Hi:
Try this:
xyplot(y ~ x1 | x2 * x3, groups = x4, data = example,
type= c(l, p), pch=c(16, 4), col=c(blue, red),
col.line = c('blue', 'transparent'),
xlim = c(1.04, 1.26), ylim = c(0, 1.1), xlab = X2, ylab = Y,
key = list(space = top, text = list(A),
Hi:
One answer is eval(substitute(...)): [Note: it's a good idea NOT to use the
same name on different objects - R is pretty good about distinguishing TEST
the function from TEST the data frame, but you could be playing with fire if
both objects have the same class but have different contents.
Hi:
It would help if you named your variables such that alphanumeric ordering
doesn't disturb your variable ordering. Having been burned by this a few
times, I've learned the basics of sprintf() :) Here's your example
revisited, along with an alternative stacking/unstacking display with
package
Hi:
This problem came up the other day - see
http://stats.stackexchange.com/questions/7884/fast-ways-in-r-to-get-the-first-row-of-a-data-frame-grouped-by-an-identifier/7985#7985
Dennis
On Sat, Mar 12, 2011 at 3:20 AM, Vincy Pyne vincy_p...@yahoo.ca wrote:
Dear R helpers
Suppose I have a
Hi:
The value of your title is a(n attached) data frame - perhaps you meant
'HMF' instead...
HTH,
Dennis
On Sat, Mar 12, 2011 at 2:41 AM, derek jan.kac...@gmail.com wrote:
Hello R users, Im having this strange problem.
http://r.789695.n4.nabble.com/file/n3350024/bad2.png
I have txt file of
Hi:
Since you didn't provide any data, you're stuck with my example instead.
# Generate three sine curves on (0, 2*pi)
x - seq(0, 2 * pi, length = 200)
df - data.frame(x, y1 = sin(x), y2 = sin(2 * x), y3 = sin(4 * x))
# Create a small function to type (n), where n is a number to be input
mytext
Hi:
(1) The first argument to get.hist.quote() is instrument, not instruments. I
concur with David that get.hist.quote()
takes a single character string as an argument.
(2) I tried running this with lapply() but got a download error on the last
one:
getStockData(PBR-B)
trying URL '
Hi:
Here are a few one-liners. Calling your data frame dd,
aggregate(cbind(height, weight, age) ~ sample, data = dd, FUN = mean)
sample heightweight age
1 A 12.2 0.503 6.00
2 B 12.75000 0.715 4.50
3 C 11.35250 0.5125000 3.75
4 D 14.99333
Hi:
This is probably not quite what you had in mind, but both work for your
example - I think the second is probably easier.
x - c(a, b)
y - c(x, y)
as.vector(outer(x, y, function(x, y) paste(x, y, sep = '')))
[1] ax bx ay by
apply(expand.grid(x, y), 1, paste, collapse = '')
[1] ax bx ay by
Hi:
Perhaps something like
f1(x, y) * I(z 0) + f2(x, y) * I(z = 0) ??
HTH,
Dennis
On Fri, Mar 11, 2011 at 3:10 AM, Otto Kässi otto.ka...@gmail.com wrote:
Dear r-helpers,
This might be an elementary question, but I have a hard time getting
my head around it, so all help is much
Hi:
df - read.table(textConnection(
VAR DATETIME CONC COVAR
1 NOV20.2510 group1
1 NOV20.5 20 group1
1 NOV21 5 group1
1 NOV22 1 group1
1 NOV23 0.1 group1
2 NOV20.2510
Hi:
To get the samples, here's one approach:
df - data.frame(gender = rep(c('F', 'M'), c(165, 42)), y = rpois(207, 20))
# Sampling function to take 20 F and 20 M with replacement
# The sample function operates on the rows of df to get idx and then takes
# the y's corresponding to those rows
sfun
Hi:
Assuming that z constitutes the last two elements of your input vector,
try2 - function(x) {
n - length(x)
y - x[1:(n - 2)]
z - x[n - 1] + x[n]
sum(y) * z
}
x0 - 1:4# result should be 21
x1 - 1:8# z = 15, result = 21 * 15 = 315
try2(x0)
try2(x1)
HTH,
Dennis
On Thu,
Hi:
A box plot is based on a five number summary, so you need at a minimum five
observations (and preferably at least twice that) to make a box plot a
viable summary measure for a continuous variable. Consider other graphical
summaries for these data - perhaps a strip chart or a simple
Hi:
I'm not quite sure what you have in mind, but...
Input data strdat.txt:
probeID
rc_AI104113_at
rc_AI178259_f_at
rc_AI179134_i_at
rc_AI179134_f_at
rc_AI104113_at
rc_AA819429_f_at
strg - paste(scan('strdat.txt', what = '', skip = 1), collapse = ',')
strg
[1]
Hi:
Here's another approach:
purchase_amount - transform(purchase_amount,
diff = with(purchase_amount, ave(amount, customer, FUN =
function(x) c(0, diff(x)
purchase_amount
HTH,
Dennis
On Wed, Mar 9, 2011 at 7:27 AM, rens_1112 tibid...@rocketmail.com wrote:
Dear all,
Probably a
Hi:
Perhaps something like this?
m - matrix(rnorm(100, m = 10, s = 2), ncol = 5)
colnames(m) - paste('V', 1:5, sep = '')
# Summary function:
summs - function(x) c(mean = mean(x), sd = sd(x), range = diff(range(x)))
# Apply to columns of m and transpose the result:
t(apply(m, 2, summs))
For
Hi:
Here's one approach, although I imagine there are more efficient ways.
# A function to strip spaces and return the first three non-blank elements
of a string
keyset - function(x) substr(gsub(' ', '', x)[1], 1, 3)
# Apply the function to the data frame to generate the key:
a$key -
df - read.table(textConnection(
category type values
1 treat_A AA 0.38200018
2 treat_A0.10068056
3 treat_A B0.59648427
4 treat_A AAA 0.89910581
5 treat_A BB 0.88460952
6 treat_A 0.95846431
7 treat_ABBB
David, David, David...you forgot the solid and dashed lines :) It's OK, it
gives me an excuse to look at this from a slightly different angle. [The
intersection() function is a *really* good trick, BTW - thanks for the
reminder.]
Back to the OP. Let's re-read the data:
dat=data.frame(Age =
, 2011 10:50 AM
*To:* Umesh Rosyara
*Cc:* Jorge Ivan Velez; Dennis Murphy; sarah.gos...@gmail.com; R mailing
list
*Subject:* Re: [R] Data lebals xylattice plot: RE: displaying label
meeting condition (i.e. significant, i..e p value less than 005) in plot
function
This is easy to do
Hi:
Here's one way to piece it together. All we need is the first variable, so
I'll manufacture a vector of Start.action's and go from there.
w - data.frame(Start.action = c(rep('Start.setting', 3),
rep('Start.hauling', 4),
rep('Start.setting', 4),
Hi:
One approach is through the data.table package:
library(data.table)
df - as.data.frame(data.table(iris, key = 'Species')[Sepal.Length == 6.7])
str(df)
Using Species as a key variable automatically sorts by Species; the
bracketing allows you to subset on Sepal.Length == 6.7.
ddply() in the
Hi:
This should get you most of the way there; I'll let you figure out how to
assign the BLOCK and RUN numbers.
tx - Subject ID,ExperimentName,2010-04-23,32:34:23,Version 0.4, 640 by
960 pixels, On Device M, M, 3.2.4,
zz_373_462_488_...@9z.svg,592,820,3.35,zz_032_288_436_...@9z.svg
Hi:
This is basically Scott's idea with a few added details.
Let's assume your files have similar names - e.g., they differ only by
number.
The example below creates ten files of similar structure to yours. There are
then two paths one can follow: (1) put all the files into a specific
directory,
Also see the example in the ScaleTests function from package coin, which I
found with
library(sos)
findFn('Lepage test')
Dennis
On Fri, Mar 4, 2011 at 12:00 PM, mike vassalosmich...@uky.edu wrote:
Hey everyone,
I am interest in running a Lepage multi sample test and i would like to ask
if
Hi:
Perhaps you had something like this in mind:
df-data.frame(a1=1:10, a2=11:20, a3=21:30, b1=101:110, b2=111:120,
b3=121:130)
avars - names(df)[grep('^a', names(df))]
bvars - names(df)[grep('^b', names(df))]
cvars - paste('c', 1:length(avars), sep = '')
df - within(df, for(i in
Hi:
You need the second variable in D1 to be named fGRAZE - the variable names
in the newdata data frame (D1) have to be the same as the variable names on
the RHS of the model formula, in this case L.AREA and fGRAZE.
HTH,
Dennis
On Thu, Mar 3, 2011 at 7:43 AM, Heike Schmitz
Hi:
Here's an example:
x - rnorm(300)
hist(x, breaks = 15, yaxt = 'n', ylab = 'Relative frequency')
axis(2, at = seq(0, 50, by = 10), labels = round(seq(0, 50, by =
10)/length(x), 3))
HTH,
Dennis
On Thu, Mar 3, 2011 at 8:29 AM, jpmaroco jpmar...@gmail.com wrote:
Dear David,
Thanks for your
Hi:
On Wed, Mar 2, 2011 at 1:52 PM, John Smith zmr...@gmail.com wrote:
Hello All,
I try to use the attached code to produce a cross over plot. There are 13
subjects, 7 of them in for/sal group, and 6 of them in sal/for group. But
in
xyplot, all the subjects are listed in both subgraphs.
# from pair M3a and M3b
4 # from pair M4a and M4b
to the end of the file
Thank you in advance
Umesh R
--
*From:* Dennis Murphy [mailto:djmu...@gmail.com]
*Sent:* Friday, February 18, 2011 12:28 AM
*To:* Umesh Rosyara
*Cc:* r-help@r-project.org
Hi:
On Tue, Mar 1, 2011 at 8:22 AM, Bodnar Laszlo EB_HU
laszlo.bod...@erstebank.hu wrote:
Hello there,
I have a problem concerning bootstrapping in R - especially focusing on the
resampling part of it. I try to sum it up in a simplified way so that I
would not confuse anybody.
I have a
...and this is where we cue the informative article on least squares
calculations in R by Doug Bates:
http://cran.r-project.org/doc/Rnews/Rnews_2004-1.pdf
HTH,
Dennis
On Tue, Mar 1, 2011 at 10:52 AM, AjayT ajaytal...@googlemail.com wrote:
Hey thanks alot guys !!! That really speeds things up
Hi:
1. expression() in plotmath ignores control characters such as \n.
2. The workaround, discussed a couple of times on this list (hence in the
archives), is to use the atop function, so try something like
plot(0:1,0:1,xaxt=n)
axis(side=1,at=.3,expression(paste(IFN-, gamma, \n, TNF-, alpha)))
Hi:
As far as I can see, the problem has to do with the way you wrote f and g:
f(2)
f for 2
[1] 4
g(5)
g for 5
[1] 15
You output an unsaved character string with a numeric result, but the
character string is not part of the return object since it is neither saved
as a name or an attribute. If
Hi:
On Mon, Feb 28, 2011 at 8:42 AM, John Sorkin jsor...@grecc.umaryland.eduwrote:
R 2.10.0
Windows XP
I am trying to run lme. I receive the following error message:
My lme code is:
fitRandom - lme(values ~ factor(subject),
data=withindata)
Where's the random factor? Perhaps you mean
Hi:
There has to be a better way to do this, but one option is to use pmin() and
pmax().
# Function to apply to pairs of columns:
ordPair - function(x, y) data.frame(pmin(x, y), pmax(x,y))
u - with(odataframe, cbind(ordPair(X1, X2), ordPair(X3, X4), ordPair(X5,
X6))
u
pmin.x..y. pmax.x..y.
Hi:
Using package sos:
# install.packages('sos') # if necessary
library(sos)
findFn('conditional logistic regression')
the following appear to be reasonable candidates to start investigating:
* clogit() in package survival
* clogistic() in package Epi
* clogistCalc()in package saws
And in that vein, the recently released MethComp package by Bendix
Carstensen may be of service.
HTH,
Dennis
On Fri, Feb 25, 2011 at 5:39 AM, Marc Schwartz marc_schwa...@me.com wrote:
On Feb 24, 2011, at 4:50 PM, Denis Kazakiewicz wrote:
Dear R people
Could you please help with following
Hi:
Here's another way:
c1-c(1,2,3,2,2,3,1,2,2,2)
c2-c(5,6,7,7,5,7,5,7,6,6)
c3-rnorm(10)
x - data.frame(c1 = factor(c1), c2 = factor(c2), c3)
x - transform(x, mean = ave(c3, c1, c2, FUN = mean))
Yet another with function ddply() in package plyr:
ddply(x, .(c1, c2), transform, mean = mean(c3))
Hi:
On Fri, Feb 25, 2011 at 12:09 PM, Christopher R. Dolanc
crdol...@ucdavis.edu wrote:
I'm trying to use tapply to output means and SD or SE for my data but
seem to be limited by how many times I can subset it. Here's a snippet
of my data
stems353[1:10,]
Time DataSource Plot
Hi:
Here's one way:
d - read.table(textConnection(
ind xloc yloc gonad
1 15 2 12
2 15 4 25
3 13 3 30
4 15 4 10
5 11 4 08
10 11 4 11), header = TRUE)
closeAllConnections()
library(plyr)
dsum - ddply(d,
Hi:
df - read.table(textConnection(
+ xloc yloc yield
+ 1 101295
+ 2 111081
+ 3 121120
+ 4 121110), header = TRUE)
with(df, xtabs(yield ~ yloc + xloc))
xloc
yloc 10 11 12
10 0 81 0
11 0 0 30
12 95 0 0
HTH,
Dennis
On Tue, Feb 22, 2011 at
0 0 30
12 95 0 0
Any hints?
THANKS!!
Nico
Original Message
Subject:Re: [R] sum data from data.frame in a matrix
Date: Wed, 23 Feb 2011 00:10:52 -0800
From: Dennis Murphy djmu...@gmail.com
To: Nicolas Gutierrez nicol...@uw.edu
Hi:
Is this what you want?
x$gpvar - rep(c(1, 2), 4)
split(x, x$gpvar)
$`1`
id v1 V2 gpvar
1 1 1 9 1
3 2 3 11 1
5 3 5 13 1
7 4 7 15 1
$`2`
id v1 V2 gpvar
2 1 2 10 2
4 2 4 12 2
6 3 6 14 2
8 4 8 16 2
HTH,
Dennis
On Wed, Feb 23, 2011 at 5:16
Hi:
If by SIMCA you mean 'Soft Independent Modeling of Class Analogy', a search
with package sos indicated that a few functions in the plsRglm package have
some support for it.
HTH,
Dennis
On Sun, Feb 20, 2011 at 9:15 PM, reynolds pravindev
reynoldspravin...@gmail.com wrote:
Hi
I'm a B.E
Hi:
Here's one way with the plyr package and function d_ply(); the hist()
function itself is not very elegant, but it 'works' for this example.
Factor - rep(factor(letters[1:4]), each = 10)
Size - runif(40) * 100
library(plyr)
par(mfrow = c(2, 2))
d - data.frame(Factor, Size)
# Function to
Hi:
Here's one way with package plyr:
df-data.frame(x=c(rep(1,3),rep(2,4),rep(3,5)),
y=rnorm(12),
z=c(3,4,5,NA,NA,NA,NA,1,2,1,2,1),
w=c(1,2,3,3,4,3,5,NA,5,NA,7,8))
library(plyr)
fun - function(d) {
u - apply(d[, -1], 2, function(y)
Hi:
Assuming dd is the name of your data frame,
dd$diff - with(dd, q2 - q1)
dd
TIME ID q1 q2 diff
11 1187 3 2 -1
21 1706 3 30
31 1741 2 42
42 1187 3 2 -1
52 1706 3 30
62 1741 2 42
is one way to do it.
HTH,
Dennis
On Mon, Feb 21,
Hi:
Try the vegan and biodiversityR packages; both contain functions for
producing species accumulation curves.
On Mon, Feb 21, 2011 at 9:38 AM, Vanessa Francisco vanewa...@gmail.comwrote:
Hello! I'm a PhD student working with coral reef fish diversity in Mexico.
I want to do species
Hi:
Here's one way:
f - function(x) sin(x) + pi/4
g - function(x) {
sin(x) * (x = 0 x = pi) + sin(x) * (x = -2 * pi x = -pi) -
(pi/4) * (x -pi x 0) - pi/4 * (x pi x 2 * pi)
}
x - seq(-2 * pi, 2 * pi, length = 200)
plot(x, f(x), type = 'l', col = 'blue', ylim = c(-1, 2))
Hi:
Perhaps another possibility is ?capture.output
Dennis
On Mon, Feb 21, 2011 at 10:26 AM, Tatyana Deryugina tatya...@mit.eduwrote:
Thanks, Ista! I looked at sink() and it looks like it might work.
However, it seems as though you need to use it with the print
command. I have a lot of
Hi:
Here's one way:
# Manufacture some monthly data over 20 years
df - data.frame(year = rep(1981:2000, each = 12),
month = factor(rep(month.abb, 20), levels = month.abb),
meantemp = rnorm(240,
mean = rep(c(30, 35, 40, 50, 60, 70, 75,
Hi:
This isn't hard to do with ggplot2. Here's a toy example:
d - data.frame(gp = LETTERS[1:4], frq = c(10, 25, 30, 20))
library(ggplot2)
# The fill aesthetic colors the bars, the colour aesthetic does the same for
the borders.
# (1) Same color for both, use alpha transparency:
ggplot(d,
Hi:
These are sometimes called 'spaghetti plots'; here is a variation on an
example in the ggplot2 book by Hadley Wickham using the Oxboys data from
package nlme:
library(ggplot2)
data('Oxboys', package = 'nlme')
g - ggplot(Oxboys, aes(x = age, y = height))
g + geom_line(aes(group = Subject)) +
Hi:
Here's another approach using the reshape package:
library(reshape)
# id defines the grouping variables; the others are stacked, creating
# new variables 'variable' (containing the variable names as factor levels)
and
# 'value' (which holds the corresponding values)
dfm - melt(df, id =
This is a built-in dataset in R - see ?HairEyeColor and str() it. I smell
homework...
Dennis
On Thu, Feb 17, 2011 at 11:50 PM, Lao Meng laomen...@gmail.com wrote:
The data is in the attachment.
What I wanna get is:
, , Sex = Male
Eye
HairBrown Blue Hazel Green
Black32
Hi:
On Fri, Feb 18, 2011 at 2:49 AM, Jan jrheinlaen...@gmx.de wrote:
Hi,
I am not a statistics expert, so I have this question. A linear model
gives me the following summary:
Call:
lm(formula = N ~ N_alt)
Residuals:
Min 1Q Median 3Q Max
-110.30 -35.80 -22.77
Hi:
A couple more approaches to consider:
# Utility function to extract two rows from a data frame
# Meant to be applied to each data subset
sampler - function(d) if(nrow(d) 2) d[sample(1:nrow(d), 2, replace =
FALSE), ] else d
library(plyr)
ddply(x, 'id', sampler)
id v1 V2
1 1 2 13
2 1
Hi Hannah:
You have a few things going on, but the bottom line is that numer and denom
are both double integrals.
On Thu, Feb 17, 2011 at 1:06 PM, li li hannah@gmail.com wrote:
Hi all,
I have some some problem with regard to finding the integral of a
function containing an indicator
Hi:
This is as far as I could get:
df - read.table(textConnection(
Individual Parent1 Parent2 mark1 mark2
10 0 12 11
20 0 11 22
30 0 13 22
40 0 13 11
51 2
Hi:
There are some reasonably efficient ways to do this if you're a little
clever. If you use 16 histograms per page, you can get 928 of them onto 58
pages - with 4 per page, with one left over, that would generate 233 pages
of output. I could think of something like the following (pseudocode):
Hi:
If you look at the error message, you'll see that you removed s before
evaluating f, and since an element of s is called in the function
Try
s - c( 0.2, 0.45, 0.38, 0.9)
f - lapply(1:10, function(i)local({ force(i) ; function(x)x^2+s[i]}))
f[[1]](s)
[1] 0.2400 0.4025 0.3444 1.0100
f
Hi:
Try the following for examples of the use of MDS in R:
http://www.statmethods.net/advstats/mds.html
http://cran.r-project.org/web/packages/HSAUR/vignettes/Ch_multidimensional_scaling.pdf
http://cran.r-project.org/web/packages/HSAUR2/vignettes/Ch_multidimensional_scaling.pdf
# (2nd ed.)
Hi:
Wouldn't ifelse() work here?
tco - with(df, ifelse(TargetColor == 'B', CannonOriB, CannonOriR))
ifelse() is vectorized, so there should be no need for a loop.
Test:
df - data.table(TargetColor = c('B', 'R'), CannonOriB = c(5, 5),
+ CannonOriR = c(3, 3), stringsAsFactors =
Hi:
One (of many) ways is to use the aggregate() function in base R:
df - data.frame(g = rep(LETTERS[1:5], each = 5), y = rnorm(25))
aggregate(y ~ g, data = df, FUN = sum)
g y
1 A -1.3165949
2 B 1.9850015
3 C -0.6460323
4 D -1.6459293
5 E 6.7001807
HTH,
Dennis
On Sun, Feb 13,
Hi:
This seems to work:
mykey - list(space = 'top',
columns = 4,
text = list(as.character(unique(src$s)), col = colors),
points = list(pch = 1, col = colors)
)
xyplot(v~t, groups=s, type='o', data=src, col=colors, key = mykey)
HTH,
Dennis
Hi:
One way:
x[sample(nrow(x)), ]
HTH,
Dennis
On Wed, Feb 9, 2011 at 10:45 PM, Diogo Almeida dala...@gmail.com wrote:
Hi,
I need to permute the rows of a matrix, where each row is independently
rearranged. A simple solution is this:
shuffled - datamatrix - matrix(1:24, ncol = 4)
for (i
Hi:
There are a couple of things here. Firstly, if you intend to predict from a
model in R, make sure your input to the model is a data frame. You can fit a
model piecemeal with a set of vectors, but there's a distinct possibility
you'll end up frustrated when trying to make predictions with new
Hi:
On Thu, Feb 10, 2011 at 10:50 AM, Hui Du hui...@dataventures.com wrote:
Hi all,
I have a dataset. Each time I want to sample N(i) elements
from it and I want to repeated sampling M times. N(i) is varied from time to
time. For example,
dataset = 1:50;
My apologies to Eik - I didn't see his solution and essentially replicated
it. Sorry for the noise..
Dennis
On Thu, Feb 10, 2011 at 3:46 PM, Dennis Murphy djmu...@gmail.com wrote:
Hi:
On Thu, Feb 10, 2011 at 10:50 AM, Hui Du hui...@dataventures.com wrote:
Hi all,
I have
Hi:
It's not a perfect solution, but an 'easy' way out is to recognize that some
symbols are open and some are filled. If you want fill, use a filled symbol
and then change the color.
p - rep(16:18, 4)
x-c(1:12)
y-c(rpois(12,4))
grp-c(rep(c(3,4), each=6))
z-c(rep(c(1,2), each=6))
dd -
Hi:
Try
summary(myprobit)$coefficients[, 4]
HTH,
Dennis
On Thu, Feb 10, 2011 at 3:46 PM, Allie818 alice...@gmail.com wrote:
I can get this summary of a model that I am running:
summary(myprobit)
Call:
glm(formula = Response_Slot ~ trial_no, family = binomial(link = probit),
data =
Hi:
And yet another: :)
library(plyr)
top2 - function(x) {
y - sort(x[-c(1, 2)], decreasing = TRUE)[1:2]
area - rep(x[1], 2)
type - rep(x[2], 2)
data.frame(area, type, value = y, variable = names(y))
}
v - ldply(apply(df, 1, top2), rbind)
The last line does the
Hi:
Here's one way:
plot(1~1,ylab=expression(Areas (~mu*m^2~)))
The tildes incorporate space between the math and text elements; they're
optional, but useful. Another way that also works is
plot(1~1,ylab=expression(paste(Areas (, mu*m^2, ), sep = ' ')))
HTH,
Dennis
On Thu, Feb 10, 2011 at
Hi:
dat - read.table(textConnection(
+ response pred1 pred2
+ 1 1 01
+ 2 0 00
+ 3 1 00
+ 4 1 11
+ 5 1 01
+ 6 0 11
+ 7 1 10
+ 8 1 01
+ 9 0 11
+ 10 0 01
+ ),
Hi:
Here's one approach to the problem you posed (don't know if it is what you
want for the problem you intend you use it on...)
df1 - read.table(textConnection(
id sexage area
01 male adult LP
01 male adult LP
01 male
Hi:
On Wed, Feb 9, 2011 at 6:29 AM, Claudio Carneiro
claudio_carneiro...@hotmail.com wrote:
Hi, all
I'm trying to solve an optimization problem with two
variables, in wich Iuse the following functions(they all are working
perfectly):
fxx1=function(x){
Hi:
Did you try putting the offset in the model formula, as in
bigglm( y ~ offset(z) + x, ...) ?
I haven't tried bigglm() personally (BTW, it's in the biglm package, which
wasn't mentioned), but this syntax works in the standard glm() function, so
perhaps it maps to bigglm() as well... (?)
Hi:
Try some modification of the following toy example:
df - data.frame(names = as.character(replicate(100, paste(sample(letters,
3), collapse = ''))),
x1 = rnorm(100), x2 = rnorm(100))
tgts - sample(df$names, 20)# target list of names
df[df$names %in% tgts, ] #
Hi:
One approach would be to use dlply() from the plyr package to generate the
models and assign the results to a list, something like the following:
library(plyr)
# function to run the GLM in each data subset - the argument is a generic
data subset d
gfun - function(d) glm(Stems ~ Time, data =
Hi:
You don't state the test for which you want the p-value, and to reiterate
what Dr. Ligges asked in response to your earlier post, how do you propose
to define a single R^2 measure? One may be able to answer your question re
an overall significance test using the anova() function:
Hi:
y - rnorm(50)
kd - sm.density(y, model = Normal)
str(kd)
List of 10
$ eval.points: num [1:100] -3.79 -3.72 -3.64 -3.57 -3.5 ...
$ estimate : num [1:100] 0.00124 0.00169 0.00225 0.00295 0.00381 ...
$ h : num 0.53
$ h.weights : num [1:50] 1 1 1 1 1 1 1 1 1 1 ...
$ weights:
Hi:
You need to check that you have the sufficient statistics necessary to
obtain an ANOVA table corresponding to the model you intend to fit. If you
have that, you can use the vectorization and matrix capabilities in R to
manually get the ANOVA table, perform the F tests, etc. The only thing
Hi:
For this particular task, the aggregate() function and the doBy package
provide nicely formatted output, but you may have to do some renaming.
Let's try a more expansive toy example with which one can do a bit more.
df - data.frame(grp1 = rep(c('x', 'y'), each = 40),
grp2
Hi:
Another option is to provide a default value of an argument that can be
changed by the user:
add - function(x, y = 3) x + y
add(4)
[1] 7
add(4, 2)
[1] 6
There are several options at one's disposal - the trick is to figure out
what you want to provide to the user of the function.
HTH,
Hi:
There's definitely something amiss with aggregate() here since similar
functions from other packages can reproduce your 'control' sum. I expect
ddply() will have some timing issues because of all the subgrouping in your
data frame, but data.table did very well and the summaryBy() function in
Hi:
This is FAQ 7.22 (Why do lattice/trellis graphics not work?)
You need to print() the plot, as in
print(xyplot(a ~ b))
This problem also extends to ggplot2 graphics.
(BTW, the legend statement after xyplot() won't work because the legend()
function is meant for use in base graphics. Lattice
Hi:
The problem you have, IMO, is the multiplicity of tests and the p-value
adjustments that need to be made for them. Here's a little simulation using
normal and exponential distributions of the size you're contemplating.
# Normal data
d - matrix(rnorm(40), nrow = 2)
# Function to
Hi:
u - matrix(sample(LETTERS[1:24]), nrow = 8)
u
[,1] [,2] [,3]
[1,] W J M
[2,] K V P
[3,] X E U
[4,] T B N
[5,] I Q F
[6,] G R S
[7,] A C O
[8,] D H L
t(apply(u, 1, function(x) sort(x)))
[,1] [,2] [,3]
[1,] J M W
[2,] K P V
[3,] E U X
[4,] B N T
[5,] F I Q
Hi:
This also works, except that the result is of class 'table':
xtabs(COUNTS ~ USER_ID + SITE, data = RAW)
SITE
USER_ID SITE1 SITE2 SITE3
1101322
2101212
31344 0
4 0 099
If you need a data frame, then the earlier
Hi:
Using df as your data frame,
max(subset(df, condition == 'GPR119a', select = responce))
[1] 0.6451204
You asked this question yesterday with several correct responses. What was
wrong with them?
Dennis
On Wed, Feb 2, 2011 at 12:48 AM, Asan Ramzan asanram...@yahoo.com wrote:
Hello
I am
Hi:
Try package elrm - its description file says that it performs exact logistic
regression by MCMC. I found this as the first set of hits I got from
# install.packages('sos') # if necessary
library(sos)
findFn('exact logistic regression')
HTH,
Dennis
On Wed, Feb 2, 2011 at 8:43 AM, Den
Hi:
Try this:
S1 - sample(10001:98999, 1000)
S2 - sample(10001:98999, 1000)
intersect(S1, S2) # lists have some overlap
[1] 72484 82292 93508 50438 29603 28965 68063 79598 16497 43016 22119
# test list
x - sample(10001:98999, 1000)
# tests first if x is in S1; if not, test whether it's in
Hi:
Try this:
d - read.table(textConnection(
+ procedure property sensor_data sensor_date
+ 1S_10 nord626821.0 2002-09-30T00:00:00+0200
+ 2 S_10 nord626821.0 2002-12-05T00:00:00+0100
+ 3S_10 nord626821.1 2008-07-31T00:00:00+0200
+ 4
Hi:
Not sure, but perhaps you had something like this in mind:
y - runif(100)
w - matrix(y, nrow = 10) # fills in column-wise
v - apply(w, 2, cumsum)
plot(1:100, y, type = 'l', col = 'blue')
lines(1:100, as.vector(v), type = 'l', col = 'red')
Reshaping into a matrix and using apply() to
Hi:
Here are a few ways to do this.
A useful approach is to use replicate() to generate the samples, flatten the
resulting matrix into a data frame and call one or more packages that are
well capable of handling multiple outputs per data subset.
Step 1: Generate the samples and rearrange into a
Hi:
Here are three more 'solutions' using packages doBy, plyr and data.table.
### For illustration: maxima for all mouce/GPR* conditions
(1) doBy:
library(doBy)
dfsub - subsetBy( ~ 1, subset = condition != 'con', data = df)
summaryBy(responce ~ mouce + condition, data = dfsub, FUN = max)
Hi:
This also works on your example data. Using the 'x' data frame from Jim
Holtman's post,
subset(transform(x, diffAB = A - B, diffCD = C - D), select = c('diffAB',
'diffCD'))
diffAB diffCD
1 -0.6 0.10
2 -0.4 0.10
3 -0.4 -0.06
transform() allows you to do the subtractions in one
Hi:
There is an additional proviso:
If the year is divisible by 100, then the next test is whether it's
divisible by 400. If both conditions are met, then the century year is a
leap year. Hence, 2000 was a leap year because it is divisible by 400, but
2100, 2200 and 2300 will not be. For Perl
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