tly writing in
> an entirely different context, it shouldn't really make much
> difference.
>
> Jim
>
>
> On Wed, May 27, 2015 at 4:35 PM, Daisy Englert Duursma
> wrote:
> > Greetings,
> >
> > I am trying to identify at which point during the year
lot(density(obsDay))
--
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Department of Biological Sciences
Room W19F 135
Macquarie University, North Ryde, NSW 2109
Australia
Tel +61 2 9850 1302
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next
} else {
pres100km[[ii]]<-cells[ii] #100kmcells >= 30% coverage of 10kmcells
}
} else {
if(nrow(subdat) < 5) {
next
} else {
pres100km[[ii]]<-cells[ii] #more than 5 obs
}
}
}
On Mon, Feb 23, 2015 at 1:54 PM, Daisy Englert Duursma
} else {
pres100km[[ii]]<-cells[ii] #100kmcells >= 50% coverage of 10kmcells
}
}
}
--
Daisy Englert Duursma
Department of Biological Sciences
Room W19F 135
Macquarie University, North Ryde, NSW 2109
Australia
Tel +61 2 9850 1302
;;)
Error in curlPerform(curl = curl, .opts = opts, .encoding = .encoding) :
embedded nul in string:
and
a<-httpPOST("
http://biocache.ala.org.au/ws/occurrences/download?q=Banksia+ericifolia";)
Error: Internal Server Error
Any help would be appreciated.
Daisy
--
Daisy Englert Duursm
Hello,
I have a question about interfacing with the Atlas of Living Australia
website: http://biocache.ala.org.au/#tab_simpleSearch
The following code works and I am able to download species observations.
The problem I am having is that I would like additional fields of data. If
I manually downlo
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> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
--
Daisy Englert Duursma
Department of Biological Sciences
R
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> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
--
Daisy Englert Duursma
Department of Bio
n(x2)
})
do.call(rbind, dat2)
Cheers,
Daisy
--
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Department of Biological Sciences
Room E8C156
Macquarie University, North Ryde, NSW 2109
Australia
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Your question is very unclear.Can you provide a better question or at
least a column of outputs you expect and an example with the dataframe
you provided?
On Thu, Aug 9, 2012 at 8:29 AM, Abraham Mathew wrote:
> I have the following data and am trying to find the percentage of bid
> values purchas
Thanks, problem solved.
On Thu, Jun 7, 2012 at 1:58 PM, Daisy Englert Duursma
wrote:
> Hello and thanks for helping.
>
> #some data
> L3 <- LETTERS[1:3]
> dat1 <- data.frame(cbind(x=1, y=rep(1:3,2), fac=sample(L3, 6, replace=TRUE)))
>
>
> #When x==1 and y==1 I wan
not need to subset
perhaps using lapply?
Thanks,
Daisy
--
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Department of Biological Sciences
Room E8C156
Macquarie University, North Ryde, NSW 2109
Australia
Tel +61 2 9850 9256
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of the list
bg<- z_nonna[sample(1:length(z_nonna), 5000, replace=FALSE)]
Thanks for the help,
Daisy
--
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Department of Biological Sciences
Room E8C156
Macquarie University, North Ryde, NSW 2109
Australia
Tel +61 2 9850 9256
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ernative HTML version deleted]]
>
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> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, s
Thanks, that worked like a charm.
Daisy
On Wed, Jun 29, 2011 at 11:47 AM, David Winsemius
wrote:
>
> On Jun 28, 2011, at 9:18 PM, Daisy Englert Duursma wrote:
>
>> Hello,
>>
>> I think this is a simple problem but I am not coming up with a simple
>> solution. I
and values to be replaced in the matrix
dfr <- data.frame(cbind(x=seq(120,125,by=0.05), y=-25, var.1=1))
#insert the values from the dfr into the matrix
bb[dfr$x,dfr$y]<-d$var.1
Thanks for your help,
Daisy
--
Daisy Englert Duursma
Department of Biological Sciences
Room E8C156
Macquarie U
Hello,
I think this is a simple problem but I am not coming up with a simple
solution. I think it just an indexing problem.
I can easily replace values in a matrix from a dataframe when the
dataframe has row and column numbers. In the example below I use row
and column names and I can not get it
e posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
--
Daisy Englert Duursma
Department of Biological Sciences
Room E8C156
Macquarie University, North Ryde, NSW 210
Australia
Tel +61 2 9850 9256
manually-tp3263786p3263786.html
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>
> ______________
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> PLEASE do read the posting guide http://www.R-project.o
0)), NA)
>> g <- function(x) ifelse(sum(x > 0) == 2L, names(which(x == 0L)), NA)
>>> apply(df1[, 3:5], 1, f)
>> [1] "ANN" "CTA" "GLM" NA "ANN" NA NA NA NA "CTA"
>>> apply(df1[, 3:5], 1, g)
>> [1] N
quot;NA","NA","NA","CTA")
one_absence<-c("NA","NA","NA","NA","NA","NA","GLM","CTA","NA","NA")
The end result should look like
df2<-(cbind(df1,one_presence,
;
>
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> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
Thanks for the help. Easy as..
On Tue, Oct 26, 2010 at 3:33 PM, David Winsemius wrote:
>
> On Oct 25, 2010, at 8:56 PM, Daisy Englert Duursma wrote:
>
>> Hello,
>>
>> If I have a dataframe:
>>
>> example(data.frame)
>>
>> zz<-c("aa
he column named group by the "_", how would I do this?
so instead of the first row being
x y fac char group
1 1 Ca aa_bb
it should be:
x y fac char group_agroup_b
1 1 Ca aa bb
I know for a vector I can:
x1 <- c("a_b","
),"w")
cat(paste(group.vars[gv]),file=zzz)
cat(paste(species[sp]),file=zzz)
close(zzz)
}
}
I know it is a silly script test but the division is similar to my
real task which is repeated 459 times and each run takes 18 hours
At this point I am confused what should be
check out ?par for all the details on plotting
‘mgp’ The margin line (in ‘mex’ units) for the axis title,
axis labels and axis line. Note that ‘mgp[1]’ affects
‘title’ whereas ‘mgp[2:3]’ affect ‘axis’. The
default is ‘c(3, 1, 0)’.
On Tue, Sep 14, 2010 at 8:56 AM,
Hello and thanks in advance,
Using the dataset volcano:
ascols =
colorRampPalette(c("gray","yellow","darkgoldenrod1","orange","red"),interpolate="spline")
x <- 10*(1:nrow(volcano))
y <- 10*(1:ncol(volcano))
image(x, y, volcano, col = ascols, axes = FALSE)
In the example above, how would I change
Thanks for the advice, problem solved.
-daisy
On Wed, Aug 25, 2010 at 1:35 PM, David Winsemius wrote:
>
> On Aug 24, 2010, at 10:45 PM, Daisy Englert Duursma wrote:
>
>> Hello help,
>>
>> I have changed around some graphing code and made it into a function.
>&g
so:
text(1,1,paste(b_unit),cex=1)
and
b_unit<-expression(~degree~C)
This now inserts ~degree~C instead of the symbol.
Any advice?
Thanks,
Daisy
--
Daisy Englert Duursma
Room E8C156
Dept. Biological Sciences
Macquarie University NSW 2109
Austra
--
Daisy Englert Duursma
Room E8C156
Dept. Biological Sciences
Macquarie University NSW 2109
Australia
Tel +61 2 9850 9256
10A Carrington Rd
Hornsby, NSW 2077
Mobile: 0421858456
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Thanks to all,
I did eventually come up with a solution which involved sub-setting
the data first into categories, making an empty plot, and then
plotting rectangles. I really like the ggplot2 and I will use that
next time.
My ugly script is below
Thanks,
Daisy
plot(1, xlim=c(5,45),ylim=c(0,1
d min and max. With that
I could make the lines thick and achieve the effect I am looking for.
Hopefully I have missed an obvious method to graph this.
Thanks,
Daisy
--
Daisy Englert Duursma
Room E8C156
Dept. Biological Sciences
Macquarie University NSW 2109
-
> Think before you print - do you really need to print this email?
>
>
> --------
> Scanned by MailMarshal - Mars
me and not the position.
Any advice would be useful.
Thanks,
Daisy
Daisy Englert Duursma
Room E8C156
Dept. Biological Sciences
Macquarie University NSW 2109
Australia
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packages went, the
updated package went to "C:\\RLIBRARY" . I have messed something up
and I do not know how to fix it.
Any advice would be welcome.
Thanks,
Daisy
--
Daisy Englert Duursma
Room E8C156
Dept. Biological Sciences
Macquarie University NSW 2109
Australia
_
Sorry, that should be
qplot(x,y,data=yourdataset, colour= z)
On Thu, May 20, 2010 at 3:07 PM, Daisy Englert Duursma
wrote:
> the package ggplot2 makes it really really easy!
>
> qplot(x,y,data=yourdataset, colour= color)
>
--
Daisy Englert Duursma
Room E8C156
Dept. Biologi
the package ggplot2 makes it really really easy!
qplot(x,y,data=yourdataset, colour= color)
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an
Hello,
Thanks for your help so far, I am still having the same problem but I
think I am getting closer. Just to recap, since setting up R on my
server, so that everyone shares a common Library, I have had issues
getting packages to load. Here is an example with the package
survival. I have the sur
er correctly.
Thanks,
Daisy
--
Daisy Englert Duursma
Room E8C156
Dept. Biological Sciences
Macquarie University NSW 2109
Australia
Tel +61 2 9850 9256
10A Carrington Rd
Hornsby, NSW 2077
Mobile: 0421858456
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(strsplit(as.character(x$GCM), ""), function(zzz)zzz[5] ==
> y),]
}
Basically this loops through and subsets the rows when the 5th
character has the defined y value (1, 2,or 3). The problem is that y
can occur anywhere in the GCM value.
Thanks for the help.
Daisy
--
Daisy Eng
ot;)
for (y in year) {
sdat2 <- sdat[sapply(strsplit(as.character(sdat$GCM), ""),
function(zzz)zzz[5] == y),]
Thanks
Daisy Englert Duursma
Bioclimatic Modeller
Macquarie University
Sydney, NSW, Australia
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