hello, I am very new to R.
My current data set is a mix of values and categories. It is a geoscience
data set, with values per rock sample. Case in point, each sample belongs to
a lithology class, and each sample has several physical property
measurements (density, porosity...).
I want to be able
Hello,
I have one question regarding dotplots and one question about the strip
function in lattice. In the following function I wish to use two
different line types in a dotplot. Specifying lty =c(1,2) does not
work. No matter what line type is specified, solid lines are produced.
The other
I ve installed wordnet 2.1 and R 2.11.0 on windows 7.
Whenver i try to load wordnet in R, i get an error
initDict()
FALSE
cannot find wordnet 'dict' directory, Please set the WNHOME variable to its
parent.
I have tried setting WNHOME to C:\Program Files
(x86)\WordNet\2.1\dict,C:\\Program Files
Dear Colleagues,
I have a data set that looks as below. I'd like to count the number of dates in
a series of arbitrary ranges (breaks) i.e. not pre-defined breaks such as
months, quarters or years. table(format()) produces ideally formatted output,
but table() does not appear to accept
Hi,
First a good tip when you ask on the R list is to provide data in a way
that we can readily use it. I think the best way to do it is to copy the
output of dput(tc) into the email you write.
There might be better ways to do what you want, but here is what I would do:
#first subset what
Hi,
The solution with lapply() should work; it seems that the problem was
with the file name.
mylist - lapply(a, read.table, header = TRUE, sep = '\n')
What is a? Is it an object containing a path+file name? Maybe not.
Second, I'm not sure sep should be \n. I think , would be better.
Just a
You could try useing '127.0.0.1' instead of 'localhost' and see if that
works.
--
View this message in context:
http://r.789695.n4.nabble.com/Accessing-MySQL-Database-in-R-tp3221264p3221291.html
Sent from the R help mailing list archive at Nabble.com.
Hi:
On Mon, Jan 17, 2011 at 10:10 PM, Byerly, Mike M (DFG)
mike.bye...@alaska.gov wrote:
Hello,
I have one question regarding dotplots and one question about the strip
function in lattice. In the following function I wish to use two
different line types in a dotplot. Specifying lty
On Mon, Jan 17, 2011 at 12:02:21PM -0800, Aaron Polhamus wrote:
Dear list,
I'm writing a function to re-grid a data set from finer to coarser
resolutions in R as follows (I use this function with sapply/apply):
gridResize - function(startVec = stop(What's your input vector),
to =
Hi:
Perhaps you were looking for something like this:
table(cut(mydata[[1]], breaks=seq(from = as.Date(2008-06-26),
to =
as.Date(2009-06-26), by = 'month')))
2008-06-26 2008-07-26 2008-08-26 2008-09-26 2008-10-26 2008-11-26 2008-12-26
Yes, of course. I think all the posts up to Bert's addressed the coding
question as asked - how to calculate a particular version of the median (not
the mean) rather than any underlying, unstated, statistical or scientific
question.
IIRC there hasn't been any indication that non-positive
Hi
I have a long data set on which I want to do Bland-Altman style plots for each
rhythm type
Using ggplot2, when I use geom_hline with facet_grid I get an extra set of
empty panels.
I can't get it to do it with the Diamonds data supplied with the package so
here is a (much abbreviated)
Hi all,
Does anyone knows how to handle ordered preferences applying the R
package mlogit (multinomial logit model)? My data set provides for each
customer preferences (given as percentages) for 6 different brands. I
would like to use for model calibration not just that brand with
maximum stated
On Mon, Jan 17, 2011 at 8:24 PM, S Ellison s.elli...@lgc.co.uk wrote:
I was distracted enough by the possibility of hijacking hist() for this
to give it a go.
The following code implements a basic hanging rootogram based on a
normal density with hist() breaks used as bins and bin midpoints
Hi
It is rather unclear how do you want sample your data. It seems to me that
you can sample row numbers and choose sampled rows from data frame.
I probably direct output to a list (mylist) and use
result - lapply(mylist, GFD)
Regards
Petr
r-help-boun...@r-project.org napsal dne
Hi R,
The dates when exported from Excel to R by Put R Var, gives one less
day.
For example,
1.Let 1/1/2011 be the cell A1 in date format.
2.Right click and select Put R Var and give a name to it say,
StartDate (R in foreground process)
3.In the R console see the value
Hi:
Your intention isn't crystal clear to me, but I'll give it a shot...
On Mon, Jan 17, 2011 at 10:48 PM, Ben Harrison
b.harris...@pgrad.unimelb.edu.au wrote:
hello, I am very new to R.
My current data set is a mix of values and categories. It is a geoscience
data set, with values per rock
Thank you very much got something running now based on this.
Joh
jim holtman wrote:
building on the previous responses, does this give you what you want:
x
A B
1 1 1
2 2 NA
3 NA NA
4 NA 4
# determine where the NAs are
row.na - apply(x, 1, is.na)
# now convert to list of
On 2011-01-17 22:48, Ben Harrison wrote:
hello, I am very new to R.
My current data set is a mix of values and categories. It is a geoscience
data set, with values per rock sample. Case in point, each sample belongs to
a lithology class, and each sample has several physical property
measurements
Dear R community,and especially Giovanni Millo,
For my master's thesis i need to simulate a panel data with the fixed effects
correlated with the predicor, so i run the
the following code:
set.seed(1970)
###Panel data simulation with alphai correlated with
Hi,
I have a variable (current speed direction) which is circular (0=360 degrees),
and I'd like my GLM to include the variable as a circular variable. Can I do
this? And what is the code?
I'm actually doing a GLM-GEE using the 'geepack' package, so want to use it in
that, but also
Dear R,
Is there an efficient way to make a list that each element is from the
corresponding column of a matrix. For example, if I have a matrix a
a - matrix(1:10, 5, 2)
a
[,1] [,2]
[1,]16
[2,]27
[3,]38
[4,]49
[5,]5 10
I would like to have a list b
Try this:
c(unname(as.data.frame(a)))
On Tue, Jan 18, 2011 at 10:53 AM, Feng Li m...@feng.li wrote:
Dear R,
Is there an efficient way to make a list that each element is from the
corresponding column of a matrix. For example, if I have a matrix a
a - matrix(1:10, 5, 2)
a
[,1]
hi feng,
a possible solution is
b1-apply(a,2,list)
and possibly
lapply(b1,unlist)
if you want exactly the output equal to list(a[, 1], a[, 2])
best,
vito
Il 18/01/2011 13.53, Feng Li ha scritto:
Dear R,
Is there an efficient way to make a list that each element is from the
corresponding
I think it is called time zones. Is IST India standard time, if
so that accounts for the half hour. What is your time zone and what
is the time zone of the data coming from Excel? Exactly how are you
importing it?
On Tue, Jan 18, 2011 at 5:35 AM, Shubha Vishwanath Karanth
Hi
Thanks for you suggestion. I used histograme in lattice packages. If
it is possible lattice can do it, it will be better.
On Mon, Jan 17, 2011 at 3:28 PM, Peter Ehlers ehl...@ucalgary.ca wrote:
On 2011-01-17 02:26, Fabrice Tourre wrote:
Hi all,
How to plot as the coordinate as in my
if you only want to count the number of days in an arbitrary time
period, you may try:
install.packages(hydroTSM, dependencies=TRUE)
library(hydroTSM)
dip(from=2007-05-01, to=2009-09-10, out.type=nmbr)
I hope this helps.
Kinds,
Mauricio
--
===
Linux user
Hi,
I have this character vector:
A-c(Tell me how many different letter this vector has?)
Is there a way with R that it can let me know how many different letters I
have on this vector?
If I use nchar(A) que gives me the number 50. With this function he is
counting all the letters present and
I understand that plot.dendrogram() plots a dendrogram vertically, so the root
is to the left and leaves to the right. I also understand that the horiz=TRUE
option plots it horizontally so the root it at the top and the leaves at the
bottom.
My question is can these be plotted so the root is
From: simonjk...@yahoo.ca
Date: Tue, 18 Jan 2011 02:16:37 -0500
To: r-help@r-project.org
Subject: [R] Counting dates in arbitrary ranges
Dear Colleagues,
I have a data set that looks as below. I'd like to count the number of dates
in a series
Try this:
table(strsplit(A, ?)[[1]])
On Tue, Jan 18, 2011 at 10:51 AM, ADias diasan...@gmail.com wrote:
Hi,
I have this character vector:
A-c(Tell me how many different letter this vector has?)
Is there a way with R that it can let me know how many different letters I
have on this
Hi
r-help-boun...@r-project.org napsal dne 18.01.2011 13:51:05:
Hi,
I have this character vector:
A-c(Tell me how many different letter this vector has?)
Is there a way with R that it can let me know how many different letters
I
have on this vector?
A-c(Tell me how many different
Many thanks Greg!
I try to use tcltk2 and tclTaskSchedule function but in argument expr
is possible to insert a R script?
Have you an example?
Alessandro
Il 17 gennaio 2011 22.02.51 UTC+1, Greg Snow greg.s...@imail.org ha scritto:
You could write a batch file and then have your OS schedule to
Hi I was wondering whether anyone can help me with this problemit's been
driving me nuts, I've been trying to figure it out for months and months
without success!! Basically I have a group of participants who attended 2
experimental sessions a few months apart. I took measures of the way they
Thanks Dennis
Yes the plot you sent was what I was expecting. Unfortunately I do not get the
same thing using the same commands whn I try it.
I notice that I am using a slightly earlier version of R than you. Given that
it worked on the Diamonds data I doubt that this the problem but I will
It appears that you have a 2x2 table coming from paired binary data..
If this is the case the McNemar test is appropriate.
See
?mcnemar.test
or even better the package exact2x2, function mcnemar.exact() for an
exact approach,
vito
Il 18/01/2011 14.40, debz ha scritto:
Hi I was
Hi
r-help-boun...@r-project.org napsal dne 17.01.2011 15:59:37:
days=Sys.Date()-1:70
price=abs(rnorm(70))
regular=rep(c(0,0,0,0,1,0,1,0,0,1),c(7,7,7,7,7,7,7,7,7,7))
y=data.frame(cbind(days,price,regular))
y is like
days price regular
1 14990 0.16149463 0
2 14989
Hi Vito
Thank you so much for your help I really appreciate it! I have a 4x4 table
for one task and a 4x3 Table for the second task.
Thanks
Debbie
On Tue, Jan 18, 2011 at 2:18 PM, Vito Muggeo (UniPa)
vito.mug...@unipa.itwrote:
It appears that you have a 2x2 table coming from paired binary
Dear R Help,
I'd like to be able to use gamlss to generate distributions that are
both truncated and have censoring. It doesn't look as though it is
possible to do this at the moment:
gen.trun(par=c(0),family=NO,name=tr,type=left)
A truncated family of distributions from NO has been generated
Please do not cross post: http://stackoverflow.com/q/4720076/271616
At the minimum, it would be polite to respond here with the answer you
accepted on Stack Overflow.
--
Joshua Ulrich | FOSS Trading: www.fosstrading.com
On Mon, Jan 17, 2011 at 8:59 AM, Daniel Wu daniel_w...@163.com wrote:
Here is how you should be creating your dataframe so that each element
has the proper mode:
days=Sys.Date()-1:70
price=abs(rnorm(70))
regular=rep(c(0,0,0,0,1,0,1,0,0,1),c(7,7,7,7,7,7,7,7,7,7))
y=data.frame(cbind(days,price,regular))
str(y)
'data.frame': 70 obs. of 3 variables:
$ days :
Dear all,
how can I perform a string operation like strsplit(x, ) on a column
of a dataframe, and put the first or the second item of the split into a
new dataframe column?
(so that on each row it is consistent)
Thanks
Boris
__
Hi,
I guess it's not the nicest way to do it, but it should work for you:
#create some sample data
df - data.frame(a=c(A B, C D, A C, A D, B D),
stringsAsFactors=FALSE)
#split the column by space
df_split - strsplit(df$a, split= )
#place the first element into column a1 and the second into
Toby Marthews toby.marthews at ouce.ox.ac.uk writes:
Dear R HELP,
ABOUT glmmPQL and the anova command. Here is an example of a
repeated-measures ANOVA focussing on the way
starling masses vary according to (i) roost situation and
(ii) time (two time points only).
[snip]
Hello!
I hope, i'm doing right, writing here.
I have a question about the R statistical formulas: were can I look,
what are they? E. g. I'd like to know which exactly mathematical formula
is used in Wilcoxon test, and others. Is there somewhere on the Internet
information about that?
Thanks a
I was running a sampling syntax based on a data frame (ago) of 160 rows and
25 columns. Below are the column names:
names(ago)
[1] SubID AGR1 AGR2 AGR3 AGR4 AGR5 AGR6 AGR7
AGR8
[10] AGR9 AGR10 WAGR1 WAGR2 WAGR3 WAGR4 WAGR5 WAGR6
WAGR7
[19] WAGR8 WAGR9 WAGR10 ocbi
how can I perform a string operation like strsplit(x, ) on a column of a
dataframe, and put the first or the second item of the split into a new
dataframe column?
(so that on each row it is consistent)
Have a look at str_split_fixed in the stringr package.
Hadley
--
Assistant Professor /
Hi,
use the help files (e.g. ?wilcox.test) to see references. There you can see
articles and book parts where you can find the original formula (and
background theory).
Also, analyzing the source code of a function may help.
Regards,
Gergely
On Tue, Jan 18, 2011 at 5:23 PM, ilya
Hi Ilya,
If you're looking for general information about statistical tests, etc., you'll
probably need to buy yourself a textbook. There are online pages
(http://homes.msi.ucsb.edu/~byrnes/rtutorial.html is a good one), but a good
textbook is probably better. It's a bit old, but I still
Is the variable to be used as a predictor? If so mgcv::gam has a couple of
circular smoother built in...
s(x,bs=cc) or s(x,bs=cp)
(if the range of x is less than the full [0, 360] then you can supply a
`knots' argument to gam to force the x range to go from 0 to 360, otherwise
s(min(x),
Hello all,
Hoping that there is a fairly simple solution to my query...
I'm trying to overlay a line plot of some data onto a barplot of different
data. The y-axes are different for each set of data but the x-axes are the same
(1:12, corresponding with 12 months of observations). The
In case anyone is interested in this issue, I found a solution. The psych
package has a function named score.items which will calculate raw scores.
You need to know the survey item loadings and the raw responses to each
item, but it will do it. See Grice's 2001 paper on factor scores for more
On Mon, Jan 17, 2011 at 10:37:42PM +, Monica Pisica wrote:
Hi,
I've got 2 very good solutions, thank you very much. One, from Henrique
Dallazuanna using the library reshape and one line of code - although it will
take me quite some time to understand it. Here it is what he sent:
From: toby.marth...@ouce.ox.ac.uk
To: flya...@gmail.com; r-help@r-project.org
Date: Tue, 18 Jan 2011 16:42:51 +
Subject: Re: [R] Statistical formulas
Hi Ilya,
If you're looking for general information about statistical tests, etc.
Hello,
I have a set of data for which I am making linear model. I would like to
restrict this model for a subset of the data.
I have 100 independent variables labeled x1-x100. I would like to only like
an lm() that is only accounts for x1-x20.
I have tried restricting my plot:
On Wed, 19 Jan 2011, Kohske Takahashi wrote:
Thanks for the reply.
I got that there is no general way to generate PDF with ttf or otf
fonts. Thanks, it's enough.
Also, cario_pdf is useful (I'm using OSX).
But I found it cannot correctly(?) work with CJK.
The Mac OS X (why do Mac users not
Hello fellow R users,
I am trying to read a 6.9 million row text file with 26 columns separated by
spaces into R using ff. When I specify a small number for first.rows,
next.rows and nrows it is read with no issue. However, when I try to specify
larger next.rows values and no nrows parameter to
On 11-01-18 12:49 PM, Karen Aanensen wrote:
Hello,
I have a set of data for which I am making linear model. I would like to
restrict this model for a subset of the data.
I have 100 independent variables labeled x1-x100. I would like to only like
an lm() that is only accounts for x1-x20.
I
Thanks for the reply.
I got that there is no general way to generate PDF with ttf or otf
fonts. Thanks, it's enough.
Also, cario_pdf is useful (I'm using OSX).
But I found it cannot correctly(?) work with CJK.
I will play with them some more.
Thanks again.
--
Kohske Takahashi
Hi I need some help trying to restructure a data frame:
I have:
foodtransaction quantity
pizza BUY 5
pizza SELL 3
apple BUY 2
orange SELL 1
pizza BUY 2
And the data frame I want to
Dear R-help and Prof. Harrell:
My question concerns the baseline state for continuous variable in lrm()
within the RMS package.
I have a model which can be reduced to:
lrm(FT ~ rcs(V1, c(0, 1,5))
The model makes perfect sense if the baseline state is where V1=5 but
the model makes no sense
Try this:
xtabs(quantity ~ food + transaction, DF)
On Tue, Jan 18, 2011 at 5:50 PM, chris99 chea...@hotmail.com wrote:
Hi I need some help trying to restructure a data frame:
I have:
foodtransaction quantity
pizza BUY 5
pizza SELL 3
On 11-01-18 16:42, Toby Marthews wrote:
Hi Ilya,
If you're looking for general information about statistical tests, etc., you'll
probably need to buy yourself a textbook. There are online pages
(http://homes.msi.ucsb.edu/~byrnes/rtutorial.html is a good one), but a good
textbook is probably
It's not immediately evident to me just looking at this, but I do have
one suggestion. Run this again and see if you get the same error and
if you do, use this command to help you find where -
traceback()
traceback is extremely helpful when you are debugging these things, it
will go through
I should rephrase my question.
When using the notation rcs(V1, N) within the lrm function in the rms
package, the associated model reports out N-1 values for the associated
variable. Fine.
The rcs-constructed temporary variables are denoted V1, V1', V1'' when
one looks at the model output,
On Jan 18, 2011, at 2:51 PM, Rob James wrote:
Dear R-help and Prof. Harrell:
My question concerns the baseline state for continuous variable in
lrm() within the RMS package.
I have a model which can be reduced to:
lrm(FT ~ rcs(V1, c(0, 1,5))
The model makes perfect sense if the baseline
On 2011-01-18 04:51, ADias wrote:
Hi,
I have this character vector:
A-c(Tell me how many different letter this vector has?)
Is there a way with R that it can let me know how many different letters I
have on this vector?
If I use nchar(A) que gives me the number 50. With this function he is
Hi
Trying to work out that:
Each person can take from 1 to 9 drugs. And data frame looks like
id drug
1 d
2 g
2 d
2 r
3 e
3 a
...
Q: How many drugs one person takes on average?
Thank you
__
BTW. There are help pages coming with R installation. Do you have them
corrupted? There is strsplit in see also section of nchar help page.
yes I do. But I have many dificulties in finding what I need. And on top of
that R has a very specific way of working that is quite diferent from
I have done like this to get the result I need more directly
A-c(Tell me how many different letter this vector has?)
prop.table(table(strsplit(A,)))
?acdefhilmnors
t
0.16 0.02 0.04 0.02 0.02 0.14 0.04 0.06 0.04 0.06 0.04 0.04 0.04
Hi all,
Gabor came up with a very nice of code for my request:
Lines - DateTimeOpenHighLow Close
1/2/200517:05 1.3546 1.3553 1.3546 1.35495
1/2/200517:10 1.3553 1.3556 1.3549 1.35525
1/2/200517:15 1.3556 1.35565 1.35515 1.3553
1/2/2005
I tried it several times. it still gives me the same error.
I also used traceback()
here is what I got:
traceback()
3: `[.data.frame`(df, -s1$SubID, )
2: df[-s1$SubID, ]
1: crossed1(agr)
But I don't think there is anything wrong with it..
--
View this message in context:
Greetings
I have a bunch of NAs in a column of categorical variables designating
the size classes (e.g., smallest to largest: 1,2,3,4) of cave crickets.
I'd like to substitute U (for unknown) for the NAs. Can anyone give me
an idea how to do this? Thanks in advance.
Cheers
Kurt
Well, (have you read An Intro to R, which I think might have enabled
you to figure this out for yourself?)
Convert the factor to character, use is.na() to substitute, convert
back to factor. e.g.
z - factor (c(1,2,3,NA))
z- as.character(z)
z[is.na(z)] - U
factor(z)
[1] 1 2 3 U
Levels: 1
Homework exercise ??
-- Bert
On Tue, Jan 18, 2011 at 2:03 PM, Denis Kazakiewicz
d.kazakiew...@gmail.com wrote:
Hi
Trying to work out that:
Each person can take from 1 to 9 drugs. And data frame looks like
id drug
1 d
2 g
2 d
2 r
3 e
3 a
On 2011-01-18 13:06, wangwallace wrote:
I tried it several times. it still gives me the same error.
I also used traceback()
here is what I got:
traceback()
3: `[.data.frame`(df, -s1$SubID, )
2: df[-s1$SubID, ]
1: crossed1(agr)
But I don't think there is anything wrong with it..
I
On Jan 18, 2011, at 5:25 PM, kurt_h...@nps.gov wrote:
Greetings
I have a bunch of NAs in a column of categorical variables
designating
the size classes (e.g., smallest to largest: 1,2,3,4) of cave
crickets.
I'd like to substitute U (for unknown) for the NAs. Can anyone
give me
an
On Jan 18, 2011, at 5:03 PM, Denis Kazakiewicz wrote:
Hi
Trying to work out that:
Each person can take from 1 to 9 drugs. And data frame looks like
id drug
1 d
2 g
2 d
2 r
3 e
3 a
...
Q: How many drugs one person takes on average?
On 2011-01-18 08:14, Ivan Calandra wrote:
Hi,
I guess it's not the nicest way to do it, but it should work for you:
#create some sample data
df- data.frame(a=c(A B, C D, A C, A D, B D),
stringsAsFactors=FALSE)
#split the column by space
df_split- strsplit(df$a, split= )
#place the first
Amy
It would have been helpful if you had sent your R code of how you
constructed the sab object.
If you have a data.frame, the subset command you are having trouble with
should work fine. See below.
# Working Example
sab = data.frame(group=c('Group A', 'Group A', 'Group C', 'Group B', 'Group
On 2011-01-18 08:14, Ivan Calandra wrote:
Hi,
I guess it's not the nicest way to do it, but it should work for you:
#create some sample data
df- data.frame(a=c(A B, C D, A C, A D, B D),
stringsAsFactors=FALSE)
#split the column by space
df_split- strsplit(df$a, split= )
#place the first
Assuming every row is split into exactly two values by whatever string
you choose as split, one fancy exercise in R data structures is
dfsplit = function(df, split)
as.data.frame(
t(
structure(dim=c(2, nrow(df)),
unlist(
Or:
read.table(textConnection(as.matrix(df)), sep = )
On Tue, Jan 18, 2011 at 11:02 PM, Waclaw Kusnierczyk w...@idi.ntnu.nowrote:
Assuming every row is split into exactly two values by whatever string you
choose as split, one fancy exercise in R data structures is
dfsplit =
On 01/18/2011 07:05 PM, Henrique Dallazuanna wrote:
Or:
read.table(textConnection(as.matrix(df)), sep = )
no, that's too simple: you can't use regular expressions. (well, i
guess it's enough for the original problem.)
vQ
On Tue, Jan 18, 2011 at 11:02 PM, Waclaw Kusnierczyk
Hi all:
Here's a question about result of loglinear analysis.
There're 2 factors:area and nation.The raw data is in the attachment.
I fit the saturated model of loglinear with the command:
glm_sat-glm(fre~area*nation, family=poisson, data=data_Analysis)
After that,I extract the coefficients:
On Jan 18, 2011, at 8:45 PM, Lao Meng wrote:
Hi all:
Here's a question about result of loglinear analysis.
There're 2 factors:area and nation.The raw data is in the attachment.
I fit the saturated model of loglinear with the command:
glm_sat-glm(fre~area*nation, family=poisson,
Thanks for the reply Peter.
On 18 January 2011 22:52, Peter Ehlers ehl...@ucalgary.ca wrote:
Since you don't provide data, let's borrow from the
help(droplevels) page:
I had no joy with my R install finding droplevels exactly, but found this
instead:
??droplevels
gdata::drop.levels
I am coming to R from Fortran and I used to use fixed size arrays in
named common. common /name1/array(100)
The contents of array can be accessed/modified if and only if this
line occurs in the function. Very helpful if different functions need
different global data (can have name2, name3 etc.
Peter,
I found it. There is a missing value under column SubID. I filled it out
with an ID number, and the error message never occur again. Although I found
the problem, I have no idea why would this be associated with such an error
message. If anybody else have some input, I would really
Dennis, thank you for the response!
Sorry for lack of clarity, I'll explain a little more below...
plot(Depth[LithClass=='sand'], Conductivity[LithClass=='sand'])
(ad nauseum... how can I loop through them all?)
I have several lithology classes - sand, clay, limestone, etc... I wish to
On 18 January 2011 22:52, Peter Ehlers ehl...@ucalgary.ca wrote:
Since you don't provide data, let's borrow from the
help(droplevels) page:
As an aside, is it normal practice then to attach data files to questions on
this mailing list? I might do that in future if it's possible and
Nicos,
there are certainly better / faster methods to compare a time range which I
simply don't know of, but the condensed code below should do it. Your main
problem was that you can't compare time(x) to 7:55, since the
latter expression denotes a sequence from 7 to 55 in R. Consequently,
On Jan 18, 2011, at 6:09 PM, wangwallace wrote:
Peter,
I found it. There is a missing value under column SubID. I filled it
out
with an ID number, and the error message never occur again. Although
I found
the problem, I have no idea why would this be associated with such
an error
Hi,
I'm trying to make a ts object that has both NA values and a frequency other
than 1 (so I can use stl). I've tried all permutations I can think of, but
cannot get the desired (expected?) results.
The values live in x and the corresponding semi-regular time stamps are in
t:
library('zoo')
On Jan 18, 2011, at 7:32 PM, Ben Harrison wrote:
On 18 January 2011 22:52, Peter Ehlers ehl...@ucalgary.ca wrote:
Since you don't provide data, let's borrow from the
help(droplevels) page:
As an aside, is it normal practice then to attach data files to
questions on
this mailing list? I
Hi all,
I analysed my data with lme and after that I spent a lot of time for
mean separation of treatments (post hoc). But still I couldnât make
through it. This is my data set and R scripts I tried.
replication fertilizer variety plotheight
1 level1 var1150452
On Tue, 18 Jan 2011, analys...@hotmail.com wrote:
I am coming to R from Fortran and I used to use fixed size arrays in
named common. common /name1/array(100)
The contents of array can be accessed/modified if and only if this
line occurs in the function. Very helpful if different functions
Just learning so excuse me if I'm being too basic here. But I'm wondering how
should I know that as.ts would be needed for lag ? Is there a thought
process or way to inspect that I should have gone through to know that log
would work on y[,5] but lag would not work on [,5] ?
Is the general rule
Hi - I'm up against a complicated reshape problem. I have data of the form
X1,Y1,hr1,hr2,hr3
X1,Y2,hr1,hr2,hr3
X1,Y3,hr1,hr2,hr3
X2,Y1,hr1,hr2,hr3
X2,Y2,hr1,hr2,hr3
X2,Y3,hr1,hr2,hr3
where X and Y are factors and the hr(1,2,3) are values. I need it as
,X1, X2
Y1,hr1,hr1
Y1,hr2,hr2
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 call
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