plot(1,1,main=expression(-70*degree*C%+-%10*degree*C/Ambient))
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Fri, Feb 27, 2015 at 7:27 PM, li li hannah@gmail.com wrote:
Hi all,
I would like to add -70°C ± 10°C/Ambient as the title of my plot.
Could anyone give some help on this?
Many thanks. Unfortunately, I cannot work directly on these expressions since
they’re only created from other strings. Would I first have to transform these
strings to unevaluated expressions?
Von: William Dunlap [mailto:wdun...@tibco.com]
Gesendet: Freitag, 27. Februar 2015 23:39
An: Alrik
Hi all,
I would like to add -70°C ± 10°C/Ambient as the title of my plot.
Could anyone give some help on this?
Thanks.
Hanna
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Thanks very much.
Also How do add an empty space when using expression()?
When I do the following, it returns an error message.
plot(1,1,main=expression(-70*degree*C%+-%10*degree*C/Ambient Condition))
Hanna
2015-02-27 23:03 GMT-05:00 William Dunlap wdun...@tibco.com:
So, Duncan, do I understand you correctly:
When I use x$x6, R doesn't know if it's TRUE or FALSE, so it returns
a logical value of NA.
When this logical value is applied to a row, the R says: hell, I don't
know if I should keep it or not, so, just in case, I am going to keep
it, but I'll replace
Hi,thanks all for the answer.I am using mclapply to call the lapply many times
as needed. My function returns only a value if the fit is succesful.For testing
if the fit is sucessfuly my code works like that
On Feb 27, 2015, at 4:49 AM, marKo mton...@ffri.hr wrote:
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Gee. That simple. I knew that!
Thanks a lot.
Essentially, I needed only the diagonal elements.
Easily solved by:
diag(outer( X=v1,Y=v2, FUN= fV)
I am sure that there are
Dear all,
I have a Mac, OSX 10.7.5, run R 3.1.2 (2014-10-31) and Matlab R2013a
(8.1.0.604).
I try to call R from within Matlab to run a function in Batch mode.
To do this, I followed this example:
From: www.mathworks.com/matlabcentral/newsreader/view_thread/163726
In an m-file you would have:
I Try to run this script on order to get all the protein named Delta 9 acyl
CoA desaturase.
library(seqinr)
choosebank(swissprot)
D9aCd - query(Delta 9 acyl-CoA desaturase,KD=Delta 9 acyl-CoA
desaturase)
Nothing happen
i be glad to know what i do wrong...
Thnks
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Sorry Jim I forgot to reply on what I am trying to do.I have many data sets
that contain some numbers: I am trying to fit those with a mixture
distribution. For that I am using the mix function that returns me back the
fitted parameters and the chi square (which is a first performance
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Yes. That's it.
Thanks, a lot, really.
Marko
On 02/27/2015 02:46 PM, David Winsemius wrote:
On Feb 27, 2015, at 4:49 AM, marKo mton...@ffri.hr wrote:
Gee. That simple. I knew that! Thanks a lot. Essentially, I needed
only the diagonal
Use Jeff’s solution. This doesn’t account for ties.
On Feb 26, 2015, at 1:11 PM, Don McKenzie d...@u.washington.edu wrote:
Kate — here is a transparent solution (tested but without NA treatment).
Doubtless there are cleverer faster ones, which later posters will present.
HTH
#
I know how to get the output I need, but I would benefit from an
explanation why R behaves the way it does.
# I have a data frame x:
x = data.frame(a=1:10,b=2:11,c=c(1,NA,3,NA,5,NA,7,NA,NA,10))
x
# I want to toss rows in x that contain values =6. But I don't want
to toss my NAs there.
On 27/02/2015 9:04 AM, Dimitri Liakhovitski wrote:
I know how to get the output I need, but I would benefit from an
explanation why R behaves the way it does.
# I have a data frame x:
x = data.frame(a=1:10,b=2:11,c=c(1,NA,3,NA,5,NA,7,NA,NA,10))
x
# I want to toss rows in x that contain
This should be a simple question, but I am at my wits end.
dt-data.table(a=rep(1:10, 26), b=1:260, c=rep(1:2, 130))
sumvar - 'mysum'
bvar - 'b'
dt_min - dt[, list(sumvar = sum(get(bvar))), by=list(a)]
print(dt_min)
I want the function to return two variables, a and mysum. However, it
instead
This usually has to do with the caller (Matlab) and the callee (R) being
dynamically lined with different versions of dynamic libraries. On Linux
I work around this sort of thing by setting LD_LIBRARY_PATH to exactly
what I want when invoking R. E.g., instead of making the sh command
Thank you very much, Duncan.
All this being said:
What would you say is the most elegant and most safe way to solve such
a seemingly simple task?
Thank you!
On Fri, Feb 27, 2015 at 10:02 AM, Duncan Murdoch
murdoch.dun...@gmail.com wrote:
On 27/02/2015 9:49 AM, Dimitri Liakhovitski wrote:
So,
You could define functions like
is.true - function(x) !is.na(x) x
is.false - function(x) !is.na(x) !x
and use them in your selections. E.g.,
x - data.frame(a=1:10,b=2:11,c=c(1,NA,3,NA,5,NA,7,NA,NA,10))
x[is.true(x$c = 6), ]
a b c
7 7 8 7
10 10 11 10
Bill Dunlap
Here is another approach
maxv - apply(df, 2, max) # Get the column maximums
maxv0 - ifelse(maxv == 0, -1, maxv) # Replace 0 maximums with -1
Sum - rowSums(sweep(df, 2, maxv0, ==))
data.frame(df, Sum)
A B C D Sum
1 0 1 0 7 1
2 0 2 0 7 1
3 0 3 0 7 1
4 0 4 0 7 1
5 0 1 0 0 0
6
On 27/02/2015 10:27 AM, Dimitri Liakhovitski wrote:
Thank you very much, Duncan.
All this being said:
What would you say is the most elegant and most safe way to solve such
a seemingly simple task?
If you have NA values, test for them explicitly, e.g. your original
x[(x$c6) | is.na(x$c),]
All,
I am trying to read the SP 500 constituents from the iShares website using the
following code:
URL - http://www.ishares.com/us/239726/fund-download.dl;
setInternet2(TRUE)
download.file(url=URL, destfile=temp.xls)
out - readWorksheetFromFile(file=temp.xls, sheet=Holdings,
On 27/02/2015 9:49 AM, Dimitri Liakhovitski wrote:
So, Duncan, do I understand you correctly:
When I use x$x6, R doesn't know if it's TRUE or FALSE, so it returns
a logical value of NA.
Yes, when x$x is NA. (Though I think you meant x$c.)
When this logical value is applied to a row, the R
Thank you very much guys!
On Fri, Feb 27, 2015 at 11:04 AM, William Dunlap wdun...@tibco.com wrote:
You could define functions like
is.true - function(x) !is.na(x) x
is.false - function(x) !is.na(x) !x
and use them in your selections. E.g.,
x -
On 27/02/2015 1:18 PM, Brian Trautman wrote:
This should be a simple question, but I am at my wits end.
dt-data.table(a=rep(1:10, 26), b=1:260, c=rep(1:2, 130))
sumvar - 'mysum'
bvar - 'b'
dt_min - dt[, list(sumvar = sum(get(bvar))), by=list(a)]
print(dt_min)
I want the function to
This works:
Change the destination directory to suit you.
MyURL1 = http://www.ishares.com/us/239726/fund-download.dl;
download.file(MyURL1,paste(C:/Data/Rtest1,date1,r.xls,sep=),method=wget,quiet=TRUE,mode=wb,
extra=--header=\User-Agent: Mozilla/5.0
(X11; Linux x86_64;
Hi all,
I am wanting to model some patch dynamics using raster objects in R and am
trying to figure out the best way to do it.
I want to treat each patch (pixel, cell, grid) and assign multiple attributes
(for example biomass, temperature, precipitation, etc). I know that I can do
this using
This worked perfectly, thank you!
On Fri, Feb 27, 2015 at 10:47 AM, Duncan Murdoch murdoch.dun...@gmail.com
wrote:
On 27/02/2015 1:18 PM, Brian Trautman wrote:
This should be a simple question, but I am at my wits end.
dt-data.table(a=rep(1:10, 26), b=1:260, c=rep(1:2, 130))
sumvar -
On Fri, Feb 27, 2015 at 10:01 AM, Bos, Roger roger@rothschild.com
wrote:
All,
I am trying to read the SP 500 constituents from the iShares website
using the following code:
URL - http://www.ishares.com/us/239726/fund-download.dl;
setInternet2(TRUE)
download.file(url=URL,
What ***on earth does the content of your message have to do with the
subject line?
Get your expletive deleted act together if you want to ask a question
of this list!!!
cheers,
Rolf Turner
On 27/02/15 08:40, macfire wrote:
I have a problem with statistics, I think this forum is not
If your string will always represent an R expression, you could work with
the expression directly with functions like all.names() and substitute().
f - function (expr)
{
toReplace - setdiff(all.names(expr), c(pmin, pmax))
toReplace - grep(value = TRUE, [a-z], toReplace)
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Gee. That simple. I knew that!
Thanks a lot.
Essentially, I needed only the diagonal elements.
Easily solved by:
diag(outer( X=v1,Y=v2, FUN= fV)
I am sure that there are simpler options, but that works like a charm.
Thanks a lot.
Cheers,
Marko
Dear R-help list,
I would like to replace all lower-case letters in a string that are not part
of certain fixed expressions. For example, I have the string:
pmin(pmax(pmin(x1, X2), pmin(X3, X4)) == Y, pmax(Z1, z1))
Where I would like to replace all lower-case letters that do not belong to
the
Perhaps you could circulate this?
*Online courses in R from Statistics.com*
*Basic programming courses: *
- R Programming - Intro 1 http://www.statistics.com/R-Prog-Intro-1 (Dr.
Paul Murrell, instructor - core development team for R and author of *R
Graphics *and *Introduction to Data
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