.
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Muhammad Rahiz
(za) +27 071 719 0104
(skype) muhammad.rahiz
[[alternative HTML version deleted]]
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Dear all,
Does anyone know of a function in R which allows for correlation of
zero-truncated timeseries such as x and y as illustrated below?
x - rnorm(100)
x[which(x0)] - 0
y - abs(rnorm(100))
Thanks
--
Muhammad
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R-help@r-project.org mailing
thanks petr.
--
Muhammad
On Wed, 7 Mar 2012, Petr Savicky wrote:
On Wed, Mar 07, 2012 at 09:42:15AM +, Muhammad Rahiz wrote:
Dear all,
Does anyone know of a function in R which allows for correlation of
zero-truncated timeseries such as x and y as illustrated below?
x - rnorm(100)
x
Hi all,
I'm trying to get the min and max of a sequence of number using a loop
like the folllowing. Can anyone point me to why it doesn't work.
Thanks.
type- c(min,max)
n - 1:10
for (a in 1:2){
print(type[a](n)) }
--
Muhammad
__
Perhaps something like this?
par(oma=c(0,0,2,0))
par(mar=c(1,1,1,1))
par(mfcol=c(3,1))
plot(rnorm(10))
mtext(title)
plot(rnorm(10))
plot(rnorm(10))
--
Muhammad Rahiz
On Thu, 10 Nov 2011, Kevin Burton wrote:
I can get multiple plots on a page like:
op - par(mfcol = c(3, 1))
What I
at 2:51 PM, Muhammad Rahiz
muhammad.ra...@ouce.ox.ac.uk wrote:
Hi everyone,
I'm trying to determine the significance of a trendline. From my internet
search months ago, I came across the following post. I modified tim and
dat for simiplicity.
tim - 1:10
dat - c(0.17, 1.09 ,0.11, 0.82, 0.23, 0.38
Hi everyone,
I'm trying to determine the significance of a trendline. From my internet
search months ago, I came across the following post. I modified tim and
dat for simiplicity.
tim - 1:10
dat - c(0.17, 1.09 ,0.11, 0.82, 0.23, 0.38 ,2.47 ,0.41 ,0.75, 1.44)
fstat -
Dear all,
I'm having issues with the installation of the ncdf package. It returns a
non-zero exit status. Can anyone suggest what I should do next? FYI, I do
not have problems installing other packages.
Thanks.
Muhammad
* installing *source* package ‘ncdf’ ...
checking for nc-config...
Hi,
Can someone advise why the followind did not produce the array, given
the condition specified?
s - 1
a1 - array(dim=c(1,4))
a2 - array(dim=c(2,4))
arr - ifelse(s==0,a1,a2)
Thanks.
Muhammad
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Understood now. Thanks Duncan.
Muhammad
On Sat, 18 Jun 2011, Duncan Murdoch wrote:
On 11-06-18 10:45 AM, Muhammad Rahiz wrote:
Hi,
Can someone advise why the followind did not produce the array, given
the condition specified?
s- 1
a1- array(dim=c(1,4))
a2- array(dim=c(2,4))
arr- ifelse(s==0
Hi all,
I have the following script which fills the values which are less than
the mean of a given timeseries.
If you look closely, the colored regions are out of line.
Any suggestions how I can rectify this?
Thanks
Muhammad
# -
#rm(list=ls())
x - abs(rnorm(100))
tt - 1:100
m -
(y),col=red)
legend(topright,legend=c(x,y),col=c(black,red),lty=1,bty=n)
x1 - fitdistr(x,gamma)$estimate # get scale and rate
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Muhammad Rahiz
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PLEASE do read
Hi Jason,
This is one way;
c1 - seq(2,20,2)
c2 - seq(1,19,2)
c3 - cbind(c1,c2)
c3[,1][which(c3[,1]12)] - -1
c3[,2][which(c3[,2]10)] - -1
Muhammad
On Fri, 3 Jun 2011, Jason024 wrote:
I have a data frame like this:
col1 col2
r1 21
r2 43
r3 65
r4 87
r5
Try using a loop like the following
dat0 - read.table(time1.dat)
id - c(e1dq, e1arcp, e1dev, s1prcp, s1nrcp,s1ints,a1gpar, a1pias,
a1devt)
for (a in 1:length(id)) {
dat0[dat0$id[a]==-999.,as.character(id[a])] - NA
}
--
Muhammad Rahiz
Researcher DPhil Candidate (Climate Systems
Scarlet,
If the mfrow is being overridden, perhaps the rimage package might be
able to piece the individual plots...
--
Muhammad Rahiz
Researcher DPhil Candidate (Climate Systems Policy)
School of Geography the Environment
University of Oxford
On Thu, 17 Mar 2011, scarlet wrote:
Jim
Hi all,
I'm trying to get the value of y when x=203 by using the intersect of
three curves. The horizontal curve does not meet with the
other two. How can I rectify the code below?
Thanks
Muhammad
ts - 1:10
dd - 10:1
ts - seq(200,209,1)
dd - c(NA,NA,NA,NA,1.87,1.83,1.86,NA,1.95,1.96)
Josh's recommendation to use predict works. At the same time, I'll work on
your suggestions, David.
Thanks.
Muhammad Rahiz
Researcher DPhil Candidate (Climate Systems Policy)
School of Geography the Environment
University of Oxford
On Mon, 3 Jan 2011, David Winsemius wrote:
On Jan 3
That works - Thanks Ravi.
Muhammad Rahiz
On Tue, 19 Oct 2010, Ravi Varadhan wrote:
You can do this.
dsm = c(800,600,NA,525,NA,450,400,NA,NA,NA,0)
s3 = seq(0.05,1.05,0.1)
plot(s3,dsm,col=blue,las=1,xlab=fraction,ylab=distance (km))
fc - function(x,a,b){a*exp(-b*x)}
fm - nls(dsm~fc(s3,a,b
Hi all,
I'm plotting to get the intersection value of three curves. Defining
the x-axis as dsm, the following code works;
dsm = c(800,600,NA,525,NA,450,400,NA,NA,NA,0)
s3 = seq(0.05,1.05,0.1)
plot(dsm,s3,col=blue,las=1,ylab=fraction,xlab=distance (km))
fc - function(x,a,b){a*exp(-b*x)}
fm -
Hi all,
I'd like to find how many sets of 1s there are in the following example;
x - rep(c(1,2,1,3,5), each=5)
I know that there are two sets of 1s, visually. Any function in R that
allows me to automate the process?
Thanks.
Muhammad
__
Dear Arnaud,
Wrap a pair of curly brackets around the command line and remove the + like
the following;
zz-{merge(transform(merge(value,allcon,all.y=T),SHORTDESCRIPTION=NULL,
VALUE=NULL,PL=-VALUE*pl,quantity=NULL),PosB,all.x=T,sort=F )}
Muhammad
arnaud Gaboury wrote:
Dear group,
I have
Hi all,
I have a file, say, test.txt, which contains the following information.
I'm trying to read in the file and specifying the missing values as NA
so that each column has the same number of rows.
I've tried all sorts of manipulation but to no avail.
r1 r2 r3
1 3
2 3
3 2 3
4 2 3
5 2
, ... Then read.csv should handle it
fine.
On Thu, May 6, 2010 at 12:10 PM, Muhammad Rahiz
muhammad.ra...@ouce.ox.ac.uk mailto:muhammad.ra...@ouce.ox.ac.uk
wrote:
Hi all,
I have a file, say, test.txt, which contains the following
information. I'm trying to read in the file and specifying
This could work
out - c()
for (a in 1:10){
out[a] - paste(loci,a,sep=)
write.table(out[a],file=out[a],row.names=FALSE,col.names=FALSE)}
Muhammad
karena wrote:
I have a question about the write.table
I have 100 data.frames, loci1, loci2, loci3.,loci100.
now, I want to
000/000 returns NaN, which is no different than NA unless you want it as
string i.e. 000/000
Muhammad
Lanna Jin wrote:
Try: x[which(is.na(x)),] - 000/000, where is x is your data frame
-
Lanna Jin
lanna...@gmail.com
510-898-8525
__
Hi Nevil,
You can try a method like this
x - c(rnorm(5),rep(NA,3),rnorm(5)) # sample data
dat - data.frame(x,x) # make sample dataframe
dat2 - as.matrix(dat) # conver to matrix
y - which(is.na(dat)==TRUE) # get index of NA values
dat2[y] -
1) I believe you wanted a scree plot which shows the percentage of
variance explained (y-axis) versus the principal components (x-axis). To
do that, all you needs a simple plot(x,y) function where x = pc and y =
variance. It is called a scree plot because the plot looks like a scree
on the
Hi Ozan,
The {calibrate} package allows you to do that.
install.packages(calibrate)
library(calibrate)
df=data.frame(year=c(2001,2002,2003),a=c(8,9,7),b=c(100,90,95))
df
plot(b~a,data=df)
textxy(df$a,df$b,df$year)
Muhammad
On 05/02/2010 08:25 PM, Ozan Bakis wrote:
Hi R users,
I would like
(0,120))
abline(v=in1$minimum,col=red,lty=2)
abline(h=n,col=red,lty=2)
Muhammad
David Winsemius wrote:
On Apr 23, 2010, at 8:06 PM, Muhammad Rahiz wrote:
Thanks David Peter,
The locator() works but not practical as I have to repeat the
process many times.
Does the code works
Does anyone know of a method that I can get the intersection where the red and
blue curves meet i.e. the value on the x-axis?
x - 1:10
y - 10:1
plot(x,y)
abline(lm(y~x),col=blue)
abline(h=2.5,col=red)
Muhammad
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PM, Peter Ehlers wrote:
On 2010-04-23 11:46, David Winsemius wrote:
On Apr 23, 2010, at 1:06 PM, Muhammad Rahiz wrote:
Does anyone know of a method that I can get the intersection where the
red and blue curves meet i.e. the value on the x-axis?
x- 1:10
y- 10:1
plot(x,y)
abline(lm(y~x
Hi all,
I would like to get the array index for a range of values, say 0 x
1.5. I'm wondering if there is an alternative for the following which
I've done
x0 - rnorm(100)
x1 - ifelse(x0 0 x0 1.5,t,f)
x2 - which(x1==t,arr.ind=TRUE)
x0[x2]
Thanks.
--
Muhammad
Hello everyone,
I'm trying to draw the non-linear regression curve to look at the
distance decay relationship between distance and correlation values. But
I'm not able to plot the curve as I'm struggling with the following
error (parameters without starting value in 'data': x) and found no
Hi Gabriele,
This is one way but I'm sure that there is an optimal way of doing so...
x - c(A,B,C,C,C,C,C,B,B)
name - unique(x) # get unique characters
freq - c()
for (a in 1:length(name)){
freq[a] - sum(x==name[a])}# get frequency
out - cbind(name,freq)
Muhammad
It now works. Thanks both.
Muhammad
Peter Ehlers wrote:
On 2010-04-22 11:09, Muhammad Rahiz wrote:
Hello everyone,
I'm trying to draw the non-linear regression curve to look at the
distance decay relationship between distance and correlation values. But
I'm not able to plot the curve
Hello listeRs,
I'm trying to make a square radius around a given reference point. So
given the following array, how can I manipulate it so that
x0 - array(1,dim=c(5,5))
x0
1 1 1 1 1
1 1 1 1 1
1 1 *1* 1 1
1 1 1 1 1
1 1 1 1 1
becomes
into
3 3 3 3 3
3 2 2 2 3
3 2 *1* 2 3
3 2 2 2 3
3 3 3 3 3
Dear useRs,
I'm having a slight problem with plotting on 2 axes. While the following
code works alright on screen, the saved output does not turn out as
desired i.e. the secondary y-axis does not display fully.
Just run the code and look at image output. Suggestions please...
thanks,
Thanks Ivan, Jim and Petr.
The output turns out as desired after I've taken your suggestions.
Muhammad
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I've been calling R from shell using the following (as example) ...
#!/bin/bash
for dir in $(ls *.txt); do
R CMD BATCH script.R
done
Muhammad
Tsjerk Wassenaar wrote:
Hi,
That seems quite neat. To make it a bit more flexible, and maybe do
some argument acrobatics with bash, you could
in the same file.
HTH
Jannis
Muhammad Rahiz schrieb:
Hello all,
I'd like some advise on this. When I read my files, I pass the
argument, skip=6, to skip 6 lines of header information. After
performing the desired calculations, I have the output. Now I want to
copy the 6 lines of skipped
Yes, it works. Just wondering if the technique can be optimized...
David Winsemius wrote:
On Mar 28, 2010, at 12:51 PM, Muhammad Rahiz wrote:
Dear Jannis,
Thanks for the tip. It works but I'd like to improve on the way I
did it.
x - array(1:50,dim=c(10,10)) # data
h1 - c(ncols
290.00
-20.00
-20.00
5000.00
-.00
1 3 5
2 4 6
Muhammad Rahiz | Doctoral Student in Regional Climate Modeling
Climate Research Laboratory, School of Geography the Environment
Oxford University Centre for the Environment, University of Oxford
South Parks Road, Oxford, OX1 3QY
Hello all,
I'd like some advise on this. When I read my files, I pass the argument,
skip=6, to skip 6 lines of header information. After performing the
desired calculations, I have the output. Now I want to copy the 6 lines
of skipped information to the output.
What I've been doing so far
The following functions may help;
strptime()
ISOdate()
Muhammad
Hosack, Michael wrote:
R community:
Hello, I would to like to convert a character date variable from %m/%d/%Y to
%m/%d/%y. Any advice would be greatly appreciated. I have tried functions for
changing the formatting and
Slightly longer method, but works as well.
z - c(-12,-9)
e - c(-2,0)
k - c(NA,NA)
x - c(z,e,k)
x1 - which(x!=NA,arr.ind=TRUE) # get elements which are not NA
x2 - x[x1]
sum(x2)
[1] -23
--
Muhammad
tj wrote:
Hi all,
May I request for your help if you have time and if you have an idea on
this the solution I gave??
On Fri, Jan 29, 2010 at 9:43 AM, Muhammad Rahiz
muhammad.ra...@ouce.ox.ac.uk mailto:muhammad.ra...@ouce.ox.ac.uk
wrote:
Thanks David Dennis,
I may have found something.
Given that the object xx is the product of unlist(x), to create a
2x2 matrix
Hello all,
I'm trying to create a 2x2 matrix, 32 times after unlist() so that I can
convert the list to matrix. I've looked through the R archive but
couldn't find the answer. There is what I've done.
f - system(ls *.txt, intern=TRUE)
x - lapply(f, read.table)
x
[[1]]
V1V2
1
Thanks David Dennis,
I may have found something.
Given that the object xx is the product of unlist(x), to create a 2x2
matrix with subsets, I could do,
y - matrix(xx[c(1:4)], 2, 2).
This returns,
[,1] [,2]
[1,] -27.3 14.4
[2,] 29.0 -38.1
If I do,
y2 - matrix(xx[c(5:8)],2,2)
this right?
Muhammad
--
Muhammad Rahiz wrote:
Thanks David Dennis,
I may have found something.
Given that the object xx is the product of unlist(x), to create a 2x2
matrix with subsets, I could do,
y - matrix(xx[c(1:4)], 2, 2).
This returns,
[,1] [,2]
[1,] -27.3 14.4
[2
Dear all,
My script returns the following error:
Error in storage.mode(test) - logical :
(list) object cannot be coerced to type 'double'
I've looked around and some suggest that it could be the way data have been
imported. For my case, I don't think it is such. Rather I think it has to be
.
How do I make the data in Excel's Column C in R? Or in other words, how
do I update the result of the conditional statements into the dummy
file, d, which I created (which apparently didn't work).
I hope my explanation makes sense.
thanks.
Muhammad
--
Muhammad Rahiz | Doctoral Student
Does anyone have the code to calculate the drought severity index?
--
Muhammad Rahiz | Doctoral Student in Regional Climate Modeling
Climate Research Laboratory, School of Geography the Environment
Oxford University Centre for the Environment, University of Oxford
South Parks Road, Oxford
)7854-625974
Email: muhammad.ra...@ouce.ox.ac.uk
Jim Lemon wrote:
On 01/12/2010 07:21 PM, Muhammad Rahiz wrote:
Does anyone have the code to calculate the drought severity index?
Hi Muhammad,
If you mean does anyone have the algorithm? it seems pretty hard to
find. Even that old
Thanks Steve!
Muhammad Rahiz | Doctoral Student in Regional Climate Modeling
Climate Research Laboratory, School of Geography the Environment
Oxford University Centre for the Environment
South Parks Road, Oxford, OX1 3QY, United Kingdom
Tel: +44 (0
This works.
m - matrix(1:3,3,3)
x1 - list(m, m+1, m+2, m+3, m+4)
out - list()
for (i in 1:4){
t[[i]] - Reduce(+, x1[c(i:i+1)])
}
Muhammad
--
Muhammad Rahiz | Doctoral Student in Regional Climate Modeling
Climate Research Laboratory, School
)])
}
Muhammad
Muhammad Rahiz | Doctoral Student in Regional Climate Modeling
Climate Research Laboratory, School of Geography the Environment
Oxford University Centre for the Environment
South Parks Road, Oxford, OX1 3QY, United Kingdom
Tel: +44 (0)1865
Yes, should be
1. out[[i]], instead of t[[i]].
2. x1[c(i:(i+1))] # For this case, I was trying out the rolling sum.
Muhammad Rahiz | Doctoral Student in Regional Climate Modeling
Climate Research Laboratory, School of Geography the Environment
[c(2:3)]
x1[c(3:4)]
where x1 =
m - matrix(1:3,3,3)
x1 - list(m, m+1, m+2, m+3, m+4)
Thanks.
Muhammad
--
Muhammad Rahiz | Doctoral Student in Regional Climate Modeling
Climate Research Laboratory, School of Geography the Environment
Oxford
Thanks Gabor,
It works alright. But is there alternative ways to perform rollmean
apart from permutating the data?
Possibly combining the Reduce and rollmean functions?
The easiest but traditional way is
(x[[1]]+x[[2]]) / 2
(x[[2]]+x[[3]]) / 2
Muhammad Rahiz | Doctoral Student
= (1+4)/2 , (4+7)/2
2,5,8 = ..
3,6,9 = ..
Thanks
Muhammad
--
Muhammad Rahiz | Doctoral Student in Regional Climate Modeling
Climate Research Laboratory, School of Geography the Environment
Oxford University Centre for the Environment, University of Oxford
South Parks Road, Oxford, OX1 3QY
]] ) / 2
( x[[2]] + x[[3]] ) / 2
... and so on.
The desired output will return
2.5 2.5 2.5
3.5 3.5 3.5
4.5 4.5 4.5
5.5 5.5 5.5
6.5 6.5 6.5
7.5 7.5 7.5
Muhammad Rahiz | Doctoral Student in Regional Climate Modeling
Climate Research Laboratory, School of Geography the Environment
Oxford
something like the following;
1
10
19
2
11
20
3
12
21
and so on..
Thanks..
--
Muhammad Rahiz | Doctoral Student in Regional Climate Modeling
Climate Research Laboratory, School of Geography the Environment
Oxford University Centre
write.table(x2, filename.txt, col.name=FALSE, row.name=FALSE)
# where x2 = variable you want to save
Muhammad Rahiz | Doctoral Student in Regional Climate Modeling
Climate Research Laboratory, School of Geography the Environment
Oxford University
data but not for my actual dataset.
At the moment, I'm thinking of something in the lines of a for loop
function i.e.
for (i in ...){
statement...
}
Is there a syntax in the for loop that allows me to select the
column/row in the file?
Thanks.
--
Muhammad Rahiz | Doctoral Student
and only the first element will be used
The problem is solved if I use the ifelse function but I want to specify
the code as following;
if (/condition/) {
/statement 1a
statement 1b
/} else {
/statement 2a
statement 2b
/}
Thanks.
Muhammad
--
Muhammad Rahiz | Doctoral Student in Regional
(test.txt))
Original code;
if (x 20){
y - x + 100 # statement 1a
... # statement 1b
} else {
y - x -100 # statement 2a
... # statement 2b
}
New code:
n0 - x + 100
n1 - n0 + 0
m0 - x - 100
m1 - m0 - 0
ifelse(x 20, n1, m1)
Muhammad
Muhammad Rahiz | Doctoral Student in Regional Climate
the following, the error is gone but I'm not getting the output for
each individual file I require
d2 - f[[i]] - m
The issue is, how do I make d2 subsettable?
Thanks.
Muhammad
Muhammad Rahiz | Doctoral Student in Regional Climate Modeling
Climate Research
Thanks Petr, Barry David for your useful comments.
Given
d2[[i]] - f[[i]] - m
The problem lies in how I define the object, d2[[i]], which I managed to
resolve by
d2 - list()
--
Muhammad Rahiz | Doctoral Student in Regional Climate Modeling
objects()
list
[1] 2006.01.txt.h 2006.02.txt.h 2006.12.txt.h
The listed files are the last files in the sequence.
Am I doing the loop correctly?
--
Muhammad Rahiz | Doctoral Student in Regional Climate Modeling
Climate Research Laboratory, School
))
myfile1 - paste(strsplit(seq[[i]], \\.)[[1]][1], sum.txt, sep=.)
write.table(mean, file=myfile1, row.names=FALSE, col.names=FALSE)
}
-
Muhammad Rahiz wrote:
Hi all,
I've got a list of files from 1914 to 2000. For each file, I can call the
read.table function as follows.
file - read.table
.
for (j in 1:3) print (file[[j]]-mean)
Thanks.
Muhammad
--
Muhammad Rahiz | Doctoral Student in Regional Climate Modeling
Climate Research Laboratory, School of Geography the Environment
Oxford University Centre for the Environment
South Parks
Code works fine. Thanks!
Muhammad Rahiz | Doctoral Student in Regional Climate Modeling
Climate Research Laboratory, School of Geography the Environment
Oxford University Centre for the Environment, University of Oxford
South Parks Road, Oxford, OX1 3QY, United Kingdom
Tel: +44 (0)1865
.
--
Muhammad Rahiz | Doctoral Student in Regional Climate Modeling
Climate Research Laboratory, School of Geography the Environment
Oxford University Centre for the Environment
South Parks Road, Oxford, OX1 3QY, United Kingdom
Tel: +44 (0)1865-285194 Mobile
cell in the file
(I'm using gridded data). The desired output is
Output
3 3 3
3 3 3
3 3 3
Thanks.
Muhammad
--
Muhammad Rahiz | Doctoral Student in Regional Climate Modeling
Climate Research Laboratory, School of Geography
Muhammad
--
Muhammad Rahiz | Doctoral Student in Regional Climate Modeling
Climate Research Laboratory, School of Geography the Environment
Oxford University Centre for the Environment
South Parks Road, Oxford, OX1 3QY, United Kingdom
Tel: +44 (0)1865
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