I'm not exactly sure how you want the file to look, but try this code and
see if it works for you.
# since I don't have the data file, I'll just set the values here
No_GWPMax <- 8
NbpolicyClass1 <- 5
NbpolicyClass2 <- 4
NbpolicyClass3 <- 3
# output1 is the first row of headers
Output1 <- paste("
two matrices, I want to end up with this list:
result <- list(c("A", "b", "D", "f"), c("C", "E"), c("D", "e"))
result
I'm using R for Windows version 2.9.0.
Thanks for your help.
Jean
`·.,, ><(((º&
(I'm resending this, because there seemed to be a problem with my previous
attempt.)
I've been scratching my head over this one for too long. I'm hoping
someone out there can solve this riddle.
I have two vectors of characters, v1 and v2, both of length L, and two
matrices of logicals, m1 and
Just wait until after you have the forecasts before you create the plot.
# Sample dates
xValues <- seq.Date(as.Date("1990-01-31"), to=as.Date("1992-12-31"),
by="month")
# Sample y value
yValues <- seq(0.1, length=length(xValues))
# Sample forecast one year from xValue's end point
fcastDate <- s
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 <
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
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
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
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
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=""))
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
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
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
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
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]))
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,
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
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
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.
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
>
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.
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
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
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).
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$resid, m
ion(x) plot(x, fit$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
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
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
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
:
> 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
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
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
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
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
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
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
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
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,
>
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
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,
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
;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
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 =
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)
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
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
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&#
Alison,
Check out the options for the function bxp(), they include control over
the colors of all parts of the boxplot, e.g., whiskcol for whisker color.
Jean
`·.,, ><(((º> `·.,, ><(((º> `·.,, ><(((º>
Jean V. Adams
Statistician
U.S. Geological Survey
Great La
at <- c(50.50, 52.35, 54.12, 54.91, 53.17, 52.21, 54.06, 53.61, 50.71,
50.71, 50.50)
# use the map() function to set up the long/lat projection without drawing
anything
map("world", xlim=range(long), ylim=range(lat), type="n")
lines(long, lat)
Jean
`·.,, ><(((º> `·
Try this:
q <- z[match(p, y)]
Jean
`·.,, ><(((º> `·.,, ><(((º> `·.,, ><(((º>
Jean V. Adams
Statistician
U.S. Geological Survey
Great Lakes Science Center
223 East Steinfest Road
Antigo, WI 54409 USA
715-627-4317, ext. 3125 (Office)
715-216-8014 (
j] - mat[i+1, j]
c(c1, c2, c3)
}
compare(a, 1, 1)
compare(a, 1, 2)
Jean
`·.,, ><(((º> `·.,, ><(((º> `·.,, ><(((º>
Jean V. Adams
Statistician
U.S. Geological Survey
Great Lakes Science Center
223 East Steinfest Road
Antigo, WI 54409 USA
From:
kokavolchko
2_mrna,
data=Data_pp2_mrna, start=list(k1=3.3, v2_Kd=2.4, v2_h=1,
v5_Kd=1.2, v5_h=1))
summary(fm_pp2_mrna)
Jean
`·.,, ><(((º> `·.,, ><(((º> `·.,, ><(((º>
Jean V. Adams
Statistician
U.S. Geological Survey
Great Lakes Science Center
223 East Steinfest Road
Ant
A search of the R-help archives may provide some help.
See, for example,
http://r.789695.n4.nabble.com/including-tabular-information-with-a-plot-in-R-Graphics-tt885431.html
Jean
`·.,, ><(((º> `·.,, ><(((º> `·.,, ><(((º>
Jean V. Adams
Statistician
U.S. Ge
1, outer=TRUE, text=smry2, cex=0.75, line=9, adj=0,
family="mono")
Jean
`·.,, ><(((º> `·.,, ><(((º> `·.,, ><(((º>
Jean V. Adams
Statistician
U.S. Geological Survey
Great Lakes Science Center
223 East Steinfest Road
Antigo, WI 54409 USA
From:
Greg S
1 0.2 setosa
45 32 0.2 setosa
55 41 0.2 setosa
65 42 0.4 setosa
Jean
`·.,, ><(((º> `·.,, ><(((º> `·.,, ><(((º>
Jean V. Adam
rdiellipse(mod, Management, kind="se", conf=0.95, label=T,
font=2, cex=1.5, col=i, show.groups=groupz[i])
}
Jean
`·.,, ><(((º> `·.,, ><(((º> `·.,, ><(((º>
Jean V. Adams
Statistician
U.S. Geological Survey
Great Lakes Science Center
223 East S
from T to X, because T has
special meaning in R.
Jean
`·.,, ><(((º> `·.,, ><(((º> `·.,, ><(((º>
Jean V. Adams
Statistician
U.S. Geological Survey
Great Lakes Science Center
223 East Steinfest Road
Antigo, WI 54409 USA
From:
"Johannes Radinger"
To:
r-hel
ecessary."
There is an argument in read.csv() called check.names. Try setting this
to FALSE and see if that works.
read.csv(file="Something.csv", check.names=F)
Jean
`·.,, ><(((º> `·.,, ><(((º> `·.,, ><(((º>
Jean V. Adams
Statistician
U.S. G
Following the suggestion by Duncan Murdoch, this should work for you.
X <- runif(length(lT), lT, uT)
Jean
From:
"Johannes Radinger"
To:
Jean V Adams
Cc:
r-help@r-project.org
Date:
08/10/2011 08:40 AM
Subject:
Re: [R] function runif in for loop
Jean,
thank you for your answe
ound(y, 2) else y))
Jean
`·.,, ><(((º> `·.,, ><(((º> `·.,, ><(((º>
Jean V. Adams
Statistician
U.S. Geological Survey
Great Lakes Science Center
223 East Steinfest Road
Antigo, WI 54409 USA
Liviu Andronic wrote on 08/10/2011 10:26:43 AM:
> [image removed]
uot;, dimnames(m)[[1]], fill=seq(mut.no),
title="Mutation_Status")
Jean
`·.,, ><(((º> `·.,, ><(((º> `·.,, ><(((º>
Jean V. Adams
Statistician
U.S. Geological Survey
Great Lakes Science Center
223 East Steinfest Road
Antigo, WI 54409 USA
r-help-boun..
unlist(x)
r-help-boun...@r-project.org wrote on 08/10/2011 01:58:57 PM:
> [image removed]
>
> [R] convert 'list' to 'vector'?
>
> Liviu Andronic
>
> to:
>
> r-help@r-project.org Help
>
> 08/10/2011 02:02 PM
>
> Sent by:
>
> r-help-boun...@r-project.org
>
> Dear all
> How does one conver
Shouldn't the "i" in your theta() function refer to the selected rows (a
"vector of indices" as referred to in the help file for boot) of the data
used by boot()?
theta <- function(data, i) {
data <- data[i, ]
data.cov <- cov(data)
data.eigen <- eigen(data.cov)
data.eigen$values[1]/sum(d
uld you want your merged data frame to
look?
>
> Does anyone have ideas about how to accomplish this?
> Thank you,
>
> Matthew Keller
>
> --
> Matthew C Keller
> Asst. Professor of Psychology
> University of Colorado at Boulder
> www.matthewckeller.com
>
Je
ince the two sides are identical, there is little value in having
> both displayed at the same time. Moreover, it considerably slows down
> the inspection of the results.
>
> Thank you
> Liviu
>
Jean
`·.,, ><(((º> `·.,, ><(((º> `·.,, ><(((º
> [R] Increase the size of the boxes but not the text in a legend
> Jürgen Biedermann
> to:
> r-help
> 08/21/2011 06:02 PM
>
> HI there,
>
> I want to add a legend to a plot using the density and angle argument,
> so patterns with lines in different angles are used in the plot and
> should be
> [R] Selecting cases from matrices stored in lists
> mdvaan
> to:
> r-help
> 08/22/2011 07:24 AM
>
> Hi,
>
> I have two lists (c and h - see below) containing matrices with similar
> cases but different values. I want to split these matrices into multiple
> matrices based on the values in h. So
> Re: [R] Selecting cases from matrices stored in lists
> mdvaan
> to:
> r-help
> 08/22/2011 09:46 AM
>
> Jean V Adams wrote:
> >
> >> [R] Selecting cases from matrices stored in lists
> >> mdvaan
> >> to:
> >> r-help
> >>
> [R] Counting Elements Conditionally
> Edward Patzelt
> to:
> r-help
> 08/22/2011 02:33 PM
>
> R -
>
> I have 3 variables with data below. Variable "Rev" is a vector that
changes
> from 1 to 2, 2 to 3, etc Variable "FF" is a binary variable with
1's
> and 0's. Variable "bin" is a diffe
> Re: [R] Counting Elements Conditionally
> Jean V Adams
> to:
> Edward Patzelt
> 08/22/2011 03:53 PM
>
> > [R] Counting Elements Conditionally
> > Edward Patzelt
> > to:
> > r-help
> > 08/22/2011 02:33 PM
> >
> > R -
> >
&
So, using the full data set, what should the result look like?
c(NA, NA, NA, 3, NA,NA, NA, 2) ?
Jean
Edward Patzelt wrote on 08/22/2011 03:58:38 PM:
> [image removed]
>
> Re: [R] Counting Elements Conditionally
>
> Edward Patzelt
>
> to:
>
> Jean
that is exactly correct, assuming we did not start at the beginning,
> but started at the first transition (this is the correct way to
> think about it)
> On Mon, Aug 22, 2011 at 4:08 PM, Jean V Adams wrote:
>
> So, using the full data set, what should the result lo
ashz wrote on 08/23/2011 03:25:57 AM:
>
> Hi,
>
> I am using this script to read a xlsx file to a data frame:
> library(xlsx)
> File <- file.path("d:", "car ", "car95-99.xlsx")
> B_car <- read.xlsx(File, "raw_data")
> Car2x <- data.frame(month = B_car$Date,Ch = B_car$Ch.des,
> lat=B_car$Latitude
So, you are looking for confidence intervals for each factor level?
You can use the predict() function to do that.
fit <- aov(values ~ ind, data=otestme)
newdat <- data.frame(ind=factor(levels(otestme$ind)))
cbind(newdat, predict(fit, newdata=newdat, interval="confidence"))
Jean
Anna Dunietz wr
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 nu
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" whi
uot;8630.11"
> [25] "8803.11" "9186.11" "9453.11" "10132.11" "10669.21" "10720.61"
> [31] "10755.13" "11326.11" "11440.13" "11486.11" "11508.11" "11711.11"
> [37] "11
)
activeR <- append(activeR, lastR)
}
Jean
> 2011/8/24 Jean V Adams
>
> 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
You could try using the numeric representation of date, and split the data
frame using that variable. For example:
src$date.num <- as.numeric(src$date)
Jean
Franc Lucas wrote on 08/24/2011 02:42:58 PM:
>
>Hello everyone,
>I want to split a data.frame by the column date . The data fram
Try this:
require(zoo)
lvd <- tapply(df$visit_date, df$unique_id, max)
index <- tapply(df$visit_date, df$unique_id)
df$last_visit_date <- as.Date(lvd[index])
Jean
Kathleen Rollet wrote on 08/24/2011 04:15:45 PM:
>
> Dear R users,
>
> I am encoutering the following problem: I have a dataset wit
I'm assuming that you meant to write
s <- smooth.spline(x, y)
You can look at the code for the predict method for smooth.spline by
typing
getAnywhere(predict.smooth.spline)
If a new value is provided (i.e., the 500 in your example) then the
predict method for smooth.spline.fit is applied to t
>From the help for barplot
?barplot
The argument "... space may be specified by two numbers, where the first
is the space between bars in the same group, and the second the space
between the groups. If not given explicitly, it defaults to c(0,1) if
height is a matrix and beside is TRUE ..."
S
Have you tried searching the R-Help archives?
I found some answers to a similar question by Googling
r multivariate uniform distribution
Jean
Soberon Velez, Alexandra Pilar wrote on 08/25/2011 04:15:11 AM:
>
> Hello,
>
>
>
> I want to create two random variables (x1,x2) both with unifo
It's not clear to me what problem you're having.
tan(data$slope)
should work. As should
log(data$uparea/data$slope)
I suggest that you provide a small subset of example data along with an
example of any code you've tried, and the output that you'd like.
Jean
Tom Vanwalleghem wrote on
I don't see a quick solution to this.
You could contact the maintainer of the rpart.plot package, Stephen
Milborrow
maintainer("rpart.plot")
or you could try to modify the rpart.plot() function yourself to meet your
needs
rpart.plot
Jean
Jay wrote on 08/25/2011 05:30:25 AM:
>
>
Check out the maps package for a choropleth map. Specifically the map()
function (and its argument col=).
?map
Check out the image() function for a heat map which can be overlaid.
?image
Jean
Reza Salimi-Khorshidi wrote on 08/25/2011 09:00:04 AM:
>
> Hi all,
> I would like to use
Try the dir() function.
?dir
# for example
dir("c:/", pattern="foo.pdf", full.names=T, ignore.case=T, recursive=T)
Jean
Tyler Rinker wrote on 08/25/2011 11:54:28 AM:
>
> I am not a programmer and am self-taught so I may lack the
> language to ask this appropriately (perhaps why an rseek searc
for the US + a couple of
> other country that unfortunately UK is not one of them. What I want
> is a solution that takes a list of coordinates/postcodes/cities and
> a list of values and gives me a colourful UK map. Any thoughts?
> Cheers
> On Thu, Aug 25, 2011 at 3:50 PM, Jean
By the "null distribution" do you mean that the assignment of each
observation to a column is equal? If so, the function sample() might
serve your needs. For example:
rows <- 3
cols <- 4
rowtot <- 100
m <- matrix(NA, nrow=rows, ncol=cols)
for(i in seq(rows)) {
m[i, ] <- tabulate(samp
Try this
data <- eval(parse(text=paste(study, level, ".", population, sep="")))
Jean
-
dbateman wrote on 08/31/2011 17:44:44:
I have several datasets that come from different studies (fv02 and fv03),
they represent different levels (patients and lesions), and they have
different patient
Anna Dunietz wrote on 09/02/2011 07:16:45 AM:
>
> Hi All!
>
> Please find code and the respective lists below. My problem: I specify
the
> case that lilwin[[p]] is not an NA and want the code found in iwish to
be
> returned ONLY for that case. Why do I get a list of length 2 (and why
is
> NU
Look at the function daisy in the package cluster.
require(cluster)
?daisy
Jean
Lorenzo Isella wrote on 09/02/2011 11:50:04 AM:
>
> Dear All,
> I will be confronted (relatively soon) with the following problem:
> given a set of known statistical indicators {s_i} , i=1,2...N for a N
> countries
Try the table() function.
?table
For example,
df <- data.frame(real=sample(0:1, 20, replace=T), pred=sample(0:1, 20,
replace=T))
table(df)
pred
real 0 1
0 3 7
1 4 6
Jean
Doussa wrote on 09/02/2011 08:46:42 PM:
>
> hi users
> I have a data frame in with there are two colomns real
Try this:
funa <- function(n, y, a, rate, samp) {
lambda <- a * n
dexp(n, rate) * do.call(paste("d", samp, sep=""), y, lambda)
}
funb <- function(y, a, rate, samp) {
integrate(f1, 0, Inf, y, a, rate)
}
funb(1, 0.1, 0.1, "pois")
Jean
>
> Hello guy
Varsha Agrawal wrote on 09/07/2011 05:18:10 AM:
>
> The code looks like this:
> L1=list(a=1,b=2,c=3)
> f1=as.factor(c)
> L1[[f1]] returns 1
>
> What happens if we give a factor as an index at a list?
>
L1=list(a=1,b=2,c=3)
f1=as.factor(L1$c)
L1[[f1]]
When you use a factor (e.g., f1, corrected
If you read the help file on the step() function
?step
you will see a reference to BIC under the description of the k= argument.
This suggests that you could try:
BIC.fitted = step(glm.fit, k=log(dim(dat)[1]))
Jean
Andra Isan wrote on 09/07/2011 06:12:19 PM:
>
> Hi All,
> After fitt
francogrex wrote on 09/08/2011 08:08:19 AM:
>
> Hi, anyone has experience with 3D plot (ex: in package RGL) I have a
> question, I draw a 3D plot of country, year and sales in z axis but when
the
> type is "h" then it's ok but when I want to link the points and type is
'l'
> lines it's a mess Is
Mohan L wrote on 09/08/2011 12:35:18 PM:
>
> Hi All,
>
> I have txt file like :
>
> $ cat data.txt
> US 10
> UK 12
> Ind 4
> Germany 14
> France 8
>
> > rawdata <- read.table(file='data.txt',sep='\t' , header=FALSE)
>
> > rawdata
>V1 V2
> 1 US 10
> 2 UK 12
> 3 I
Rainer provided an example of subsetting by the value of a variable in the
data frame. Below is an example of subsetting by the value of the row
name of the data frame.
df <- data.frame(var1=1:10, var2=letters[1:10], var3=sample(10),
row.names=month.abb[1:10])
subset(df, subset = row.name
maxbre wrote on 09/09/2011 06:28:15 AM:
>
> This is my reproducible example:
>
> example<-structure(list(SENSOR = structure(1:6, .Label = c("A", "B",
"C",
> "D", "E", "F"), class = "factor"), VALUE = c(270, 292.5, 0, 45,
> 247.5, 315), DATE = structure(1:6, .Label = c(" 01/01/2010 1",
> " 01/
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