12 PM:
>
> Hi list,
> I am writing a for loop that looks like this:
> samples<-rep(NA,10)
> x <- rep(c(111, 225), 5)
Kirsten,
The overall model is the combination of both models. If you call the
parameter estimates from the logistic regression betas and the parameter
estimates from the linear regression alpha, you could write the predictive
equation something like this (ignoring error terms):
cover =
;1.0996;3.0952;0.94324;2.3146;0.9822;1.0752;2.
> 6336;4.7202;0.82122;1.2563;0.73988;1.7051;2.3569;1.4296;0.85812;4.
> 8422;1.9687;2.511;4.5446;1.9065;2.3899;2.2784;4.174;2.6654;4.8175;3.
> 9665;3.902;3.5763;1.337;4.0643;3.6533;0.78097;1.6724;4.957;3.7316;1.
> 7372;4.9859;4.2946;4.3697;2
Andrew,
Interesting issue. My tack would be to define an age key that
incorporates all of the different cut-points that are used in your data
tables. Then, with the use of some simple functions, you can test which
factors are "nested" within other factors, and you can broaden those
categorie
Irucka,
What is the code that you are using that results in a character matrix?
What does the character matrix look like?
If your matrix is called m, submit the following code and share the
results with us.
m[1:6, ]
as.numeric(m[1:6, ])
Jean
iembry wrote on 11/27/2012 11:35:15 PM:
>
> Hi,
Ben,
You can use the sample() function to randomly add -1, 0, or 1 to each
observation, and control for the probability of mutation at the same time.
Then you can use the match() function to make sure that any mutations in
X are carried through to Y in the same way. I wrote the function to do
Create an empty list called "result" before you run the loop. Then store
the results of the loop in the list. For example:
result <- vector(mode="list", length=1000)
for(i in 1:1000){
result[[i]] <- try(harvest(i))
}
Jean
mdvaan wrote on 11/27/2012 12:09:38 AM:
>
> Hi,
>
It could be that for some levels of your independent factor variables (WS,
SS), the response is either all zeroes or all ones. Or, for your
continuous independent variables (DV, DS), there is a clean break between
the zeroes and ones. For example, if all the CIDs are one when DS <= 18
but all
Andras,
What do you want your code to do? Give us a little explanation with your
code.
When I try to run your code, I get
Error: could not find function "genoud".
Either supply the code for the functions included in your code, or tell us
what packages we need to run it.
Jean
Andras
You've included three "I want"s in your message, but no question.
Are you looking for functions? Try searching ... in R, on google, on
http://rseek.org/, ...
Have you tried some code, but can't get it to work? Send your code with
error messages, and tell us precisely what you trying to get yo
Florian,
There are a number of different ways to handle data like this. Two that
come to my mind are shown below. You could allow each observation to be
represented by multiple rows in the data frame:
obs dishes
11 saucer
22cup
32 plate
42 saucer
53cup
63
Use the col= argument to the heatmap() function to specify the three
colors you want. For example:
x <- matrix(rnorm(50), ncol=5)
heatmap(x, col=1:3)
Jean
furor wrote on 11/14/2012 03:31:42 AM:
>
> Hi all,
>
> I've made a heatmap using '-', '=' and '+' as possible values. H
ay jagged like
> this because the rows would need to be of equal number for the array
> function, yet in a list there is not such requirement, and
> operations on matrices can target elements in specific matrices by
[[,]][,] ?
>
> Best Wishes,
>
> Ben W.
>
> U
Nicolas,
Maybe something like this would work for you. Put all of your x values in
a list (or a vector, if your x values are scalars). Use sapply to loop
through all of your x values, applying the value1() and value2()
functions, and saving the results in a data frame. Then write the data
f
:
> The format would be a 2D layout, Where every line is an effector
> gene and every column an aspect of the effector gene(value,
> expression state, fitness contribution etc.) This 2D layout of rows
> and columns is then repeated in the 3rd dimension (the z of x,y,z)
> of the
etc. This 2D layout of rows and columns is then
> repeated in the z dimension of the array for each individual. It is
> ragged in the sense each individual, each slice through the array in
> the z direction, would have different numbers of rows - different
> numbers of effectors. I c
Katrin,
I believe that error message is caused by empty lines of data in the file
being read by read.table().
What is the last value for "j" printed when you run the code?
What does the file look like at rows j to j+8?
Does your code work if you just submit these lines?
for (i in 0)
{ j=subjec
Ben,
Can you provide a small example data set for
inds
so that we can run the code you have supplied?
It's difficult for me to follow what you've got and where you're trying to
go.
Jean
"Benjamin Ward (ENV)" wrote on 11/06/2012 03:29:52 PM:
>
> Hi all,
>
> I have a list of genes p
ion(x) plot(x, fit$resid))
fits[[i]] <- fit
}
Jean
Ross Ahmed wrote on 11/06/2012 09:25:13 AM:
>
> Thanks Jean
>
> This works for the plots, but it only stores the last lm() computed
>
> Ross
>
> From: Jean V Adams
> Date: Tuesday, 6 November
Ross,
Here's one way to condense the code ...
DV <- c("mpg", "drat", "gear")
IV <- list(c("cyl", "disp", "hp"), c("wt", "qsec"), c("carb", "hp"))
for(i in seq(DV)) {
fit <- lm(formula=paste(DV[i], paste(IV[[i]], collapse="+"),
sep="~"), data=mtcars)
plot(fit$fitted, fit$resid, m
David,
How would you interpret the results of a post hoc test for sexcolor when
you have an interaction term with sexcolor in your model?
Perhaps it would be helpful to plot doy vs. predicted tle with confidence
intervals for each of the four levels of sexcolor at a fixed tl (e.g., the
mean).
Christof,
You could use single linkage clustering to separate the dates into
different groups if they are more than 14 days apart. Below is a simple
example, where x represents day.
x <- sort(sample(1:500, 100))
y <- rnorm(100)
cluster <- hclust(dist(x), method="single")
group <- cutree(clust
Using the data you provided, a combination of slope and height comes
close:
X <- seq(Y)
high <- Y > 0.6
upslope <- c(FALSE, diff(Y) > 0)
section <- rep(1, length(Y))
section[upslope==TRUE & high==TRUE] <- 2
section[upslope==FALSE & high==TRUE] <- 3
plot(X, Y, col=section)
Or you could base the
Sally,
It's great that you provided data and code. To make it even more
user-friendly for R-help readers, supply your data as Rcode, using (for
example) the dput() function.
The reason you were getting all 1s with your code, is that you had told it
to aggregate by trip, LENGTH, and species.
Hannes,
A bit inelegant, but it works. Try this:
mtext("Overall Title Row 2", outer=TRUE, line=-17)
mtext("Overall Title Row 3", outer=TRUE, line=-34)
Jean
capy_bara wrote on 10/23/2012 09:08:39 AM:
>
> Dear all,
> I have a 3x2 plot and in addition to the title of the individual plots I
>
7;d spread the "word" (ha!) about rtf.
Below is some introductory code based on examples in
http://cran.r-project.org/web/packages/rtf/vignettes/rtf.pdf
Give it a try. You may like it!
Jean
`·.,, ><(((º> `·.,, ><(((º> `·.,, ><(((º>
Jean V.
Baoqiang,
Here's an approach that should work:
(1) Make sure that the column names of trainx and testx are the same.
(2) Combine trainy and trainx into a data frame for fitting the model.
(2) Use the newdata= argument in the predict() function.
(3) Convert testx from matrix to data frame.
# some
If you provide some example data in reproducible code, I might be able to
help. Otherwise, not much I can do.
Jean
JIMonroe wrote on 10/08/2012 01:17:55 AM:
>
> Jean,
>
> It's definitely bigger now, but my axes are cut-off. As in your
example, I
> had them drawn after generating the heat
It's helpful to provide reproducible code in your posting to R help. The
dput() function can be used to share some of your data. For example, you
might have used
dput(mydata[1:10, 1:10])
# here's some data I made up as an example ...
df <- structure(list(`2000` = c(44L, 31L, 55L, 83L,
DL,
Looks like you have a typo in the expression() function.
You had only one "s".
This works for me ...
m <- 10
beta.q <- matrix(rnorm(81), ncol=9)
for (i in 1:9){
plot(seq(1/m, 1-1/m, 1/m), beta.q[, i], type="l", col=1,
ylim=range(beta.q), xlab="quantile", ylab=expression(beta[i]))
Try this
data.to.analyze$VegIndex <- cut(data.to.analyze$Veg,
breaks=seq(0, 70, 5), include.lowest=TRUE)
plot(data.to.analyze$VegIndex)
Jhope wrote on 10/04/2012 02:25:09 PM:
>
> Hi,
>
> Allow me to recap my question. In plyr I am trying to group distances of
> nests to the vegetati
Jean,
Take a look at the cut() function,
?cut
For example ...
mydf <- data.frame(nest=1:100, d2veg=runif(100, 0, 60))
mydf$dgroup <- cut(mydf$d2veg, breaks=seq(0, 70, 5), include.lowest=TRUE)
head(mydf)
(another) Jean
Jhope wrote on 10/04/2012 02:27:38 AM:
>
> Hi R listers,
>
> I am tryi
Jacob,
Try increasing the size of the pdf. For example, I can read all 919
labels in this plot ...
pdf(width=200, height=200)
plot(1:919, 1:919, axes=FALSE)
axis(1)
axis(2, at=1:919, las=1, cex=0.01)
box()
graphics.off()
Jean
JIMonroe wrote on 10/01/2012 03:42:24 PM:
>
> Hello,
> I have a
You have not specified a nonlinear formula. There are no parameters to
estimate in the formula you provide, y1~dist. What is the nonlinear
relation you are trying to fit? Look at the help file for nls to see some
examples worked.
?nls
Jean
Gyanendra Pokharel wrote on 10/01/2012
10:27:23
If your previously posted code worked with files all having the same
number of columns, and if your data is read in correctly without the
colClasses argument, then I think the following code should work.
for(i in names){
filepath = file.path("~/Silvano/Dados", paste(i, ".raw", sep=""))
Will your data be read in correctly if you do away with the colClasses
argument to read.delim (or read.table)?
Jean
"Silvano Cesar da Costa" wrote on 09/26/2012 09:11:33 AM:
>
> Hi,
>
> I have 35 data files for reading. I would like get a program for
> performing reading of 35 files at once
This might be quicker.
Para.5C.sorted <- Para.5C[order(Para.5C[, 1]), ]
Para.5C.final <- Para.5C.sorted[!duplicated(Para.5C.sorted$REQ.NR), ]
If your data are already sorted by date, then you can skip the first step
and just run
Para.5C.final <- Para.5C[!duplicated(Para.5C$REQ.NR), ]
Jean
ww
Mike,
You could use bootstrapping. See for example the function boot() in
pacakge boot.
library(boot)
?boot
Jean
Michael Eisenring wrote on 09/25/2012 04:54:29
AM:
>
> Dear R-help members.
> Maybe this is not the right platform to ask this, but I'm looking
> desperately for a test which
Try this.
A <- get(x[1])
Jean
Sri krishna Devarayalu Balanagu wrote on
09/17/2012 07:13:51 AM:
>
> a=c(1,2,3)
> b=c(23, 24, 25)
> x=c("a", "b")
> #if (length(x[1]) == 0) {cat("x[1] is having 3 elements")}
>
> Suppose I want to send the vector a into the Object A,
> um getting only "a" as th
Bryan,
Try this.
char <- paste("X", 1:2, sep="", collapse="+")
eval(parse(text=char))
Jean
Bryan Keller wrote on 09/16/2012 11:04:19 PM:
>
> Is it possible to use "paste" to write out an expression and evaluate
it?
> Suppose I want to add two vectors X1 and X2, defined as follows:
>
> X1 <
Michi,
Try putting your par() statements BEFORE your barplot() statement. For
example,
oldpar <- par(mar=c(18, 7, 2, 2.1))
oldfont <- par(font=3)
bp <- barplot(data.matrix(geo_barplot_percentage_top10),
...
Jean
Michael Eisenring wrote on 09/06/2012 12:40:27 PM:
>
> Dear R-help members
Dontrece,
Try using the all.terms argument to the plot.gam() function.
plot(gam4, shade=TRUE, pages=1, all.terms=TRUE)
Does that work?
Jean
Dontrece Smith wrote on 09/06/2012 06:50:07 AM:
>
> Hello,
> My name is Dontrece Smith. I am creating figures for my GAMs. I change
my
> qualitative v
April Lindeman wrote on 09/04/2012 01:43:06 PM:
>
> I am trying to compare Vbert growth curves from several years of
> fish data. I am following the code provided by: http://
> www.ncfaculty.net/dogle/fishR/gnrlex/VonBertalanffy/VonBertalanffy.pdf
> . Specifically the section on VBGM Comparisons
Try this:
# I created some example data
mydf <- data.frame(
Species=sample(paste0("Species", 1:3), 50, TRUE),
d13C=15+rnorm(50),
d15N=15+rnorm(50),
Year=sample(2009:2012, 50, TRUE))
attach(mydf)
Nmean <- tapply(d15N, list(Year,Species), mean)
Cmean <- tapply(d13C
Emil,
It's helpful to provide code that generates your (example) data.
You can use the function dput() to do that.
# I set up your matrix a bit differently than what you presented.
output <- structure(c(-2.3109, -0.2934, -0.5539, -2.1352, -2.6968,
-2.0429,
-0.3973, -0.6468, 0.4181, -0.6243, 2.9
Not sure why that is. You can always use manova() to get the multivariate
summary statistics and then use individual models to get predictions for
each response. For example,
mydata <- data.frame(y1=rnorm(50), y2=rnorm(50), x1=rnorm(50),
x2=rnorm(50), x3=rnorm(50))
myfit <- manova(cbind(y1,
Kev,
The dist() function handles more than 2 dimensions.
Using the example you provided ...
mydat <- structure(list(Species = c("spA", "spB", "spC", "spD"),
x = c(2.9, 5.5, 1.4, 8.3),
y = c(34.2, 46.5, 48.6, 56.1),
z = c(0.54, 0.45, 0.84, 0.48),
n = c(15.7, 19.
Try this.
# the data
y <- sample(0:1, 50, replace=T) # response
x1 <- sample(0:1, 50, replace=T) # explanatory var
x2 <- sample(0:1, 50, replace=T) # explanatory var
# the number of regressors
nregressors <- 2
# all possible combinations
comb <- as.matrix(expand.grid(rep(list(0:1), times=nregr
As Bert suggests, the help file explains the differences between princomp
and prcomp. Look under the Details section
?princomp
?prcomp
# the first PC using princomp
a <- princomp(USArrests, cor=TRUE)
a1 <- a$scores[, 1]
# the first PC using prcomp
# use center=TRUE and scale.=TRUE for the equiv
The functions diversityresult() and diversitycomp() in package
BiodiversityR call the function diversity() from package vegan. The
calculation of the indices are described in the help file.
?vegan:::diversity
Jean
Marte Lilleeng wrote on 08/22/2012 09:28:57 AM:
>
> Hello!
> I am searching
Mitra,
Try this ...
# provide the data you want to plot, and the desired range
mydat <- c(3.34, 3.28, 1.37, 1.12, 3.52, 4.07, 3.66)
myrange <- c(0, 5)
# create a data frame with the max and min as the first two rows
mydf <- data.frame(rbind(max=myrange[2], min=myrange[1], mydat))
# create a rad
Steve,
I tried to replicate your error, but couldn't do it. The code below works
just fine for me. Does it for you?
# create an example data frame
Bridges <- data.frame(CONDITION=1:5, TRAFFIC=rnorm(5), WEATHER=rnorm(5))
# create new variable "LOG_COND"
Bridges$LOG_COND = log(Bridges$CONDITION
Sapana Lohani wrote on 08/16/2012 01:41:23 PM:
>
> Hi,
>
> I have a table in which one column has the name of the objects as shown
below.
You don't provide example data, so I'm not sure what you mean by a
"table". I will assume that you mean a data frame.
> Name
>
> Budlamp-Woodcutter
I suggest that you re-post your question, but this time make it easy for
readers to help you by providing a simple reproducible example of your
problem. Provide some fake data, or a small subset of your data using,
for example, the function dput(). Trim the fat from your code. Get rid
of any
Zuki,
You did not provide any information on what table$attrib is. Using a log
scale for the y-axis doesn't make much sense to me, since the y-axis is
the cumulative proportion of observations, from 0 to 1. Perhaps you are
interested in the log scale for the x-axis, instead? In any case, log
HJ,
You don't provide any reproducible code, so I had to make up my own.
dat <- data.frame(a=letters[1:5], x=c(20110911001084, 20110911001084,
20110911001084, 20110911001084, 20110911001084),
y=c(2.10004e+12, 2.10004e+12, 2.10004e+12, 2.10004e+12,
2.10004e+12))
In my example,
Dominic,
It's great that you provided some example data, but a much smaller data
frame would have sufficed. For example, 10 randomly selected rows from
your data ...
LF <- structure(list(Serra.da.Foladoira = c(27.335652173913,
25.4632608695652,
24.464652173913, 22.550652173913, 22.217782
You would make it much easier for R-help readers to solve your problem if
you provided a small example data set with your code, so that we could
reproduce your results and troubleshoot the issues.
Jean
Naidraug wrote on 08/05/2012 09:08:25 AM:
>
> I've looked everywhere and tinkered for thre
tgodoy wrote on 08/03/2012 12:00:24 PM:
>
> Hi, I'm a new user or R and I try to concatenate a several rows
according
> with the value in a column.
>
> this is my data.frame and I want to concatenate my data.frame according
with
> the column "b" and make a new data.frame with the information
Jason,
In the help file for predict.rpart it says, "The predictors referred to in
the right side of formula(object) must be present by name in newdata."
?predict.rpart
So, that's just the way it is. There are a couple ways to work around
this, if you wish. You could create a data frame with a
Domi wrote on 08/01/2012 Jul 29, 2012; 8:59am:
>
> Hello erverybody,
>
> I have a problem with my second for-loop.
>
> 1. First i read the tables.
>
> datos.mx1 <- read.table('PETmx1.csv',head=TRUE,sep=';')
> datos.min <- read.table('PETmin.csv',head=TRUE,sep=';')
You are most likely to g
"R Heberto Ghezzo, Dr" wrote on 08/01/2012
08:06:30 AM:
>
> Hello,
> I have a big data frame where consecutive time dates and
> corresponding observed values for each subject (ID) are on a line. I
> want to compute the linear slope for each subject. I would like to
> use apply but I do
> not k
You could use the xlim argument to barplot. For example,
barplot(t(data[,2:3]), beside=TRUE, xlim=c(0, 20),
col=c(rgb(.537, .769, .933), rgb(.059, .412, .659)),
width=1, horiz=TRUE, cex.names=.9, border ="white",
las=1, cex.axis=1, cex.lab=1.2)
Jean
Manish Gupta wro
As Petr suggests, the dist() function will do much of the work for you.
For example ...
# example matrix of data
nsamples <- 40
nreadings <- 46
dat <- matrix(runif(nsamples*nreadings), nrow=nsamples)
# Euclidean distance between the ROWS of dat
distance <- dist(dat)
Jean
JenniferH wrote on
Can you provide a small example data set?
Does the number after the X and Y indicate the subject?
Are you trying to come up with a single line of code to do this ... ?
lm(Y1 ~ X1, data=DF)
lm(Y2 ~ X2, data=DF)
...
lm(Y5 ~ X5, data=DF)
Jean
"R Heberto Ghezzo, Dr" wrote on 07/
Frederico,
This is not exactly what you're after, but perhaps it will help. In this
example I fit a cluster analysis to the data, then I "cut the tree" at a
height of 3 (you would do this with your data at a height of 40). It's
not a perfect solution, but it might be good enough, depending on
?is.element
Jean
cm wrote on 07/30/2012 07:55:34 AM:
>
> I need to write an if/then statement saying something along the lines
of:
> if (VALUE is in list) {...
> How do I write that in R's language.
> Thanks!
[[alternative HTML version deleted]]
_
Look at the examples for the functions dist() and hclust(),
?dist
?hclust
Jean
eliza botto wrote on 07/30/2012 01:43:55 AM:
>
> Dear R Users,i am very new to R. I want your help on an issue
> regarding distance matrix and cluster analysis
> i had discharge data of 4 rivers(a,b,c,d) in 4 vecto
Try this:
# define the locations on the x axis for each cell in the image
nxcells <- dim(m)[2]
xlocs <- (seq(nxcells)-1)/(nxcells-1)
# define the groupings
group <- rep(1:3, c(10, 20, 25))
# calculate the range of locations on the xaxis for each group
group.ranges <- sapply(split(xlocs, group),
Philippe,
In your example, you have four unique values for Yrs (I had to change your
code a little to get it to run, so I have the modified version with my
code below), and those values are what you are referring to when you say
T1, T2, T3, T4, right? If I follow what you want to do, the code
use a single explanatory variable. My problem
> is with understanding the model when there are multiple categorical
> explanatory variables.
>
> --
> Joao.
>
> On Fri, Jul 27, 2012 at 1:04 PM, Jean V Adams wrote:
> > Joao,
> >
> > There's a very thorough
Michi,
The code that you shared was incomplete. Two objects (plant_labels and
colors) were not defined.
The pie chart would be easier to interpret by direct labeling of slices,
rather than trying to decipher a color legend. If you change the initial
angle of the first slice, you can rotate t
Joao,
There's a very thorough explanation at
http://www.ats.ucla.edu/stat/r/library/contrast_coding.htm
Jean
Joao Azevedo wrote on 07/27/2012 06:32:31 AM:
>
> Hi!
>
> I'm failing to understand the value of the intercept value in a
> multiple linear regression with categorical values. Taking
You can learn a lot from the help files. Check out the help files for the
lm() and summary.lm() functions
?lm
?summary.lm
You can extract the beta values in a few different ways.
These two will give you just the estimates in a vector:
coef(result)
result$coef
These two will give you the estim
Try changing your plot and outer margins.
par(mfrow=c(2,1), mar=c(4, 4, 2, 1), oma=c(0, 0, 2, 0))
Jean
phillen wrote on 07/26/2012 04:28:58 AM:
>
> hey guys
>
> I want two plots in one window with an overall title and with individual
> titles for each plots.
>
> my code:
>
> par(mfrow=c(2
I am having trouble using the R2wd package. The last time I used it
successfully, I was running an earlier version of R and an earlier version
of Word with an earlier Windows OS. I'm not sure which if any of these
changes might be contributing to the problem. Right now I'm using
R versio
Bert,
Is it important that you end up with a data frame? If not, it would be
very easy to generate a list with the unique values for each column. For
example:
df <- data.frame(v1 = sample(5, 20, T), v2 = sample(7, 20, T),
v3 = sample(9, 20, T), v4 = sample(11, 20, T))
lapply(df, uniq
Did you try this?
image(x, xlim=c(0, 100), ylim=c(0, 100))
Jean
li li wrote on 07/22/2012 09:28:33 PM:
> Dear all,
>I have a question regarding changing the xlim and ylim in the
function
> image().
>For example, in the following code, how can I have a heatmap with
> xlim=ylim=c(0, 100
ast 24 hours right now.
>
> Optimizing a program is not my thing.
>
> Sent from my Verizon Wireless 4GLTE smartphone
It's not mine either.
Better to ask the group.
I'm ccing R-help on this message.
Jean
> "Jean V Adams" wrote on 07/20/2012 10:05 AM:
>
>
ast 24 hours right now.
>
> Optimizing a program is not my thing.
>
> Sent from my Verizon Wireless 4GLTE smartphone
It's not mine either.
Better to ask the group.
I'm ccing R-help on this message.
Jean
> - Reply message -
> From: "Jean V Adams"
Michael Eisenring wrote on 07/20/2012 09:35:03 AM:
> Dear Jean,
> thanks for this email. I'm grateful for this instruction Just to
> make sure that I understood you properly: something like this?:
Michael,
Yes, this is perfect. Very helpful.
> My data:
>
> dput(geo_anova_nested_input):
>
Petr,
This is great.
MUCH faster than the code I provided.
And much more elegant code.
Thanks for posting!
Jean
Petr Savicky wrote on 07/20/2012 09:26:34 AM:
> On Fri, Jul 20, 2012 at 05:45:30AM -0700, wwreith wrote:
> > General problem: I have 20 projects that can be invested in and I need
Michael,
Use dput() to output your data (or perhaps a small subset). Then paste
the result of that call and your R code (just those lines of code that are
needed to reproduce the problem) right in your message to R-help. That
makes it easier for the R-help list readers to help you troubleshoo
I've had to do something similar, so I wrote a small function to help.
This runs in about 1/4 the time of your code on my machine.
Others may have a more efficient approach.
all.combs <- function(num, from=0, to=num) {
# create a matrix of all possible combinations of num items
# r
As Duncan Murdoch pointed out, the example data frame that you provided
doesn't give very interesting results (all the shock values are zero), so
I created a different shock variable for illustration. I suggest using
the interp() function in the R package akima.
df <- structure(list(c = 1:6, z
.6925
> x -0.11826047 0.09731719 89 -1.2152063 0.2275
> Correlation:
> (Intr)
> x 0.056
>
> Standardized Within-Group Residuals:
>Min Q1Med Q3Max
> -2.0882452 -0.7718563 0.1156507 0.6119178 1.7986478
>
What is the function index() that you use in this line of code?
thisIndex <- as.character(index(diveData_2008[i,]))
Is it from some package? Or a function you wrote yourself?
I'm trying to run the code you submitted, but I don't have a function
called index().
Jean
Yolande Tra wrote on 07/18
I'm not sure why, but lme() doesn't seem to like the variables to be
referenced as part of a list using [ or $.
Here's an easy workaround ...
ids <- a$id
for(i in 2:4){
for(j in 5:7){
y <- a[, j]
x <- a[, i]
lme(y ~ x , random= ~1|ids, na.action="na.exclude")
}}
J
# MCT ordered by the second column of MCT, from smallest to largest values
MCT[order(MCT[, 2]), ]
Jean
Trying To learn again wrote on 07/17/2012
04:42:24 PM:
> Hi all,
>
> I want to order a series that is included on the second column in
MCT.csv.
>
> I do but R doesn´t order, could be beca
seq(residuals)[abs(residuals) > (2*sd)]
gcm wrote on 07/16/2012 07:39:20 AM:
> But I'll need the indexes and that gives me the values. I need to
> then be able to make the indexes into a list of numbers.
>
> From: Jean V Adams [via R] [
mailto:ml-node+s789695n4636641...@n
residuals[abs(residuals) > (2*sd)]
gcm wrote on 07/16/2012 07:07:31 AM:
> Jean, how would you modify your function to grab all the selected
> and not just the first?
> Thank you so much!
>
> From: Jean V Adams [via R] [
mailto:ml-node+s789695n4636370...@n4.nabble.com]
>
The example you gave had only one split. If your real situation has three
splits, you'll have to take a look at testtree$csplit matrix and decide
how you want to define the new grouping variable. Here's one way to do it
...
Jean
library(rpart)
library(rpart.plot)
test_set <- data.frame(
I wrote a little function called first() to help with situations like
this. It returns a 1 every time an element of a vector is different from
the previous element, and a 0 otherwise.
first <- function(x) {
L <- length(x)
c(1, 1-(x[-1]==x[-L]))
}
sd <- 1
residuals <- c(1, 2.1, 3, 4
Try using the rbind() function to combine the two vectors from colSums()
into a matrix. Then assign row names and get rid of column names using
the dimnames() function. For example:
OLDMatrix <- matrix(c(22, 25, 10, 24, 27, 13, 23, 27, 15),
ncol=3, dimnames = list(NULL, c("X1", "X2",
Thanks. Very helpful.
You can use the information from the splits in the first tree, to define a
new grouping variable, which will simplify the plot:
suvar <- sort(unique(test_set$list_var))
test_set$var_grp <- as.factor(testtree$csplit[match(test_set$list_var,
suvar)])
testtree2 <- rpart ( lis
If you provide a simple example, with actual data and code so that we can
reproduce your issue, it would make it easier for readers of this list to
help.
Jean
MarkBeauchene wrote on 07/06/2012 04:38:32 PM:
> I have a class with 732 members, so using rpart.plot is giving me a tiny
plot
> in
request:
> "Non-members are not allowed to post messages to this list.
> If you think you find a bug (think twice and then) maybe ask once more
> on R-help then use the R bugs repository "
Mateus Teixeira wrote on 07/06/2012 09:07:27
AM:
> Thank you, Jean.
>
> Mate
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