> x <- lapply(1:100, function(x) summary(runif(100)))
> head(x, 4)
[[1]]
Min. 1st Qu. MedianMean 3rd Qu.Max.
0.02922 0.38330 0.58120 0.58230 0.83430 0.99870
[[2]]
Min. 1st Qu. Median Mean 3rd Qu. Max.
0.004903 0.281400 0.478900 0.497100 0.729900 0.990700
[[3]]
The first problem is that you are using a character string as the first
argument to agnes()
The help information for agnes says that its first argument, x, is
x: data matrix or data frame, or dissimilarity matrix, depending
on the value of the 'diss' argument.
Not a character
Just to start things off:
> var.name <- c("gdp","inf","unp")
> var.id <- c("w","i")
>
> x <- paste(var.name, rep(var.id, each=length(var.name)), sep="_")
> x
[1] "gdp_w" "inf_w" "unp_w" "gdp_i" "inf_i" "unp_i"
>
Now the three differences:
gdp_w - gdp_i
inf_w - inf_i
unp_w - unp_i
Can be got u
Package XLConnect appears to provide this kind of thing.
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Shi, Tao
Sent: Friday, 24 June 2011 2:42 PM
To: r-help@r-project.org
Subject: [R] extract worksheet names from an Excel file
The point I would make is that for safety it's much better to use FALSE rather
than F. FALSE is a reserved word in R, F is a pre-set variable, but can easily
be changed at any time by the user.
Secondly, doesn't this do the same as yours:
readFF.csv <- function(..., stringsAsFactors = FALSE)
..or something like that. Without more details it is hard to know just what is
going on.
Firstly in R the object is a 'data frame' (or object of class "data.frame" to
be formal). There is no standard object in R called a 'database'.
If you read in your data using read.csv, then mydata is g
> con <-
+ textConnection("13053 13068 13068 13053 14853 14853 14850 14850 13053 13053
13068 13068
+ ")
> x <- scan(con)
Read 12 items
>
> cut(x, 4)
[1] (1.31e+04,1.35e+04] (1.31e+04,1.35e+04] (1.31e+04,1.35e+04]
[4] (1.31e+04,1.35e+04] (1.44e+04,1.49e+04] (1.44e+04,1.49e+04]
[7] (1.44e+04,1.
Suppose
names(xm1) <- c("alpha", "beta", "gamma", "delta")
then
xm2 <- subset(xm1, select = alpha:gamma)
or
xm2 <- subset(xm1, select = -delta)
will do the same job as xm1[, -4]
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of
The way to guarantee a specific number of panels in the histogram, say n, is to
specify n+1 breaks which cover the range of the data. As far as I know this is
the only way.
Bill Venables.
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Be
The advice is always NOT to use Microsoft Word to edit an R file. That stuff
is poisonous. Microsoft word, typical of all Microsoft software, does not do
what you tell it to do but helpfully does what it thinks you meant to ask it to
do but were too dumb to do so.
Even notepad, gawdelpus, wou
If I understand you correctly, you are trying to find the cumulative maximum
from the end within each level of the factor. If this is what you are trying
to do, then here is one way you might like to do it. First, define the
function:
> cumMax <- function(x) Reduce(max, x, right = TRUE, accum
The two commands you give below are certain to lead to very different results,
because they are fitting very different models.
The first is a gaussian model for the response with a log link, and constant
variance.
The second is a gaussian model for a log-transformed response and identity
link.
Here ia an idea that might be useful to adapt
fixedSumCombinations <- function(N, terms)
if(terms == 1) return(N) else
if(terms == 2) return(cbind(0:N, N:0)) else {
X <- NULL
for(i in 0:N)
X <- rbind(X, cbind(i, Recall(N-i, terms-1)))
X
}
Neville's algorithm is not an "improvement" on Lagrange interpolation, it is
simply one way of calculating it that has some useful properties. The result
is still the Lagrange interpolating polynomial, though, with all its flaws.
Implementing Neville's algorithm is fairly easy using the Polynom
The score test looks at the effect of adding extra columns to the model matrix.
The function glm.scoretest takes the fitted model object as the first argument
and the extra column, or columns, as the second argument. Your x2 argument has
length only 3. Is this really what you want? I would h
Here is a one way.
> tab
fm
0 to 5 11.328000 6.900901
15 to 24 6.100570 5.190058
25 to 34 9.428707 6.567280
35 to 4410.462158 7.513270
45 to 54 7.621988 5.692905
5 to 14 6.502741 6.119663
55 to 64 5.884737 4.319905
65 to 74 5.075606 4.2
Here is one way you might do it.
> con <- textConnection("
+ characteristics_ch1.3 Stage: T1N0 Stage: T2N1
+ Stage: T0N0 Stage: T1N0 Stage: T0N3
+ ")
> txt <- scan(con, what = "")
Read 11 items
> close(con)
>
> Ts <- grep("^T", txt, value = TRUE)
> Ts <- sub("T([[:digit:]]+)N([[:digit:]]+)",
If you want to go ahead with this in cold blood, you might look at the 'nnls'
package.
It fits regressions with non-negative coefficients. This might seem like the
very opposite of what you want, but it essentially gets you there. You have to
be prepared for the coefficient to go to zero th
I thought so to. If so, here is one way you could do it
fixSeq <- function(state) {
shift1 <- function(x) c(1, x[-length(x)])
repeat {
change <- state %in% c(4,5) & shift1(state) == 3
if(any(change))
state[change] <- 3 else break
}
state
}
e.g.
> state
[1] 1 3 3 5 5 3 2
Perhaps because the timezone is specified as a character string and not a
date-time object complete with timezone.
>From the help filr for as.POSIXct.numeric:
"origin:a date-time object, or something which can be coerced by
as.POSIXct(tz="GMT") to such an object."
Note the coercion.
This is really a question about the help file for gl.
The arguments are
gl(n, k, length = n*k, labels = 1:n, ordered = FALSE)
'n' is the number of factor levels. That seems to be easy enough
'k' is called the "number of replications". This is perhaps not the best way
to express what it is.
There is an urban legend that says Indiana passed a law implying pi = 3.
(Because it says so in the bible...)
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Joshua Wiley
Sent: Monday, 30 May 2011 4:10 PM
To: Vincy Pyne
Cc: r-he
You have received suggestions about this already, but you may want to consider
something like this as an alternative:
> require(english)
> lev <- as.character(as.english(0:9))
> dat <- data.frame(f = factor(sample(lev, 500,
+ rep=TRUE), levels = lev),
+ B = rno
The key line is
prep <- .checkinput(match.call(), parent.frame())
Among other things the model matrix is built in .checkinput( ) which is not
exported from the package namespace. So you have to get rough with it and use
penalized:::.checkinput
and then you see these line of code
Oops
The first line of my template should use data.matrix() rather than data.frame()
data.matrix() is guaranteed to return a numerical matrix from a data frame,
making arithmetic always possible.
Bill Venables.
From: Venables, Bill (CMIS, Dutton Park)
S
For that kind of operation (unusual as it is) work with numeric matrices. When
you are finished, if you still want a data frame, make it then, not before.
If your data starts off as data frame to begin with, turn it into a matrix
first. E.g.
myMatrix <- data.frame(myData)
myMatrix2 <- myMatr
That only applies if you have the same factors a and b each time. If this is
the case you can do things in a much more slick way.
u <- matrix(rnorm(5000), nrow = 10) ## NB, nrow
AB <- expand.grid(a = letters[1:2], b = letters[1:5])
M <- lm(u ~ a+b, AB)
rmsq <- colSums(resid(M)^2)/M$df.resid
an
>>> Martin Maechler writes:
>
> Well, then you don't know *THE ONE* case where modern users of
> R should use attach() ... as I have been teaching for a while,
> but seem not have got enought students listening ;-) ...
>
> --- Use it instead of load() {for save()d R objects} ---
>
> Th
It depends on how big A is and how much memory you have. Here is one lazy way.
A <- aperm(aperm(A, c(2,1,3,4)) + x, c(2,1,3,4))
Bill Venables
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of huron
Sent: Thursday, 19 May 2011 1:47
I used to think like that. However I have recently re-read John Chambers'
"Software for Data Analysis" and now I'm starting to see the point.
S4 classes and methods do require you to plan your classes and methods well and
the do impose a discipline that can seem rigid and unnecessary. But I ha
Hi Bert,
I think people should know about the Google Sytle Guide for R because, as I
said, it represents a thoughtful contribution to the debate. Most of its
advice is very good (meaning I agree with it!) but some is a bit too much (for
example, the blanket advice never to use S4 classes and m
If you
?plot.stl
you will see that that the second argument, set.pars, is a list of argument
settings for par(), including a (variable) default setting for mfrow. I.e.
plot.stl overrides your external setting (which will also override any layout()
setting).
It looks like to override it ba
The documentation for simprof says, with respect to the method.distance
argument, "This value can also be any function which returns a "dist" object."
So you should be able to use the Jaccard index by setting up your own function
to compute it. e.g.
Jaccard <- function(X) vegan::vegdist(X, me
PS I should have followed the example with one using with() for something that
would often be done with attach(): Consider:
with(polyData, {
plot(x, y, pch=".")
o <- order(x)
lines(x[o], eta[o], col = "red")
})
I use this kind of dodge a lot, too, but now you can mostly use data= argument
Amen to all of that, Bert. Nicely put. The google style guide (not perfect,
but a thoughtful contribution on these kinds of issues, has avoiding attach()
as its very first line. See
http://google-styleguide.googlecode.com/svn/trunk/google-r-style.html)
I would add, though, that not enough pe
?relevel
Also, you might want to fit the models as follows
Model1 <- glm.nb(Cells ~ Cryogel*Day, data = myData)
myData2 <- within(myData, Cryogel <- relevel(Cryogel, ref = "2"))
Model2 <- update(Model1, data = myData1)
&c
You should always spedify the data set when you fit a model if at all p
Here is an alternative solution
> foo <- array(data = rnorm(32), dim = c(4,4,2),
+ dimnames=list(letters[1:4], LETTERS[1:4], letters[5:6]))
>
> ind <- which(foo > 0, arr.ind = TRUE)
> row.names(ind) <- NULL ## to avoid warnings.
>
> mapply("[", dimnames(foo), data.frame(ind))
[,1] [,2] [,
Here is one way:
df <- data.frame(Value = rnorm(30),
Group = sample(c('A','B','C'), 30,
replace = TRUE))
## make a little function to do the job
iNumber <- function(f) {
f <- as.factor(f)
X <- outer(f, levels(f), "==")+0
rowSums(X * apply(X, 2, cumsum))
}
You could use a function to do the job:
withinRange <- function(x, r = quantile(x, c(0.05, 0.95)))
x >= r[1] & x <= r[2]
dtest2 <- subset(dftest, withinRange(x))
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Robert A'gata
S
You probably want to do something like this:
> fm <- lm(y ~ x, MD)
> anova(fm)
Analysis of Variance Table
Response: y
Df Sum Sq Mean Sq F valuePr(>F)
x 2250 125.0 50 1.513e-06
Residuals 12 30 2.5
Answers to questions:
1. No.
2. Yes.
(whoever you ar
The residual deviance from a multinomial model is numerically equal (up to
round-off error) to that you would get had you fitted the model as a surrogate
Poisson generalized linear model. Here is a short demo building on your example
> set.seed(101)
> df <- data.frame(f = sample(letters[1:3], 5
The two models you fit are quite different. The first is a binomial model
equivalent to
fm <- glm(I(y == "a") ~ x, binomial, df)
which you can check leads to the same result. I.e. this model amalgamates
classes "b" and "c" into one.
The second is a multivariate logistic model that considers
Isn't all you need read.fwf?
From: r-help-boun...@r-project.org [r-help-boun...@r-project.org] On Behalf Of
Kalicin, Sarah [sarah.kali...@intel.com]
Sent: 06 April 2011 09:48
To: r-help@r-project.org
Subject: [R] Pulling strings from a Flat file
Hi,
I hav
How about simply
> df <- data.frame(id = 1:6,
+ xout = c(12.34, 21.34, 2.34, 4.56, 3.24, 3.45),
+ xin = c(NA, 34,67,87,34, NA))
>
> with(df, c(NA, xin[-1]/xout[-length(xout)]))
[1]NA 2.755267 3.139644 37.179487 7.456140NA
>
BTW You seem to
dat <- within(dat, {
X2 <- ifelse(X2 > 50, 100-X2, X2)
X3 <- ifelse(X3 > 50, 100-X3, X3)
})
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of joe82
Sent: Tuesday, 22 March 2011 7:40 AM
To: r-help@r-project.org
Subjec
You might try
dat$F1 <- format(as.Date(dat$F1), format = "%b-%y")
although it rather depends on the class of F1 as it has been read.
Bill Venables.
(It would be courteous of you to give us yor name, by the way.)
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-bo
That's not the point. The point is that R has functions which have
side-effects and hence does not meet the strict requirements for a functional
language.
-Original Message-
From: ONKELINX, Thierry [mailto:thierry.onkel...@inbo.be]
Sent: Monday, 21 March 2011 7:20 PM
To: russ.abb...@g
The idiom I prefer is
pH <- structure(c(4.5,7,7.3,8.2,6.3),
names = c('area1','area2','mud','dam','middle'))
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Gabor Grothendieck
Sent: Sunday, 20 March 2011 2:33 PM
T
subset(data, grepl("[1-5]", section) & !grepl("0", section))
BTW
grepl("[1:5]", section)
does work. It checks for the characters 1, :, or 5.
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Kang Min
Sent: Thursday, 17 March 20
It doesn't work (in R) because it is not written in R. It's written in some
other language that looks a bit like R.
> t <- 3
> z <- t %in% 1:3
> z
[1] TRUE
> t <- 4
> z <- t %in% 1:3
> z
[1] FALSE
>
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-proje
Firstly, the way you have constructed your data frame in the example will
convert everything to factors. What you need to do is actually a bit simpler:
###
dum <- data.frame(date, col1, col2)
###
One way to turn this into the kind of data frame you want is to convert the
main part of i
It is possible to do it with numeric comparisons, as well, but to make life
comfortable you need to turn off the warning system temporarily.
df <- data.frame(q1 = c(0,0,33.33,"check"),
q2 = c(0,33.33,"check",9.156),
q3 = c("check","check",25,100),
It means you have selected a response variable from one data frame
(unmarried.male) and a predictor from another data frame (fieder.male) and they
have different lengths.
You might be better off if you used the names in the data frame rather than
selecting columns in a form such as 'some.data
If you want to do a stepwise selection there is a function in the klaR package
to do it. This is not what you are asking for, though. You want a way of
finding the successive error rates as additional variables are added in the
forward selection process. As far as I can see you have to do thi
Xonly <- XY[, grep("^X", dimnames(XY)[[2]])]
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Nixon, Matthew
Sent: Thursday, 10 March 2011 12:20 AM
To: r-help@R-project.org
Subject: [R] Extracting only odd columns from a matrix
H
Here is one possible way (I think - untested code)
cData <- do.call(rbind, lapply(split(data, data$prochi),
function(dat) {
dat <- dat[order(dat$date), ]
while(any(d <- (diff(dat$date) <= 3)))
dat <- dat[-(min(which(d))+1), ]
Here is possibly one method (if I have understood you correctly):
> con <- textConnection("
+ xloc yloc gonad indEneW Agent
+ 123 20 516.74 1 0.02 20.21 0.25
+ 223 20 1143.20 1 0.02 20.21 0.50
+ 321 19 250.00 1 0.02 20.21 0.25
+ 422 15
Erin
You could use
as.vector(t.test(buzz$var1, conf.level=.98)$conf.int)
Bill Venables.
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Erin Hodgess
Sent: Monday, 7 March 2011 3:12 PM
To: R help
Subject: [R] attr question
Dear
No. That's not answering the question. ALL surveys are for collecting
information.
The substantive issue is what purpose do you have in seeking this information
in the first place and what are you going to do with it when you get it?
Do you have some commercial purpose in mind? If so, what i
You can probably simplify this if you can assume that the dates are in sorted
order. Here is a way of doing it even if the days are in arbitrary order. The
count refers to the number of times that this date has appeared so far in the
sequence.
con <- textConnection("
01/01/2011
01/01/2011
02/
Here is a start
> x <- as.data.frame(runif(2000, 12, 38))
> length(x)
[1] 1
> names(x)
[1] "runif(2000, 12, 38)"
>
Why are you turning x and y into data frames?
It also looks as if you should be using if(...) ... else ... rather than
ifelse(.,.,), too.
You need to sort out a few issues, it se
Here is one way.
1. make sure y.test is a factor
2. Use
table(y.test,
factor(PredictedTestCurrent, levels = levels(y.test))
3. If PredictedTestCurrent is already a factor with the wrong levels, turn it
back into a character string vector first.
-Original Message-
From: r-help-boun
This is a purely statistical question and you should try asking it on some
statistics list.
This is for help with using R, mostly for data analysis and graphics. A glance
at the posting guide (see the footnote below) might be a good idea.
-Original Message-
From: r-help-boun...@r-pro
The "probability OF the residual deviance" is zero. The significance level for
the residual deviance according to its asymptotic Chi-squared distribution is a
possible criterion, but a silly one. If you want to minimise that, just fit no
variables at all. That's the best you can do. If you wa
You can compute the logarithm of it easily enough
> lchoose(54323456, 2345)
[1] 25908.4
Now, what did you want to do with it?
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Jim Silverton
Sent: Monday, 28 February 2011 10:38 AM
T
One way to do it is to use the 'abind' package
> NCurvas <- 10
> NumSim <- 15
> dW <- replicate(NumSim, matrix(rnorm(NCurvas * 3), NCurvas, 3),
+ simplify = FALSE)
> library(abind)
> DW <- do.call(abind, c(dW, rev.along = 0))
> dim(DW)
[1] 10 3 15
-Original Message-
From: r-h
Here is the party line, perhaps
by(data, data$TYPE, function(dat)
with(dat, weighted.mean(MEASURE, COUNT)))
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Mike Schumacher
Sent: Thursday, 24 February 2011 9:40 AM
To: r-help@r-p
The solution is probably to make the data-time columns POSIXct:
> x <- read.table(textConnection("
+ ID event.date.time
+ 1 '2009-07-23 00:20:00'
+ 2 '2009-08-18 16:25:00'
+ 3 '2009-08-13 08:30:00'
+ "), header = TRUE)
> y <- read.table(textConnection("
+ ID event.date.time
+ 4
This is a very good question. You have spotted something that not many people
see and it is important.
The bland assertion that the "deviance can be used as a test of fit" can be
seriously misleading.
For this data the response is clearly binary, "Admitted" (success) or
"Rejected" (failure) a
Your complaint is based on what you think a factor should be rather than what
it actually is andhow it works. The trick with R (BTW I think it's version
2.12.x rather than 12.x at this stage...) is learning to work *with* it as it
is rather than making it work the way you would like it to do.
With R it is always possible to shoot yourself squarely in the foot, as you
seem keen to do, but R does at least often make it difficult.
When you predict, you need to have values for ALL variables used in the model.
Just leaving out the coefficients corresponding to absent predictors is
equiv
Hi Alice,
You can use
pvals <- summary(myprobit)$coefficients[, "Pr(>|z|)"]
Notice that if the p-value is very small, the printed version is abbreviated,
but the object itself has full precision (not that it matters).
Bill Venables.
-Original Message-
From: r-help-boun...@r-project.o
!is.null(my.obj...@my.data.frame$my.var)
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Kushan Thakkar
Sent: Friday, 11 February 2011 10:30 AM
To: r-help@r-project.org
Subject: [R] if a variable is defined
I have an object type m
Using complex names, like res[, 3+i] or res$var, in the formula for a model is
a very bad idea, especially if eventually you want eventualluy to predict to
new data. (In fact it won't work, so that makes is very bad indeed.) So do
not use '$' or '[..]' terms in model formulae - this is going t
You want advice?
1. Write sentences that contain a subject and where appropriate, an object as
well. This makes your email just that bit more polite. This list is not a
paid service.
2. The "sheets" may have variables in common, but do they have the same name in
both, and the same class, and
The function factor.scores does not "inherit" anything. It is a generic
function that provieds methods for a number of classes, including those you
mention. (The terminology is important if you are to understand what is going
on here):
> library(ltm)
Loading required package: MASS
Loading re
> yearLength <- function(year) 365 + (year %% 4 == 0)
> yearLength(1948:2010)
[1] 366 365 365 365 366 365 365 365 366 365 365 365 366 365 365 365 366 365
365 365 366
[22] 365 365 365 366 365 365 365 366 365 365 365 366 365 365 365 366 365 365
365 366 365
[43] 365 365 366 365 365 365 366 365 36
You do have missing values. Setting xlim does not subset the data.
How about
link <- "http://processtrends.com/files/RClimate_CTS_latest.csv";
cts <- read.csv(link, header = TRUE)
scts <- subset(cts, !is.na(GISS) & !is.na(cts)) ## remove defectives
plot(GISS ~ yr_frac, scts, type =
lag and as.ts are separate operations (which in fact commute)
> lag(as.ts(1:10), 1)
Time Series:
Start = 0
End = 9
Frequency = 1
[1] 1 2 3 4 5 6 7 8 9 10
> as.ts(lag(1:10, 1))
Time Series:
Start = 0
End = 9
Frequency = 1
[1] 1 2 3 4 5 6 7 8 9 10
>
You do NOT need to ca
plot(y~x, type="p", xlim = x[c(2,4)])
?
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of wangxipei
Sent: Tuesday, 18 January 2011 1:27 PM
To: r-help
Subject: [R] plot continuous data vs clock time
Dear R users,
I have a question a
Here is one way
Here is one way:
> con <- textConnection("
+ ID TIMEOBS
+ 001 220023
+ 001 240011
+ 001 320010
+ 001 450022
+ 003 3900 45
+ 003 5605
nls in the stats package.
?nls
From: r-help-boun...@r-project.org [r-help-boun...@r-project.org] On Behalf Of
Erik Thulin [ethu...@gmail.com]
Sent: 15 January 2011 16:16
To: r-help@r-project.org
Subject: [R] Weighted least squares regression for an exponen
If you can specify the omitted columns as numbers there is a quick way to do
it. e.g.
> d
d1 d2d3d4
1 9.586524 4.833417 0.8142588 -3.237877
2 11.481521 6.536360 2.3361894 -4.042314
3 10.243192 5.506440 2.0443788 -3.478543
4 9.969548 6.159666 3.0449121 -4.827
I very much doubt your first example does work. the value of plot() is NULL
which if you plot again will give the error message you see in your second
example.
What where you trying to achieve doing
p <- plot(hc)
plot(p) ### this one is trying to plot NULL
?
Here is an example (s
lm() and aov() are not fully equivalent. They both fit linear models, but they
use different algorighms, and this allows aov, for example, to handle some
simple multistratum models. The algorithm used by lm does not allow this, but
it has other advantages for simpler models.
If you want to fi
That rather depends on what kind of plot you want to use.
Here is one option that you can use without any changes:
##
con <- textConnection("
Months Prec
1 Jan 102.1
2 Feb69.7
3 Mar44.7
4 Apr32.1
5 May24.0
6 Jun18.7
7 Jul14.0
8 Aug
If you use ::: to access non-exported functions, as Frank confesses he does,
then you can't complain if in the next release of the package involved the
non-exported objects are missing and things are being done another way
entirely. That's the deal.
On the other hand, sometimes package authors
You don't give us much to go on, but some variant of
country <- c("US", "France", "UK", "NewZealand", "Germany", "Austria", "Italy",
"Canada")
result <- read.csv("result.csv", header = FALSE)
names(result) <- country
should do what you want.
From: r-hel
Here is an alternaive approach that is closer to that used by lm and friends.
> df <- data.frame(x=1:10,y=11:20)
> test <- function(col, dat) eval(substitute(col), envir = dat)
> test(x, df)
[1] 1 2 3 4 5 6 7 8 9 10
> test(y, df)
[1] 11 12 13 14 15 16 17 18 19 20
>
There is a slight a
library(nlme)
lmList(y ~ x | factor(ID), myData)
This gives a list of fitted model objects.
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Entropi ntrp
Sent: Wednesday, 29 December 2010 12:24 PM
To: r-help@r-project.org
Subject:
Dear 'analyst41' (it would be a courtesy to know who you are)
Here is a low-level way to do it.
First create some dummy data
> allDates <- seq(as.Date("2010-01-01"), by = 1, length.out = 50)
> client_ID <- sample(LETTERS[1:5], 50, rep = TRUE)
> value <- 1:50
> date <- sample(allDates)
> clien
I find this function useful for digging out months from Date objects
Month <- function(date, ...)
factor(month.abb[as.POSIXlt(date)$mon + 1], levels = month.abb)
For this little data set below this is what it gives
> with(data, tapply(value, Month(date), median, na.rm = TRUE))
Jan Feb M
Dear Lurker,
If all you art trying to do is to plot something, isn't all you need something
like the following?
x <- c( 30, 50, 80, 90, 100)
y <- c(160, 180, 250, 450, 300)
sp <- spline(x, y, n = 500)
plot(sp, type = "l", xlab = "x", ylab = "y",
las = 1, main = "A Spline Interpolation")
?unique
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of zhiji19
Sent: Wednesday, 8 December 2010 2:57 PM
To: r-help@r-project.org
Subject: [R] Summing up Non-numeric column
Dear All
If I have the following dataset
V1 V2
x y
y
For a substantial calculation like this the algorithms will likely be in C or
Fortran.
You will need to download the source for the stats package from CRAN (as a
tar.gz file), expand it, and look at the source code in the appropriate
sub-directories.
You can get a bit of a road map in R by
>
A new fortune is born?
"Sharing LaTeX documents with people using word processors only is no more
difficult than giving driving directions to someone who is blindfolded and has
all 4 limbs tied behind their back. Collaboration with people who insist on
using programs that process their words m
I think all you need is
?split
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Eduardo de Oliveira Horta
Sent: Sunday, 28 November 2010 8:02 AM
To: r-help@r-project.org
Subject: [R] Two time measures
Hello!
I have a csv file of
perhaps you need something like this.
par(yaxs = "i")
plot(runif(10), type = "h", ylim = c(0, 1.1))
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Sebastian Rudnick
Sent: Tuesday, 23 November 2010 10:37 AM
To: r-help@r-project.
It's actually not too difficult to write the density function itself as
returning a function rather than a list of x and y values. Here is a no frills
(well, few frills) version:
### cut here ###
densityfun <- local({
normd <- function(value, bw) {
force(value); force(bw)
function(z)
What you show below is only a representation of the matrix to 7dp. If you look
at that, though, the condition number is suspiciously large (i.e. the matrix is
very ill-conditioned):
> txt <- textConnection("
+ 0.99252358 0.93715047 0.7540535 0.4579895
+ 0.01607797 0.09616267 0.2452471 0.308
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