Hi:
Try this:
INDIVIDUAL - transform(INDIVIDUAL, IDHOUS = IDPERS %/% 100)
merge(INDIVIDUAL, HOUSHOLD, by = c('IDHOUS', 'T'))
IDHOUS T IDPERS SEX PC02 GHS Single COM2 NBPERS NBKID
1 41 1 4101 19 1 NA2 5 3
2 41 1 4102 20 0 NA2 5 3
There are a few ways to do this. One is to use the lmList() function
in the nlme package and use conditioning in the model formula. Another
is to use the plyr package to create a list of models from which you
can extract pieces of output from each model fit to the data subsets;
for example,
Are you trying to separate the substrings in cat? If so, one way is to
use the colsplit() function in the reshape2 package, something like
(untested since you did not provide a suitable data format with which
to work):
library('reshape2')
splitcat - colsplit(mydata$cat, ' ', names = c('fat',
Hi:
On Mon, Oct 24, 2011 at 2:01 AM, Giovanni Azua brave...@gmail.com wrote:
Hello,
Suppose I have the dataset shown below. The amount of observations is too
massive to get a nice geom_point and smoother on top. What I would like to do
is to bin the data first. The data is indexed by Time
See the count() function in the plyr package; it does fast summation.
Something like
library('plyr')
count(passengerData, c('ORIGIN_WAC', 'DEST_WAC'), 'npassengers')
HTH,
Dennis
On Mon, Oct 24, 2011 at 8:27 AM, asindc siiri...@eastwestcenter.org wrote:
Hi, I am new to R so I would appreciate
Hi:
bestglmtest is your input data frame, is it not? From the names() line,
you can see that it has no variable named BestModel that corresponds
to a list containing a component named coefficients.
Were you perhaps looking for
output$BestModel$coefficients ??
Dennis
On Mon, Oct 24, 2011 at
Hi:
Aren't V1 and V2 factors in this data frame? If so, you should be
plotting bar charts rather than histograms (they're not the same). Do
you want separate graphs for V1 and V2, do you want them stacked, or
do you want them dodged (side-by side for each level 001, 002, 003)?
In the absence of
What kind of tree do you want? The sos() package can help you find R
functions associated with a particular topic:
# install.packages('sos')
library('sos')
findFn('tree')
found 2798 matches; retrieving 20 pages, 400 matches.
2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
Hi David:
When I try your code, I get the wireframe with the x, y, z axes sans
bounding cube and points, along with the error message
Error using packet 1
object 'pts' not found
sessionInfo()
R version 2.13.1 (2011-07-08)
Platform: x86_64-pc-mingw32/x64 (64-bit)
locale:
[1]
Hi:
You should perhaps do the following in the xyplot() call (untested
because the TDS and Cond data frames are missing):
xyplot('TDS'$quant ~ 'Cond'$quant | burns.tds.anal$site )
assuming that all three atomic objects have the same length. Caveat emptor.
Dennis
On Mon, Oct 24, 2011 at 2:17
Hi Michael:
Try this:
show.beta - function(model, x = 'x',
x1, x2, label, col=black, ...) {
abline(model, col=col, lwd=2)
xs - data.frame(c(x1, x2, x2))
names(xs) - attr(model$coefficients, 'names')[2]
ys - predict(model, xs)
lines(cbind(xs,ys[c(1,1,2)]),
Hi:
Here's one approach:
# Function to process a list component into a data frame
ff - function(x) {
data.frame(time = x[1], partitioning_mode = x[2], workload = x[3],
runtime = as.numeric(x[4:length(x)]) )
}
# Apply it to each element of the list:
do.call(rbind,
And the plyr version of this would be (using DF as the data frame name)
## transform method, mapping length(runtime) to all observations
## similar to David's results:
library('plyr')
ddply(DF, .(time, partitioning_mode), transform, n = length(runtime))
# or equivalently, the newer and somewhat
do.call(rbind, lapply(...))
HTH,
D.
On Sat, Oct 22, 2011 at 1:44 AM, Alaios ala...@yahoo.com wrote:
Dear all I have wrote the following line
return(as.vector(lapply(as.data.frame(data),min,simplify=TRUE)));
I want the lapply to return a vector as it returns a list with elements as
shown
Hi:
Here are a couple of ways, using the data snippet you provided as the
input data frame e.
Start by defining the function, which outputs a percentage:
f - function(n, mean, sd) {
s - rnorm(n, mean = mean, s = sd)
round(100 * sum(s 0.42)/length(s), 4)
}
(1) Use the plyr package
Here's another approach using the plyr package:
# Function to process each row of input:
g - function(d) {
y - unlist(d$observations)
if(length(y) 0)
data.frame(site = d$site, sector = d$sector, y = y) else NULL
}
library('plyr')
ddply(input, .(site), g)
site sector y
1
It appears that your object is currently a matrix. Here's a toy
example to illustrate how to get a stacked bar chart in ggplot2:
library('ggplot2')
m - matrix(1:9, ncol = 3, dimnames = list(letters[1:3], LETTERS[1:3]))
(d - as.data.frame(as.table(m)))
Var1 Var2 Freq
1aA1
2bA
Hi Michael:
Here's one way to get it from ggplot2. To avoid possible overplotting,
I jittered the points horizontally by +/- 0.2. I also reduced the point
size from the default 2 and increased the line thickness to 1.5 for
both fitted curves. In ggplot2, the term faceting is synonymous with
Hi:
Your approach to computing the means is not efficient; a better way
would be to use the aggregate() function. I would start by combining
the grouping variable and the three prediction variables into a data
frame. To get the groupwise mean for all three prediction variables,
you can use a
(legend.position = c(0.14, 0.885),
legend.background = theme_rect(fill = 'white'))
Dennis
On Fri, Oct 21, 2011 at 11:57 AM, Michael Friendly frien...@yorku.ca wrote:
Thanks very much, Dennis. See below for something I don't understand.
On 10/21/2011 12:15 PM, Dennis Murphy wrote:
Hi Michael
Here's another way, using the reshape2 package:
library(reshape2)
d - dcast(df, vial ~ measure, value_var = 'value')
d$diff - with(d, B - A)
d
vial A B diff
11 12 26 14
22 30 45 15
33 27 325
44 6 34 28
HTH,
Dennis
On Fri, Oct 21, 2011 at 3:31 PM, Nathan Miller
Hi:
Here's a way using the reshape2 package.
library('reshape2')
rsub - subset(rtest, concept %in% c('8.2.D', '8.3.A', '8.3.B'))
# want year ahead of concept in the variable list
rsub - rsub[, c(1:4, 9, 5:8)]
cast(rsub, id + test + subject + grade + year ~ concept, value_var = 'per_corr')
#
AFAIK, you can't 'add' two ggplot2 graphs together; the problem in
this case is that the two color scales would clash. If you're willing
to discretize the z values, then you could pull it off. Here's an
example:
d - data.frame(x = rnorm(100), y = rnorm(100), z = factor(1 +
(rnorm(100) 0)))
d1 -
Hi:
Here's one way with the plyr package. Using ds as the name of your
data frame (thank you for the dput and clear description of what you
wanted, BTW),
library('plyr')
ddply(ds, .(date), mutate, minspd = min(speed), Cmin =
C[which.min(speed)], diff = C - Cmin)
speed C house date hour
Here's one way to do it using the reshape package:
library('reshape')
cast(A, YEAR ~ TAX, value = 'NUMBER', fill = 0)
YEAR A B
1 2000 2 3
2 2001 2 4
3 2002 3 3
4 2003 1 0
5 2004 0 2
HTH,
Dennis
On Tue, Oct 18, 2011 at 7:51 PM, Michael E. Steiper
michaelstei...@gmail.com wrote:
Hi, I am a
Hi:
Your example data don't exhibit the types of behavior that you're
concerned about if you keep it simple. Here's how you can get the
*example* data plotted in both lattice and ggplot2; if the example
data are a subset of a much larger data set, then see below.
This 'works' on the example data
Prompted by David's xtabs() suggestion, one way to do what I think the
OP wants is to
* define day and unit as factors whose levels comprise the full range
of desired values;
* use xtabs();
* return the result as a data frame.
Something like
x - data.frame( day = factor(rep(c(4, 6), each = 8),
Hi:
levels(df.m2$Region)
[1] Africa Americas Asia Europe Oceania
Reorder your Region factor to the following:
df.m2$Region - factor(df.m2$Region, levels = c('Europe', 'Asia',
'Americas', 'Africa', 'Oceania'))
Then recopy the code from the definition of a
Hi:
I think the problem is that you're trying to append the predicted
probabilities as a new variable in the (one-line) data frame, when in
fact a vector of probabilities is output ( = number of ordered levels
of the response) for each new observation. Here's a reproducible
example hacked from
In the absence of a reproducible example (your example is not
reproducible as is), try this:
names(help.me) - as.character(2:4)
library('plyr')
newDF - ldply(help.me, rbind)
newDF[['.id']] - as.numeric(newDF[['.id']])
ldply will create a new column named .id that contains the name of the
list
Hi:
Try this:
ratok - data.frame(Id = rep(1:3, 3:1), value = c(2, 3, 4, 2, 1, 5))
aggregate(value ~ Id, data = ratok, FUN = mean)
Id value
1 1 3.0
2 2 1.5
3 3 5.0
aggregate() returns a data frame with the Id variable and mean(value).
HTH,
Dennis
On Sun, Oct 16, 2011 at 6:53 AM,
Hi:
I don't understand what you're attempting to do. Wouldn't courseid be
a categorical variable with a numeric label? If that is so, why are
you trying to compute an EDF? An EDF computes cumulative relative
frequency of a random variable, which by definition is numeric. If we
were talking about
, 2011 at 11:02 PM, David Winsemius
dwinsem...@comcast.net wrote:
On Oct 16, 2011, at 3:53 PM, Dennis Murphy wrote:
Hi:
I don't understand what you're attempting to do. Wouldn't courseid be
a categorical variable with a numeric label? If that is so, why are
you trying to compute an EDF
Hi:
Following the lead of others, here's a reproducible example that I
believe achieves what you want.
# Q1:
L - lapply(1:3, function(n)
data.frame(x = rnorm(6), y = rnorm(6), g = rep(1:2, each = 3)))
# Using David's suggestion:
L1 - lapply(L, function(d) subset(d, g == 1L))
L2 - lapply(L,
Hi:
One approach to this problem in plyr is to use the recently developed
mutate() function rather than ddply(). mutate() is a somewhat faster
version of transform(); when used as a standalone function, it doesn't
take a grouping variable as an argument. For this example, one could
use
In the absence of a reproducible example, a general question induces a
general response. I'd suggest creating a small data frame that
contains the x and y coordinates, a third variable consisting of
expressions representing each fitted model and an indicator of the
group to which the expression is
m - matrix(rpois(16, 10), ncol = 4)
m
[,1] [,2] [,3] [,4]
[1,]9 128 11
[2,] 127 118
[3,] 12788
[4,] 11 1148
apply(m, 2, sort)
[,1] [,2] [,3] [,4]
[1,]9748
[2,] 11788
[3,] 12 1188
[4,] 12 12
Hi:
Here's one approach:
dat - rnorm(40, 0, 2)
positions - matrix(c(3, 4, 5, 8, 9, 10, 20, 21, 22, 30, 31, 32),
ncol = 3, byrow = TRUE)
# Subdata
t(apply(positions, 1, function(x) dat[x]))
[,1] [,2] [,3]
[1,] 0.5679765 1.429396 2.9050931
Hi:
Here's one way to do it, adding the latticeExtra package:
array = rep(c(A,B,C),each = 36) # array replicate
spot = rep(1:4,27) # miRNA replicate on each array
miRNA = rep(rep(paste(miRNA,1:9,sep=.),each=4),3) # miRNA label
exprs = rnorm(mean=2.8,n = 108) # intensity
dat =
I tried this:
library(data.table)
N - 1000
T - N*10
d - data.table(gp= rep(1:T, rep(N,T)), val=rnorm(N*T), key = 'gp')
dim(d)
[1] 10002
# On my humble 8Gb system,
system.time(l - d[, split(val, gp)])
user system elapsed
4.150.094.27
I wouldn't be surprised if
Here's one way to do it with the plyr package:
library('plyr')
f - function(df) with(df, data.frame(B = B, E = seq(C, D)))
ddply(d, 'A', f)
A corresponding solution with the data.table package would be
library('data.table')
dt - data.table(d, key = 'A')
dt[, list(B, E = seq(C, D)), by = 'A']
Hi:
Please read this:
http://www.stata.com/support/faqs/stat/stepwise.html
Dennis
On Thu, Oct 6, 2011 at 6:41 AM, pigpigmeow gloryk...@hotmail.com wrote:
using AIC/BIC, I'm not know too much about this. I just know using p-value to
perform stepwise regression
if I used p-value to perform
Hi:
I have some data 'myData' in wide form (attached at the end), and
would like to convert it to long form. I wish to have five variables
in the result:
1) Subj: factor
2) Group: between-subjects factor (2 levels: s / w)
3) Reference: within-subject factor (2 levels: Me / She)
4) F:
Hi:
There are a few ways to do this. If you only have a few arrays, you
can simply add them and divide by the number of arrays. If you have a
large number of such arrays, this is inconvenient, so an alternative
is to ship the arrays into a list and use the Reduce() function. For
your example,
L
Hi:
Is this what you're after?
f - function(x) !any(x %in% terms_exclude) any(x %in% terms_include)
db[apply(db[, -1], 1, f), ]
ind test1 test2 test3
2 ind2 227 28.0
4 ind4 3 2 1.2
HTH,
Dennis
On Wed, Oct 5, 2011 at 8:53 AM, natalie.vanzuydam nvanzuy...@gmail.com
Try Googling 'Three factor ANOVA R'; it didn't take long to find a few
relevant hits.
Dennis
On Wed, Oct 5, 2011 at 10:56 AM, rafal rafalpedzi...@gmail.com wrote:
I am a student from Poland. What I am interested in is 3 factor anova with R.
Could you please help me find an example with using
Hi:
One option is the gridExtra package - run the example associated with
the tableGrob() function. Another is the addtable2plot() function in
the plotrix package. I'm pretty sure there's at least one other
package that can do this; I thought it was in the gplots package, but
couldn't find one
Thanks, Baptiste. I was looking for tableplot() or something like it
and thought textplot() was doing something different. Appreciate the
correction.
Dennis
On Wed, Oct 5, 2011 at 1:29 PM, baptiste auguie
baptiste.aug...@googlemail.com wrote:
On 6 October 2011 09:23, Dennis Murphy djmu
Hi:
It's a little tricky to read in a data frame 'by hand' without making
NA a default missing value; you've got to trick it a bit. I'm doing
this inefficiently, but if you have the two 'real' data sets stored in
separate files, read.table() is the way to go since it provides an
option for
Hi:
Here's one way:
m - matrix(rpois(100, 8), nrow = 5)
f - function(x) {
q - quantile(x, c(0.1, 0.9), na.rm = TRUE)
c(sum(x q[1]), sum(x q[2]))
}
t(apply(m, 1, f))
HTH,
Dennis
On Wed, Oct 5, 2011 at 8:11 PM, Ben qant ccqu...@gmail.com wrote:
Hello,
I'm trying to get the
Hi:
Here's a reproducible example:
d - data.frame(grp = factor(rep(c('x', 'y'), each = 5)),
ev = rnorm(10), dv = rnorm(10))
labl - list(expression(italic('x')), expression(italic('y')))
ggplot(d, aes(x = ev, y = dv, shape = grp)) + geom_point() +
scale_shape_manual('Group',
Hi:
Here's another way to do it with the plyr package, also not terribly
elegant. It assumes that rel.head is a factor in your original data
frame:
str(df)
'data.frame': 11 obs. of 2 variables:
$ ID : Factor w/ 6 levels 17100,17101,..: 1 1 2 3 4 4 5 5 5 6 ...
$ rel.head: Factor w/ 3
Hi:
The smooth is not going to replicate the quantile estimates you get
from the 'boxplots'; the smooth is estimating a conditional mean using
loess, with confidence limits associated with uncertainty in the
estimate of the conditional mean function, which are almost certainly
going to be
Hi Hadley:
When I tried your function on the example data, I got the following:
dd - data.frame(year = rep(2000:2008, each = 500), y = rnorm(4500))
g - function(df, qs = c(.05, .25, .50, .75, .95)) {
data.frame(q = qs, quantile(d$y, qs))
}
ddply(dd, .(year), g)
ddply(dd, .(year), g)
year
Hi:
INB4: if I have a nested design with treatment A and treatment B
within A, F-values are MSA/MSA(B) and MSA(B)/MSE, correct? How can I
make R give these values directly, without further coding?
This is how to get an equivalent model in lme4, but it probably isn't
what you expect
Hi:
Try this:
p - ggplot(mtcars, aes(factor(cyl), mpg))
p + geom_boxplot(aes(colour = factor(am)), fill = 'white') +
scale_colour_manual('am', values = c('0' = 'blue', '1' = 'black'))
HTH,
Dennis
On Tue, Oct 4, 2011 at 1:56 PM, Brian Smith bsmith030...@gmail.com wrote:
Hi,
I wanted to
Hi:
A small toy example:
fakedata - data.frame(group = factor(rep(1:3, each = 10), labels =
paste('Therapy', 1:3)),
city = factor(rep(c('Amsterdam', 'Rotterdam'), each = 5)),
pressure = rnorm(30))
with(fakedata, interaction.plot(group, city, pressure,
Hi:
I would think that, at least in principle, this should work:
a - ggplot(mtcars, aes(qsec, wt))
a + geom_point() + stat_smooth(fill=blue, colour=darkblue, size=2,
level = 0.9, alpha = 0.2) +
stat_smooth(fill = 'blue', colour = 'darkblue', size = 2,
Hi:
Here's the problem:
str(fun3)
'data.frame': 4 obs. of 3 variables:
$ sector:'data.frame': 4 obs. of 1 variable:
..$ gics_sector_name: chr Financials Financials Materials Materials
$ bebitpcchg: num -0.567 0.996 NA -42.759
$ ticker: chr UBSN VX Equity LLOY LN Equity
Let's make it a data frame instead:
# Read the data from your post into a data frame named d:
d - read.table(textConnection(
0.98 2
0.2 1
0.01 2
0.5 1
0.6 6))
closeAllConnections()
# Use the ave() function and append the result to d:
d$sumprob - with(d, ave(V1, V2, FUN = sum))
Hi:
This would be a lot easier to check with a reproducible example, but
here's a simplified version of your problem:
testd - data.frame(gps = rep(c(ADHALP,ADLCON,ADLARC,BDALAT,BDPARC),
each = 15),
trt = rep(LETTERS[1:3], each = 5),
Hi:
One way is to create a vector of pch values that you can pass into xyplot, e.g.,
dd - data.frame(x = 1:10, y = 1:10, pch = c(rep(1, 5), 16, rep(1, 4)))
library('lattice')
xyplot(y ~ x, data = dd, pch = dd$pch, col = 1)
HTH,
Dennis
On Fri, Sep 30, 2011 at 12:01 AM, Kang Min
Hi:
Here's one way to do it with the plyr package:
dd - read.table(textConnection(
yearsps cm w
200950 16 22
200950 17 42
200950 18 45
200951 15 45
200951 16 53
200951 17 73
201050 15
Hi:
Here's one approach using the reshape() function in base R:
# Read in your data:
d - read.table(textConnection(
Candidate.IDSpecialty Office Score
110002 C London 47
110002 C East 48
Hi:
The problem is that in your example, you have unequal numbers of rows
in B that match the 1's pattern in A[i, ]. The function below cycles
through the rows of A and returns, for each row of A, the rows in B
that have 1's in the same columns as A[i, ]. By necessity, this
returns a list. Notice
Hi:
The head() function is helpful here:
(i) plyr::ddply()
library('plyr')
ddply(dat, .(id), function(d) head(d, 1))
id value
1 1 5
2 2 4
(ii) aggregate():
aggregate(value ~ id, data = dat, FUN = function(x) head(x, 1))
id value
1 1 5
2 2 4
The formula version of
Hi:
Here's one approach with package reshape2. I copied the same data
frame 10 times to illustrate the idea, but as long as your data frames
have the same structure (same variables, same dimension), this should
work.
library('plyr')
library('reshape2')
# Use your example data as the template:
Hi:
Question: Do you want 37 different panels with plots of quant vs. date
by param, or two panels (one per chemical) with all 37 streams? If you
only want two of the eight chemicals, I'd suggest using subset() to
select out the pair you want and then redefine the param factor so
that the subset
I don't see the problem. AFAICT, you want plots of quant vs. sampdate
by chemicals for each of the 37 streams. Essentially, you seem to want
two superimposed time plots in each panel. A time plot is a
scatterplot with a time-ordered (usually horizontal) axis.
If you use ggplot2 or lattice to
Hi:
Using Michael's example data, here's another approach:
x = data.frame(x = 1:5)
y = data.frame(y = 1:10)
z = data.frame(z = 1:3)
# Generate a vector to name the list components
nms - c('x', 'y', 'z')
# Combine data frames into a list
L - list(x, y, z)
# name them
names(L) - nms
# Use
Hi Steven:
How about this?
d - rep(20110101,24)
h - sprintf('%04d', seq(0, 2300, by = 100))
df - data.frame(LST_DATE = d, LST_TIME = h, data = rnorm(24, 0, 1))
df - transform(df, datetime = as.POSIXct(paste(LST_DATE, LST_TIME),
format = '%Y%m%d %H%M'))
library(zoo)
X
http://cran.r-project.org/web/views/
Look for 'machine learning'.
Dennis
On Sun, Sep 11, 2011 at 11:33 PM, Jay josip.2...@gmail.com wrote:
If the answer is so obvious, could somebody please spell it out?
On Sep 11, 10:59 pm, Jason Edgecombe ja...@rampaginggeek.com wrote:
Try this:
Hi:
Here's one approach:
# date typo fixed in record 5 - changed 35 to 5
tC - textConnection(
Subject Dateparameter1
bob 3/2/99 10
bob 4/2/99 10
bob 5/5/99 10
bob 6/27/99 NA
bob 8/5/01 10
bob 3/2/02 10
steve 1/2/99 4
steve 2/2/00 7
steve 3/2/01 10
steve
Hi:
Start here:
http://www.quantmod.com/examples/
HTH,
Dennis
On Fri, Sep 9, 2011 at 10:44 PM, veepsirtt veepsi...@gmail.com wrote:
Hi
I want to bring the graph from the site www.avasaram.com
for the symbol GOOG
here is the link
Hi:
There are two ways to go about this, depending on what you want.
# 1: Jean's output format:
# I'm using the reshape package here with cast(),
# but it should work similarly with dcast() in reshape2:
cast(example, DATE ~ SENSOR, value = 'VALUE')
DATE A B C D E F
1
Hi:
Hmm, these files look familiar...Lahman database? :)
I have the 2010 set, so here's how I got this to work on my system:
files - list.files(pattern = '*.csv')
files
[1] Allstar.csv AllstarFull.csv
[3] Appearances.csv AwardsManagers.csv
[5] AwardsPlayers.csv
Hi:
Here are a couple of ways to do this. The barplot() attempt is pretty
close to the original banner plot except that the color is transparent
to the (white) background. The ggplot2 graph is similar, but the code
may be less familiar if you don't use the package. In the latter, the
variable
Hi:
You could try something like this:
# rbind the two melted data frames:
comb - rbind(mat2, tab2)
# create a group factor to distinguish the two data sets
comb - mutate(comb, gp = factor(rep(1:2, each = 1)))
# Generate the plot:
ggplot(comb) +
geom_freqpoly(aes(x = value, y =
From the plotmath help page:
Control characters (e.g. \n) are not interpreted in character strings
in plotmath, unlike normal plotting.
In other words, you can't do that.
However, since you're using grid graphics, this thread from the
ggplot2 group might be helpful:
# install.packages('sos')
library(sos)
findFn('radar plot')
gets 29 hits on my system. The two that seem to be the most relevant
are radarchart() from the fmsb package and, as mentioned in the other
reply, stars() from the autoloaded graphics package.
BTW, read the quoted text and observe how
Hi:
Your code doesn't work, but if you want to do spaghetti plots, they're
rather easy to do in ggplot2 or lattice.
The first problem you have is that one group in your test data has all
NA responses, so by() or any other groupwise summarization function is
going to choke unless you give it a
~ age, data = w, group = .id,
panel = function(x, y, groups, ...) {
panel.xyplot(x, y, groups = groups, type = 'l', col = 1, ...)
panel.lmline(x, y, col = 'blue', lwd = 2)
} )
Dennis
On Wed, Sep 7, 2011 at 2:07 PM, Dennis Murphy djmu
Hi:
For your test data, try this:
# Result of dput(NerveSurv)
NerveSurv - structure(list(Time = c(0, 0.25, 0.5, 0.75, 1, 1.25, 1.5, 0,
0.25, 0.5, 0.75, 1, 1.25, 1.5, 0, 0.25, 0.5, 0.75, 1, 1.25, 1.5
), SAP = c(1, 1.04, 1.04, 1.06, 1.04, 1.22, 1.01, 1, 1.01, 1.01,
1.06, 1.01, 0.977, 0.959, 1,
Hi:
Here are a few informal timings on my machine with the following
example. The data.table package is worth investigating, particularly
in problems where its advantages can scale with size.
library(data.table)
dt - data.table(x = sample(1:50, 100, replace = TRUE),
y =
Hi:
I tried to figure out what you were doing...some of it I think I
grasped, other parts not so much.
On Fri, Sep 2, 2011 at 8:18 PM, Maya Joshi maya.d.jo...@gmail.com wrote:
Dear R experts.
I might be missing something obvious. I have been trying to fix this problem
for some weeks. Please
Hi:
Try this:
key1 -
draw.key(list(text=list(levels(Cal_dat$Commodity)),
title=Ore type,
border = TRUE,
background = 'ivory',
points=list(pch=22, cex=1.3, fill=col.pat, col=black)),
draw = FALSE)
key2 -
Hi:
Here's one approach:
d - read.table(textConnection(
V1 V2 V3 V4 V5 V6
1 G A G G G G
2 A A G A A G), header = TRUE, stringsAsFactors = FALSE)
closeAllConnections()
# Create two vectors of variable names, one for odd numbered,
# one for even numbered
vars1 -
Hi:
Here's one approach:
X1 - sample(1:4, 10, replace = TRUE, prob = c(0.4, 0.2, 0.2, 0.2))
foo - function(x) {
m - matrix(NA, nrow = length(x), ncol = length(x))
m[, 1] - x
idx - seq_len(length(x))
for(j in idx[-1]) {
k - sample(idx, 2)
x - replace(x, k, 5)
m[, j]
From: Dennis Murphy djmu...@gmail.com
To: Alaios ala...@yahoo.com
Cc: R-help@r-project.org R-help@r-project.org
Sent: Wednesday, August 31, 2011 9:34 PM
Subject: Re: [R] ggplot2 to create a square plot
Hi:
I'd suggest using ggsave(); in particular, see its height
Hi:
Things work if x is the response and y is the covariate. To use the
approach I describe below, you need RStudio and its manipulate package
(which is only available in RStudio - you won't find it on CRAN). You
can download and install RStudio freely from http://rstudio.org/ ; it
is available
Hi:
I'd suggest using ggsave(); in particular, see its height = and width
= arguments. If you have some time, you could look at some examples of
ggplot2 themes:
https://github.com/hadley/ggplot2/wiki/themes
and some examples of how to use various opts():
Hi:
It's polite to mention from which package you extract certain
functions. For example, the function dmvnorm() exists in at least the
following packages (following a search from the sos package):
mixtools, emdbook, klaR and mvtnorm, not to mention related functions
in three other packages.
Hi:
The bars *are* ordered in the same way, but when you use coord_flip(),
the left category goes on top and the right category goes on the
bottom. Is this what you want?
ggplot(df, aes(x = name, y = value, fill = type)) +
geom_bar(position = position_dodge()) +
coord_flip() +
Hi:
integrate() is not a vectorized function. This appears to work:
sapply(1:2, function(x) func(x, 0.1, 0.1, sad = Exp))
[1] 0.250 0.125
In this case, sapply() is a disguised for loop.
HTH,
Dennis
On Mon, Aug 29, 2011 at 9:45 AM, . . xkzi...@gmail.com wrote:
Ooops,
sorry!
The problem
?tryCatch
HTH,
Dennis
On Mon, Aug 29, 2011 at 9:04 AM, s1oliver s1oli...@ucsd.edu wrote:
Hi, beginner to R and was having some problems scraping data from tables in
html using the XML package. I have included some code below.
I am trying to loop through a series of html pages, each of which
Hi:
This is straightforward to do with the plyr package:
# install.packages('plyr')
library('plyr')
set.seed(1234)
df - data.frame(clvar = rep(1:4, each = 10), yvar = rnorm(40, 10, 6),
var1 = rnorm(40, 10, 4), var2 = rnorm(40, 10, 4),
var3 = rnorm(40, 5, 2),
Hi:
Dimitris' solution is appropriate, but it needs to be mentioned that
the approach I offered earlier in this thread differs from the
lmList() approach. lmList() uses a pooled measure of error MSE (which
you can see at the bottom of the output from summary(mlis) ), whereas
the plyr approach
On Sat, Aug 27, 2011 at 11:07 AM, karthicklakshman
karthick.laksh...@gmail.com wrote:
Dear R community,
With an advantage of being NEW to R, I would like to post a very basic
query here,
Really? I found two posts with your name on it dating from October and
November of 2010.
Hi:
Try this:
require('xts')
merge.zoo(zoo(a), zoo(b), all = c(TRUE, TRUE))
ZWD.UGX SCHB.Close
2010-03-31 NA 28.02
2010-04-01 7.6343 NA
2010-04-02 7.6343 NA
2010-04-03 7.5458 NA
2010-04-04 7.4532 28.30
2010-04-05 7.4040 28.38
Hi:
Similar in vein to the other respondents, you could try something like this:
On Thu, Aug 25, 2011 at 1:17 AM, Jie TANG totang...@gmail.com wrote:
for example : I have files with the name
ma01.dat,ma02.dat,ma03.dat,ma04.dat,I want to read the data in
these files into one data.frame
#
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