On 26/03/2014 20:01, Tomassini, Letizia wrote:
I would like to understand why the fastclus procedure in SAS is affected by the
initial order of the data. So, with the same dataset, but sorted in a different
way, I get different clusters rearrangements. I find this really disturbing. R
seems
Dear R-users,
I experimented with readLine settings ('~/.inputrc') and then restored the
file, but R still behaves like if '.inputrc' is modified. '/etc/inputrc' is
also fine. I suppose there're some separate R-specific configuration files.
Could you please kindly advise how to get things
hi :)
if i using importance sampling for obtain postreior density, can i use
teachingdemos for obtain hpd interval?what are you doing with weight?
or is other package for cmpute this type of sampling?
please help me soon as soon
[[alternative HTML version deleted]]
HI Eliza,
May be this helps:
set.seed(42)
el - matrix(sample(1:20,327*365,replace=TRUE),ncol=327)
colnames(el) - as.character(interaction(df1,sep=*))
el1 - el[,colnames(el) %in% as.character(interaction(df2,sep=*))]
dim(el1)
#[1] 365 5
A.K.
On Tuesday, March 25, 2014 4:15 PM, eliza botto
Hi,
Try:
A - as.data.frame(matrix(1:20,byrow=TRUE,ncol=4))
B - as.data.frame(matrix(21:40,byrow=TRUE,ncol=4))
AB - rbind(A,B)
#if the ?row.names of both datasets are from 1:nrow(dataset)
AB1 - AB[order(as.numeric(c(row.names(A),row.names(B,]
#or
AB2 -
HI Elio,
May be this helps:
indx - order(gsub(\\d+,,colnames(res)))
res1 - res[indx,indx]
A.K.
On Thursday, March 27, 2014 6:24 AM, Elio Shijaku sel...@gmail.com wrote:
Hi Arun,
Thanks a lot for your continuous help. I am attaching a sample with the row /
column names. Basically, the
Hi,
here's my code
X - 1:100
I want to select number divisible by 3 out of them how can I select it?
( I tried following
X - 1:100
DIV - Y - X/3
But I am getting whole number and number with fractions. WHole intgers are
my number of interest from original X. How can I traceback to number
Dear R-users,
I would like to change the font of one specific axis label to bold face in
each panel. Taking one of Deepayan's plots as an example, how do I change
All Postdoctorates to bold face?
dotplot(prop.table(postdoc, margin = 1), groups = FALSE, index.cond =
function(x, y) median(x), xlab
I have a data A which looks like
author_id paper_id prob
731249431
731249431
731 6889741
731 964345.8
731 1201905.9
731 12679921
736249 .2
736 6889 1
736 94345.7
Hi,
here's my code
X - 1:100
I want to select number divisible by 3 out of them how can I select it?
( I tried following
X - 1:100
DIV - Y - X/3
But I am getting whole number and number with fractions. WHole intgers are
my number of interest from original X. How can I traceback to number
On 27-03-2014, at 10:23, Prabhakar Ghorpade dr.prabhaka...@gmail.com wrote:
Hi,
here's my code
X - 1:100
I want to select number divisible by 3 out of them how can I select it?
( I tried following
X - 1:100
DIV - Y - X/3
But I am getting whole number and number with fractions.
Hi there,
Try
X[X %% 3 == 0]
HTH,
Jorge.-
On Thu, Mar 27, 2014 at 6:46 PM, Prabhakar Ghorpade
dr.prabhaka...@gmail.com wrote:
Hi,
here's my code
X - 1:100
I want to select number divisible by 3 out of them how can I select it?
( I tried following
X - 1:100
DIV - Y - X/3
But I
Hi
One option is
seq(3,100,3)
or you can use rounded numbers for comparison
?round
Petr
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
project.org] On Behalf Of Prabhakar Ghorpade
Sent: Thursday, March 27, 2014 8:47 AM
To: r-help@r-project.org
Hello,
Something like that?
R X - 1:100
R Y - X %% 3
R Y - ifelse(Y==0, TRUE, FALSE)
R X[Y]
[1] 3 6 9 12 15 18 21 24 27 30 33 36 39 42 45 48 51 54 57 60 63 66 69 72 75
[26] 78 81 84 87 90 93 96 99
HTH,
Pascal
On Thu, Mar 27, 2014 at 6:23 PM, Prabhakar Ghorpade
dr.prabhaka...@gmail.com
Excellent Arun, it worked, many thanks.
E.
On Thu, Mar 27, 2014 at 11:55 AM, arun smartpink...@yahoo.com wrote:
HI Elio,
May be this helps:
indx - order(gsub(\\d+,,colnames(res)))
res1 - res[indx,indx]
A.K.
On Thursday, March 27, 2014 6:24 AM, Elio Shijaku sel...@gmail.com
Hello,
Inline.
Em 27-03-2014 11:32, Pascal Oettli escreveu:
Hello,
Something like that?
R X - 1:100
R Y - X %% 3
R Y - ifelse(Y==0, TRUE, FALSE)
The ifelse here is not needed, you can simply do
Y - Y == 0
Rui Barradas
R X[Y]
[1] 3 6 9 12 15 18 21 24 27 30 33 36 39 42 45 48 51 54
Hi All,
I am using R for the purpose of multilevel modelling for the first time. I am
trying to examine individuals interpersonal changes in the dependent variable
over time and how this varies between groups.
I am using the following code:
treat.lme1-lme(DependentVariable~Treatment*I(Time-1),
Dear Pascal,
Thanks for your reply. From your answer I perceived that if followings are
first three elements of a file
dput(ccc[1:3])
c(0.15912090241909, 0.167244642972946, 0.192471280694008)
then 0.15912090241909 is precipitation magnitude , 0.167244642972946 is RSTN
and 0.192471280694008 is
It's rude to ask a question both on r-help and on stackoverflow
(http://stackoverflow.com/questions/22685896), because people might
spend their time answering your question when it's already been
answered elsewhere.
Hadley
On Thu, Mar 27, 2014 at 6:01 AM, Rohit Gupta rhtgpt...@gmail.com wrote:
On Wed, 26 Mar 2014, Rich Shepard wrote:
According to the ts.union help, ‘ts.union’ pads with ‘NA’s to the total
time coverage but the problem seems to be that one series has a single NA
while the other has 2 NAs. Both series have 792 rows of (date, value).
I should have written in the
No you're not right as far as I can tell from the read_v1100r2.f90 fortran code
you can find in the download folder.
For day 1 in 1961 I think that prcp = ccc[1:25200], rstn = ccc[25201:50400],
and rsnw = ccc[50401:75600] and the same rule applies to the following days
with appropriate
Hi,
Maybe I wasn't clear enough.
Day 01 precip map of 180x140
Day 01 rstn map of 180x140
Day 01 flag map of 180x140
Day 02 precip map of 180x140
And so on
HTH
Pascal
On Thu, Mar 27, 2014 at 10:14 PM, Frede Aakmann Tøgersen
fr...@vestas.com wrote:
No you're not right as far
Dear Frede and pascal,
Thankyou very much. I am so glad that you replied positively.
Thankyou very once again.
Eliza
From: kri...@ymail.com
Date: Thu, 27 Mar 2014 22:38:42 +0900
Subject: Re: [R] reading dataset
To: fr...@vestas.com
CC: eliza_bo...@hotmail.com; r-help@r-project.org
Hi,
Dear R users,
I have created a package and would like to check it for CRAN/R-Forge
submission. I use some parallelized code with the help of foreach and doMC.
R CMD check now gives me a warning even though my code is correctly
programmed (at least I think that it is). The following dummy
A reproducible example would help us help you, but you probably need to use
the font argument, something along the lines of this...
xlab = list(some axis label,cex=1.5,font=4) # 2=bold, 4 = bold italic
Kevin
On Thu, Mar 27, 2014 at 3:57 AM, el_alisio malnama...@gmx.de wrote:
Dear R-users,
On 27/03/2014 10:29 AM, Jannis wrote:
Dear R users,
I have created a package and would like to check it for CRAN/R-Forge
submission. I use some parallelized code with the help of foreach and doMC.
R CMD check now gives me a warning even though my code is correctly
programmed (at least I think
Hi all!
I am trying to drop columns from a data frame dynamically depending on user
input. The dataframe whose columns need to be dropped is called Finaldata
So here is what I do:
V is a dataframe with columns v1 and v2 as follows
v1 v2
1 1 Shape
2 0 Length
3 0
Why aren't you using ts.intersect instead of ts.union?
Bill Dunlap
TIBCO Software
wdunlap tibco.com
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf
Of Rich Shepard
Sent: Thursday, March 27, 2014 5:45 AM
To:
There are many ways. You're making it overly complicated
Here, in an actual reproducible example (as you were requested to submit):
V - data.frame(v1=c(1,0,0), v2=c(Shape, Length, Rate),
stringsAsFactors=FALSE)
Finaldata - data.frame(Shape = runif(5), Length = runif(5), Rate = runif(5))
#
On Thu, 27 Mar 2014, William Dunlap wrote:
Why aren't you using ts.intersect instead of ts.union?
Bill,
I was wondering the same thing. The reason (perhaps excuse would be the
more appropriate term) is my emulating the example in Paul Cowpertwait and
Andrew Metcalfe's Introductory Time
Here is how you can do it with data.table:
x - read.table(text = author_id paper_id prob
+731249431
+731249431
+731 688974 1
+731 964345.8
+731 1201905.9
+731 12679921
+736249 .2
+736 6889
Hi Sarah,
Thanks! Do agree its over complicated.
However looking at the solutions I think I did not state my problem
completely.
V provides choices for only certain set of columns in Finaldata.
So v2 may not represent all columns of Finaldata.
I want to retain columns not provided as a choice for
I have tried to read the source code but since I am not a computer engineer nor
a computer programmer I was not able to fully understand it. I wonder if I
should look for somebody here on campus (Washington State University) who may
be able to read it for me. In any case, I think that David
That only requires two small changes in Sarah's first solution:
Finaldata[, !colnames(Finaldata) %in% V$v2[V$v1 == 1]]
Length Rate
1 0.53607323 0.01739951
2 0.15405615 0.11837908
3 0.04542388 0.53702702
4 0.15633703 0.68870041
5 0.35293973 0.38258981
Dear all,
I have a vector x and a vector y = f(x).
e.g.
x = c(0,1, 3,3.4, 5, 10, 11.23)
y = x**2 + rnorm(length(x))
I would like to apply a moving average to y, knowing that the x values
are not uniformly spaced. How can I do this in R?
What alternatives I have to filter out the noise from Y??
On 27.03.2014 15:58, Duncan Murdoch wrote:
I would use a slightly less dirty hack: call globalVariables() to
declare that i is global.
The foreach() function is using nonstandard evaluation to make this
work, and codetools (that does the checks) doesn't know all the ins
and outs of it.
On 27/03/2014 12:46 PM, Jannis wrote:
On 27.03.2014 15:58, Duncan Murdoch wrote:
I would use a slightly less dirty hack: call globalVariables() to
declare that i is global.
The foreach() function is using nonstandard evaluation to make this
work, and codetools (that does the checks)
Marc Schwartz marc_schwa...@me.com
on Wed, 26 Mar 2014 16:25:08 -0500 writes:
On Mar 26, 2014, at 4:14 PM, David Winsemius dwinsem...@comcast.net
wrote:
On Mar 25, 2014, at 5:31 PM, Rolf Turner wrote:
On 26/03/14 12:51, David Winsemius wrote:
On Mar
On Mar 27, 2014, at 11:58 AM, Martin Maechler maech...@stat.math.ethz.ch
wrote:
Marc Schwartz marc_schwa...@me.com
on Wed, 26 Mar 2014 16:25:08 -0500 writes:
On Mar 26, 2014, at 4:14 PM, David Winsemius dwinsem...@comcast.net wrote:
On Mar 25, 2014, at 5:31 PM, Rolf Turner wrote:
On Thu, 27 Mar 2014, Rich Shepard wrote:
I'll report back on whether that avoids the errors.
Changing from ts.union() to ts.intersect() did not make a difference. So,
I went back to bare metal by using awk to generate two-column text files,
each consisting of a date and an interger or real
On Thu, 27 Mar 2014, Rich Shepard wrote:
s95ec.z - read.zoo(s95ec.dat, sep = ., format = %Y-%m-%d
^
) didn't
copy but was present on the command line.
Trying to
Thank you! Works like a charm!
On Thu, Mar 27, 2014 at 12:19 PM, David Carlson dcarl...@tamu.edu wrote:
That only requires two small changes in Sarah's first solution:
Finaldata[, !colnames(Finaldata) %in% V$v2[V$v1 == 1]]
Length Rate
1 0.53607323 0.01739951
2 0.15405615
Pascal, Jorge, Berend, Pikal and Rui. Thanks all it helps.
On Thu, Mar 27, 2014 at 8:18 PM, Prabhakar Ghorpade
dr.prabhaka...@gmail.com wrote:
Hi,
How do I find help on %%
Regards,
Prabhakar
On Thu, Mar 27, 2014 at 4:56 PM, Jorge I Velez
jorgeivanve...@gmail.comwrote:
Hi there,
Hi,
How do I find help on %%
Regards,
Prabhakar
On Thu, Mar 27, 2014 at 4:56 PM, Jorge I Velez jorgeivanve...@gmail.comwrote:
Hi there,
Try
X[X %% 3 == 0]
HTH,
Jorge.-
On Thu, Mar 27, 2014 at 6:46 PM, Prabhakar Ghorpade
dr.prabhaka...@gmail.com wrote:
Hi,
here's my code
X -
Hello,
My main question is wheter my data is distributed normally. As the
shapiro.test doesnt work for large
data sets I prefer the ks.test.
But I have some problems to understand the completely different p-values:
ks.test (test, pnorm, mean (test), sd (test))
One-sample
i try to generate 27*5 matrix of observation using the following code but is
given error
kindly assist and correct the problem
x1-c(-1,0,1,-1,0,1,-1,0,1,-1,0,1,-1,0,1,-1,0,1,-1,0,1,-1,0,1,-1,0,1)
x2-c(-1,-1,-1,0,0,0,1,1,1,-1,-1,-1,0,0,0,1,1,1,-1,-1,-1,0,0,0,1,1,1)
Hi,
I have implemented a therapeutic intervention on two groups (one is a
control group) and tested them in two moments using some assessment
tools (with normative data). Now I want to compare the experimental
group with the control group using clinical equivalence testing. To do
this I need to
Hi Kevin,
thanks for the advice! The code is now reproducible. Sorry!
Actually I wanted to print only one x-label (All Postdoctorates) in bold
face, not all of them. This would be easily achieved, if All
Postdoctorates appeared always in the same row of each panel, but this is
not the case here.
Hello,
If I have x-c(5,9,8,3,4,5,6,7,)
and y -(a, b,c, d,e,f, g,h)
and D - data.frame(x,y)
I did this max(x)
and My expectede anwer is b.
How do I select element in y which correspond to maximum number in x?
thanks.
regards,
Prabhakar
[[alternative HTML version deleted]]
Here is one way. I took your data and applied the 'approx' function to get
evenly spaced values and then 'filter' to create a moving average of a
window of size 5.
set.seed(1)
x = c(0,1, 3,3.4, 5, 10, 11.23)
y = x**2 + rnorm(length(x))
rbind(x, y)
# create 'even' spacing using the 'approx'
ok
btw thanks
On Thu, Mar 27, 2014 at 5:55 PM, Hadley Wickham h.wick...@gmail.com wrote:
It's rude to ask a question both on r-help and on stackoverflow
(http://stackoverflow.com/questions/22685896), because people might
spend their time answering your question when it's already been
Hello,
I need to compare a polygon with a set of other polygons.
I'd like to index polygons with a grid in order not to compare definitely
distant ones.
How can I do this?
I need
1. Transform polygons to cells
2. Find cell neighbors (defined cell distance) for a polygon
Thanks
It would have been helpful it you had posted the error message you got.
Here is what I got running your code:
Error in x.matrix[i, j] - rnorm(5, mu[i], sig[i]) : object 'j' not found
No suitable frames for recover()
It seems you have not defined 'j' that you trying to index with.
Jim Holtman
First, read the Introduction to R to understand what dataframes are and how
to access data within them. Then correct you code so it does not have
error: 'x' has a trailing comma and 'y' was not call the 'c' function.
Once you have done that, you might get the following:
x-c(5,9,8,3,4,5,6,7)
Thanks, your solution using ave() works perfectly.
/johannes
-Ursprüngliche Nachricht-
Von: Bert Gunter gunter.ber...@gene.com
An: Johannes Radinger johannesradin...@gmail.com
Cc: R help r-help@r-project.org
Gesendet: Mittwoch, 26. März 2014 16:45:43 GMT+00:00
Betreff: Re: [R] dataframe
On Mar 27, 2014, at 6:53 AM, Manuel Carona wrote:
Hi,
I have implemented a therapeutic intervention on two groups (one is a
control group) and tested them in two moments using some assessment
tools (with normative data). Now I want to compare the experimental
group with the control group
On Mar 27, 2014, at 4:31 AM, IZHAK shabsogh wrote:
i try to generate 27*5 matrix of observation using the following code but is
given error
kindly assist and correct the problem
x1-c(-1,0,1,-1,0,1,-1,0,1,-1,0,1,-1,0,1,-1,0,1,-1,0,1,-1,0,1,-1,0,1)
Perhaps have a look at the sp package and especially the over function.
Br. Frede
Sendt fra Samsung mobil
Oprindelig meddelelse
Fra: Leonid Shvartser
Dato:27/03/2014 20.27 (GMT+01:00)
Til: r-help@r-project.org
Emne: [R] index polygons
Hello,
I need to compare a polygon with
On Mar 27, 2014, at 7:48 AM, Prabhakar Ghorpade wrote:
Hi,
How do I find help on %%
?'%%' # Need to quote specials
Also look at:
?'-'
?'%in%
?'function'
--
David.
Regards,
Prabhakar
On Thu, Mar 27, 2014 at 4:56 PM, Jorge I Velez
jorgeivanve...@gmail.comwrote:
Hi there,
On Mar 27, 2014, at 8:53 AM, Manuel Carona unku...@gmail.com wrote:
Hi,
I have implemented a therapeutic intervention on two groups (one is a
control group) and tested them in two moments using some assessment
tools (with normative data). Now I want to compare the experimental
group with
On Mar 27, 2014, at 8:14 AM, Hermann Norpois hnorp...@gmail.com wrote:
Hello,
My main question is wheter my data is distributed normally. As the
shapiro.test doesnt work for large
data sets I prefer the ks.test.
But I have some problems to understand the completely different p-values:
On 03/28/2014 12:53 AM, Manuel Carona wrote:
Hi,
I have implemented a therapeutic intervention on two groups (one is a
control group) and tested them in two moments using some assessment
tools (with normative data). Now I want to compare the experimental
group with the control group using
Hi,
when typing code I get like this
x -(1,2,
+
+
+
Can I delete + and comma after 2 and go back to end of comma as x -c(1,2)
Thanks.
Prabhakar
--
Dr.Ghorpade Prabhakar B.
Ph.D. Scholar ( Animal Biochemistry)
Indian Veterinary Research Institute.
India
[[alternative HTML
Or, to show the power of the R language, so to speak, and assuming that
Dave is correct, there is no need to use a for() loop:
xmat - matrix( rnorm( 27*5, rep(mu,5), rep(sig,5) ) , nrow=27)
To verify that I've created the vectors in the right order, and converted
to a matrix in the right order,
The hpd functions in the TeachingDemos package do not currently take
weights, so they will not work with the results of importance
sampling.
On Wed, Mar 26, 2014 at 10:35 PM, Baran rad baran@gmail.com wrote:
hi :)
if i using importance sampling for obtain postreior density, can i use
On 03/28/2014 06:33 AM, Prabhakar Ghorpade wrote:
Hi,
when typing code I get like this
x-(1,2,
+
+
+
Can I delete + and comma after 2 and go back to end of comma as x-c(1,2)
Hi Prabhakar,
I don't think so, as the command line reading won't pass your backspace.
I'd say just press
Or tweaking Don's solution but using the original mu and sig
vectors:
set.seed(42)
x.matrix-matrix(0, nrow=27, ncol=5)
for(i in 1:27){
+ x.matrix[i, ] - rnorm(5, mu[i], sig[i])
+ }
set.seed(42)
xmat - matrix(rnorm(27*5, rep(mu, each=5), rep(sig, each=5)),
+ nrow=27, byrow=TRUE)
Hi all,
I've spent too long in matlab land recently and seem to have forgotten
my R skillz ;-)
I'm sure I'm missing a simple way to do this...
Given a data frame
id-rep(1:5,2)
eye-c(rep('l',5),rep('r',5))
age-rep(round(runif(5,0,12)),2)
response-round(runif(10,1,10)*10)/10
I'm no expert on the adonis functions by the error you are getting
usually means the dataset is not quite what you expected. Are you sure
the event dataset is loading as you expect?
On Tue, 2014-03-25 at 16:36 -0400, Maggie Wisniewska wrote:
Hello,
I am an R noivce, so excuse my simple
It may be possible to do this in a single step, but
x1 - aggregate(response~id+age, data, mean)
x2 - data[data$eye==l, c(id, response2)]
merge(x1, x2)
id age response response2
1 1 2 4.60 High
2 2 9 2.65 High
3 3 5 3.65 High
4 4 2 7.55 High
5
Hi,
You could try:
library(dplyr)
A %.% group_by(author_id) %.% arrange(desc(prob)) %.%
summarise(paper_id=paste(paper_id,collapse=,))
A.K.
On Thursday, March 27, 2014 11:40 AM, Rohit Gupta rhtgpt...@gmail.com wrote:
I have a data A which looks like
author_id paper_id prob
Thanks David,
Neat use of merge there.
I think I can also do this with:
ddply(data,.(id),f)
On Thu, 2014-03-27 at 16:06 -0500, David Carlson wrote:
It may be possible to do this in a single step, but
x1 - aggregate(response~id+age, data, mean)
x2 - data[data$eye==l, c(id, response2)]
?%%
PLEASE read an Introduction to R (ships with R) or an online tutorial
of your choice before posting further. You then will not have to waste
nearly as much time (yours and ours) posting and waiting for answers!
Cheers,
Bert
Bert Gunter
Genentech Nonclinical Biostatistics
(650) 467-7374
On Mar 27, 2014, at 8:11 AM, el_alisio wrote:
Hi Kevin,
thanks for the advice! The code is now reproducible. Sorry!
No, it is not, since you can only read it on Nabble, yecch. There is some sort
of weird format stuff that Nabble does that prevents code from appearing on
Rhelp copies.
On Mar 27, 2014, at 2:34 PM, David Winsemius wrote:
On Mar 27, 2014, at 8:11 AM, el_alisio wrote:
Hi Kevin,
thanks for the advice! The code is now reproducible. Sorry!
No, it is not, since you can only read it on Nabble, yecch. There is some
sort of weird format stuff that Nabble
I was under the impression that
do.call(foo,list(x=x,y=y))
should yield the same result as
foo(x,y).
However if I do
x - 1:10
y - (x-5.5)^2
do.call(plot,list(x=x,y=y))
I get the expected plot but with the y-values (surrounded by c()) being
printed
You get what you wanted from
do.call(plot,list(x=quote(x),y=quote(y)))
By the time do.call() gets the arguments it doesn't know how y was
originally computed, just what values it has.
-thomas
On Thu, Mar 27, 2014 at 6:17 PM, Rolf Turner r.tur...@auckland.ac.nzwrote:
I was under the
On 27/03/2014, 7:17 PM, Rolf Turner wrote:
I was under the impression that
do.call(foo,list(x=x,y=y))
should yield the same result as
foo(x,y).
However if I do
x - 1:10
y - (x-5.5)^2
do.call(plot)
I get the expected plot but with the y-values
On 27/03/2014, 7:38 PM, Thomas Lumley wrote:
You get what you wanted from
do.call(plot,list(x=quote(x),y=quote(y)))
By the time do.call() gets the arguments it doesn't know how y was
originally computed, just what values it has.
This works, but it doesn't make sense to me. The arguments end
Hello,
I have a dataset with family data. For an analysis, I need to select one
subject per family at random.
Here is an example of what my data look like:
FamilyID IndividualID DadIDMomID Sex
1101103 104 1
1
On 28/03/14 12:49, Duncan Murdoch wrote:
On 27/03/2014, 7:17 PM, Rolf Turner wrote:
I was under the impression that
do.call(foo,list(x=x,y=y))
should yield the same result as
foo(x,y).
However if I do
x - 1:10
y - (x-5.5)^2
do.call(plot)
I get the expected plot but
Tena koe Whitney
tapply should work. Try:
tapply(yourData$IndividualID, yourData$FamilyID, sample, size=1) # untested
HTH
Peter Alspach
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Whitney Melroy
Sent: Friday, 28 March
Hi,
May be this helps:
set.seed(42)
indx - with(df,tapply(seq_along(IndividualID), FamilyID,FUN=sample,1))
df[indx,]
# FamilyID IndividualID DadID MomID Sex
#4 1 104 0 0 2
#8 2 204 202 203 2
#or
library(plyr)
ddply(df,.(FamilyID),function(x)
Dear All
I have a model of the form Approaches ~ Setup * Minute, where Setup has two
levels and Minute has three levels.
I want to test the difference between the levels of setup (Control vs Test)
within in levels minute e.g. the difference between the two setups for Minute
1, Minute 2 and
Hi. I ran some code and I am trying to access the table with the data in
it. I want in particular to delete the 31st row for example
Monthly means and totals: NA NA NA NA NA NA NA NA NA
NA NA NA NA
or remove the names of the columns (or change them). DayT TM Tm
SLPHPP VVV
Hello,
Your example leads to error. Anyway:
library(XML)
oneurl=http://www.tutiempo.net/en/Climate/FUKUSHIMA/11-2012/475950.htm;
temp.tables=readHTMLTable(oneurl, header=TRUE)
temp.tables - temp.tables[[3]]
temp.tables - temp.tables[1:30,]
HTH,
Pascal
On Fri, Mar 28, 2014 at 2:42 PM, Bill
Dear list members,
I am wondering whether there is any more efficient way to calculate
centered difference on matrix in R? Please see herewith an example:
lon - matrix(rep(seq(0,2,length.out=1e3), 1e3), 1e3, 1e3)
lat - matrix(rep(seq(0,2,length.out=1e3), each=1e3), 1e3, 1e3)
x -
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