Hi everyone,
I am using the package "raster" to interpolate a large number of
rasters (~1million) of different resolutions to a unique 1degree
resolution grid and wonder if you know if it is possible to do this in
parallel computer?.
My code (example below) works like a charm but it will
Hello,
I am using the "panel.3dpolygon" function from the LatticeExtra package. I
am graphing a matrix.
I can't find a way in which each polygon is presented in a different color.
The description of the function says that the "col" argument permits vectors
(the components of which are colors) to
Hi
Something like this.
## 4 valid zips + 4 invalid zips
zipcode <- c("22942-0173", "32601", "N9YZE6", "S7V 1J9", "0022942-0173",
"32-601", "NN9YZE6", "S7V 1J9")
tmp <- gsub("[[:space:]]", "_", zipcode)
tmp <- gsub("[[:alpha:]]", "A", tmp)
tmp <- gsub("[[:digit:]]", "N", tmp)
tmp
## [1] "
Hi
Is is something like this that you want?
mydata <- data.frame(a=1:10, b=11:20, c=21:30, d=31:40, e=41:50)
myfun <- function(a,b,c)a*b+c
mydata$cool <- do.call(myfun, mydata[,c("a", "b", "c")])
mydata
## a b c d e cool
## 1 1 11 21 31 41 32
## 2 2 12 22 32 42 46
## 3 3 13
Hi all,
I'm pretty new to R and have a question. I have a postal_code field which
can have a variety of values such as:
For US postal codes: 22942-0173 or 32601
For Canada postal codes: N9YZE6 or S7V 1J9
What I want to do is represent these as patterns, such as:
US: N- or N
Canada: AN
Hello R Mailing List Members,
My name is Peter. I am a high school student who is doing a senior
thesis in statistical analysis. As you can see that I have chosen R. So for
the project I am working with a member of the US Navy and analyzing some
data. So for this data I am using a formula to
Hi,
You could try:
mat1 <- matrix(1:(272*12),ncol=12,byrow=TRUE)
mat2 <- t(mat1)
dim(mat2) <- c(272*12,1)
#or
mat3 <- matrix(as.vector(mat2),ncol=1)
identical(mat2,mat3)
#[1] TRUE
A.K.
Hello,
I am pretty new to R and would like to transform my 272x12 matrix into a
3264X1. I'm trying to ha
Hi,
Try:
rl <- rle(is.na(a))
max(rl$lengths[rl$values])
#[1] 3
A.K.
Hi,
I'd like to detect whether a vector contains a sequence of NA values of a
certain length. So, for example, if I have a vector a =
c(1,NA,NA,4,NA,NA,NA,5); how can I find what the longest sequence of NAs is (in
this ca
On 08/01/14 15:09, Ista Zahn wrote:
On Tue, Jan 7, 2014 at 8:30 PM, Pete Brecknock wrote:
Krishia wrote
Hello,
I am pretty new to R and would like to transform my 272x12 matrix into a
3264X1. I'm trying to have the setup change from:
1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12
13,14,15,16,17,1
Hi George
If you did not want a line through the point try this
xyplot(Activity ~ Day | Subject, data = Data,pch =
ifelse(is.na(Data$EventA), 1,16), col = ifelse(is.na(Data$EventA), 4,2))
You could make a column of it otherwise change EventA to numeric - easier to
work with
Regards
Duncan
Du
On Tue, Jan 7, 2014 at 8:30 PM, Pete Brecknock wrote:
> Krishia wrote
>> Hello,
>> I am pretty new to R and would like to transform my 272x12 matrix into a
>> 3264X1. I'm trying to have the setup change from:
>>
>> 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12
>> 13,14,15,16,17,18,19,20,21,22, 23, 24
Krishia wrote
> Hello,
> I am pretty new to R and would like to transform my 272x12 matrix into a
> 3264X1. I'm trying to have the setup change from:
>
> 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12
> 13,14,15,16,17,18,19,20,21,22, 23, 24
> etc.
>
> to
>
> 1
> 2
> 3
> 4
> 5
> 6
> 7
> 8
> 9
> 10
>
Patrick,
You should cc r-help on all correspondence so that others can follow the
thread.
The color range will be matched to the "x" argument you provide to the
color.scale function (in package plotrix). So you don't need to manually
provide the min and max yourself. If for some reason you did
Look at the mvrnorm function in the MASS package.
The within or transform functions may also be of use to you in
creating the data as a combination of the factors that you create
using mvrnorm.
On Tue, Jan 7, 2014 at 2:15 PM, sevda datlı wrote:
> Hello,
>
> I am a master student in Educational
Hello,
I am a master student in Educational Measurement and Evaluation, my thesis
subject is comparison of estimation methods used for confirmatory factor
analysis. For my thesis study, I need to generate datas that provides some
features. I will generate these datas by changing correlation betw
HI,
May be this helps:
library(reshape2)
df1 <- dcast(DF,A2~A1,value.var="A3")
z <- function(a,b,c){a+2*b+c}
within(df1, newCol <- z(a,b,c))
# A2 a b c newCol
#1 m 1 4 7 16
#2 n 2 5 8 20
#3 p 3 6 9 24
On Tuesday, January 7, 2014 4:15 PM, Ron Michael
wrote:
Hi,
I have to
On 14-01-07 3:21 PM, Ron Michael wrote:
Hi,
I have to perform some formula driven calculation in a data.frame (as defined
below). Let say I have following DF:
DF <- data.frame(A1 = c('a', 'a', 'a', 'b', 'b', 'b', 'c', 'c', 'c'), A2 =
c('m', 'n', 'p', 'm', 'n', 'p', 'm', 'n', 'p'), A3 = c(1,2
Hi,
I have to perform some formula driven calculation in a data.frame (as defined
below). Let say I have following DF:
> DF <- data.frame(A1 = c('a', 'a', 'a', 'b', 'b', 'b', 'c', 'c', 'c'), A2 =
> c('m', 'n', 'p', 'm', 'n', 'p', 'm', 'n', 'p'), A3 = c(1,2,3,4,5,6,7,8,9))
> DF
A1 A2 A3
1 a
On 14-01-07 3:35 PM, Eva Prieto Castro wrote:
Hi Rich, Hi everybody
I have solved the problem!. I installed MikTeX 2.9 again, but this time I chose
basic installation (the other one I had chosen complete installation). I
noticed a diference if we compare with the previous situation: now R and
Hi Rich, Hi everybody
I have solved the problem!. I installed MikTeX 2.9 again, but this time I chose
basic installation (the other one I had chosen complete installation). I
noticed a diference if we compare with the previous situation: now R and MikTeX
are installed in the same folder (Progra
Patrick,
You were pretty close.
To fix the code you have, just change "matrix" to "mymatrix" in two places,
and either specify the argument data= or place the heat.colors bit first in
the matrix function.
Or ... you could use the array() function instead, to shorten up the code a
little.
myc
Hi again:
Rich, I see your messages but they are sent to another mail address so I have
to write from here without using your answer.
I think the problem is that inconsolata.sty is not found, and zi4.sty is not
found.
I have read several documents in www about this question but I can not solve
On 07 Jan 2014, at 19:19 , Duncan Murdoch wrote:
> I wouldn't call it a bug, but it's a documented limitation, if you know how
> to read it. As documented, the expression is evaluated with the caller's
> environment as the parent environment. But here the caller is some code in
> lapply, no
Thanks Patrick,
That did the trick.
Oddly enough (or maybe not that oddly) I was looking through "The R Inferno" to
try and find the answer. Since I often forget that there are other parameters
available to '[' I never made it to 8.1.44 on dropping dimensions.
Hopefully this lesson will stick
I think you will be okay if you change
one line to:
defMat<-sapply(defData[,-1, drop=FALSE], function(x) breakUpFun(freq, x))
In your example that doesn't work you are
ending up with a vector rather than a one
column data frame.
Pat
On 07/01/2014 17:44, Keith S Weintraub wrote:
Folks,
# I h
On 07/01/2014 10:35 AM, Pavel N. Krivitsky wrote:
Hi,
I have a list of sublists, and I want to add and/or remove elements in
each sublist in accordance with a code snippet. I had thought that an
elegant way to do that is using a combination of lapply() and within().
However, the code in the with
The image suggests that you are missing a font. Someone else might be
able to help with that.
On Tue, Jan 7, 2014 at 12:50 PM, Eva Prieto Castro wrote:
> Thanks Rich. I used your indications, but it was correctly in PATH variable.
> My path is C:\Program Files (x86)\MiKTeX 2.9\miktex\bin.
>
> I
Thanks Rich. I used your indications, but it was correctly in PATH variable. My
path is C:\Program Files (x86)\MiKTeX 2.9\miktex\bin.
I send you a print. Do you have any other idea?.
Thanks again.
Eva
El Lunes 6 de enero de 2014 23:36, Eva Prieto Castro
escribió:
Hi everybody,
I ha
Folks,
# I have the following function:
breakByFreq<-function(freq, defData) {
breakUpFun<-function(freq, defs) {
if(freq != 1) {
defs<-diff(c(0, defs))
defs<-cumsum(rep(defs/freq, each = freq))
}
defs
}
defMat<-sapply(defData[,-1], function(x) breakUpFun(freq, x))
Dear all,
I have a question regarding the package rugarch. I would like to fit a garch
model with exogenous variables using rugarch. My code is as follows, where
data.reg is a time series object where the first column corresponds to the
response variable and the remainder columns are the exoge
Hi,
I have a list of sublists, and I want to add and/or remove elements in
each sublist in accordance with a code snippet. I had thought that an
elegant way to do that is using a combination of lapply() and within().
However, the code in the within() call doesn't seem to be able to see
objects out
Chen, George roswellpark.org> writes:
> My apologies for asking this question that may have
been asked before. I am trying to plot activity
> dependent on time conditioned by the subject.
Code for sample data below.
> So I have something like this
> xyplot(Activity~Time|Subject).
> This works fi
Hi
You have got an extensive explanation by Frede Aakmann Tøgersen. I consider it
pretty strightforward and I do not have anything to add.
Do not expect that somebody can change your result, if your data does not
support it.
Regards
Petr
> -Original Message-
> From: r-help-boun...@r-
thank you sir, it works
On Tue, Jan 7, 2014 at 7:27 AM, jim holtman wrote:
> You may want to understand how your data is coming out of Excel. You
> are getting just a single character string. Here is code to convert
> the string to numerics and plot:
>
> x <- " 7.0121, -0.673354, 0.749622, -0
Hi
I usually read data from Excel this way.
open Excel file
mark values you want to copy
press Ctrl-C
type in R command window
datalist <- read.delim("clipboard")
in that case datalist shall be data.frame with probably one column so
plot(datalist[,1],type="l",ylim=c(-1,+1))
shall do what you
You may want to understand how your data is coming out of Excel. You
are getting just a single character string. Here is code to convert
the string to numerics and plot:
x <- " 7.0121, -0.673354, 0.749622, -0.549641, 0.435662, -0.328995,
0.0869976, -0.0851428, -0.191019, 0.188799, -0.373707, 0.4
I am reading only one Cell of Excel
On Tue, Jan 7, 2014 at 7:18 AM, Dániel Kehl wrote:
> Hi,
>
> are you reading only one cell of your Excel file?
> It seems your datalist is not numeric but text.
> Try saving your Excel file as csv or text, reading in data might be much
> more straightforward!
I apologies for asking this question;
I run the codes but why all the prediction for pH in 21-25 month are so
close?
station month pH
1 21 8.275635958
2 21 8.275635962
3 21 8.275635963
4 21 8.275635963
5 21 8.275635963
Hi,
are you reading only one cell of your Excel file?
It seems your datalist is not numeric but text.
Try saving your Excel file as csv or text, reading in data might be much more
straightforward!
best
daniel
Feladó: r-help-boun...@r-project.org [r-help-b
Thank you sir:
If I am tying
str(datalist)
I have this output
chr " 7.0121, -0.673354, 0.749622, -0.549641, 0.435662, -0.328995,
0.0869976, -0.0851428, -0.191019, 0.188799, -0.373707, 0.434814, "|
__truncated__
and if I try:
dput(datalist)
I have:
" 7.0121, -0.673354, 0.749622, -0.549641, 0.
>From your output, it looks like 'datalist' is a character string and
not a numeric vector as you expect. Try doing:
str(datalist)
to see the structure and to use
dput(datalist)
to post the data.
Jim Holtman
Data Munger Guru
What is the problem that you are trying to solve?
Tell me what you
thank you for your answer, I am reading my Data from an excel File. This is
my R code:
srow<-2421
wb <- loadWorkbook("C:\\users\\Babak\\Desktop\\spalte205.xls")
dat <-readWorksheet(wb, sheet=getSheets(wb)[1], startRow=srow, endRow=srow,
startCol=spalte, endCol=spalte,header=FALSE)
datalist<-dat[,1
Hi
I did not get any error with your data
plot(test)
dput(test)
c(7.0121, -0.673354, 0.749622, -0.549641, 0.435662, -0.328995,
0.0869976, -0.0851428, -0.191019, 0.188799, -0.373707, 0.434814,
-0.51979, 0.61944, -0.554766, 0.662571, -0.557779, 0.543724,
-0.452397, 0.293651, -0.248438, 0.053278
Hi all
I have such a Data and I want to plot them but I get this warning
Warning message:
In xy.coords(x, y, xlabel, ylabel, log) : NAs introduced by coercion
" 7.0121, -0.673354, 0.749622, -0.549641, 0.435662, -0.328995, 0.0869976,
-0.0851428, -0.191019, 0.188799, -0.373707, 0.434814, -0.51979
My apologies for asking this question that may have been asked before. I am
trying to plot activity dependent on time conditioned by the subject. Code for
sample data below.
So I have something like this
xyplot(Activity~Time|Subject).
This works fine, but now I want to show where on these activ
Dear all,
The latest issue of The R Journal is now available at
http://journal.r-project.org/archive/2013-2/
Many thanks to all contributors.
Hadley
--
Editor-in-chief, The R Journal
___
r-annou...@r-project.org mailing list
https://stat.ethz.ch/mail
Dear Javad Bayat
I think that people on this list has been most helpful to your with your
questions about how to use neural networks in R.
Now you have come to the point where you need a more statistical understanding
of your data before you can decide whether neural network methods is really
I use
windlog=log10(windspec)
plot(windlog)
it seems good but stupid
is there any better method ?
2014/1/7 Jie Tang
> Futher more
>
> When I write as below :,the result is still not satisfied .
> asp=spectrum(ww_spec)
> windspec=data.frame(asp)
> plot(windspec,log="dB")
>
>
> 2014/1/7 Ji
Hi
and what is wrong with e.g.
fit <- neuralnet(pH~station+month, data=yourdata)
As I said I am not an expert in neural nets but here is some explanation how
it works
http://gekkoquant.com/2012/05/26/neural-networks-with-r-simple-example/
based on that after fitting you could do
compute(fit
Futher more
When I write as below :,the result is still not satisfied .
asp=spectrum(ww_spec)
windspec=data.frame(asp)
plot(windspec,log="dB")
2014/1/7 Jie Tang
> hi
> I have a wind dataset and I want to analze its spectrum with
> the x-axis and y-axis in log .But I found that
> when I set
hi
I have a wind dataset and I want to analze its spectrum with
the x-axis and y-axis in log .But I found that
when I set log="dB" in spectrum as shown below,what I can get
is a figure by log y-axis.
How could I get a log-axis figure both in x-axis and y-axis ?
thank you .
yylab<-c("Wind Spectrum(
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