Displaying 20 results from an estimated 900 matches similar to: "calculate area of outer polygon"
2006 Dec 27
3
counties in different colours using map()
Hi,
I would like to plot a map of US counties using different colors. map()
seems to be the function to use, e.g.
library(maps); map('usa'); map('county', 'colorado', add=T,fill = T,
col=c(1:5))
plots Colorado counties using colours 1 to 5.
However, I want each color to represent a certain value - a value to be
picked from a data frame.
This code should show a
2007 Mar 19
5
order of values in vector
Dear all,
I would like to get the order of the values in a vector. I have tried
rank(), order() and searched the archive, though without success.
Here is an example of a try
x= c(20,30,50,40,60,10)
cbind(sort.list(x),x)
x
[1,] 6 20
[2,] 1 30
[3,] 2 50
[4,] 4 40
[5,] 3 60
[6,] 5 10
but I was hoping to get this:
x
[1,] 2 20
[2,] 3 30
[3,] 5 50
[4,] 4 40
[5,] 6 60
[6,] 1 10
I'm
2007 Jan 10
1
map data.frame() data after having linked them to a read.shape() object
Dear all,
I try to first link data in a data.frame() to a (polygon) read.shape()
object and then to plot a polygon map showing the data in the
data.frame. The first linking is called "link" in ArcView and "relate"
in ArcMap. I use the code shown below, though without success.
Help with this would be greatly appreciated.
Thanks!
Tord
require(maptools)
# Read shape file
2009 Oct 23
3
opposite estimates from zeroinfl() and hurdle()
Dear all,
A question related to the following has been asked on R-help before, but
I could not find any answer to it. Input will be much appreciated.
I got an unexpected sign of the "slope" parameter associated with a
covariate (diam) using zeroinfl(). It led me to compare the estimates
given by zeroinfl() and hurdle():
The (significant) negative estimate here is surprising, given
2003 Jul 21
2
bold AND italic as font in text()
Dear all,
Is it possible to somshow plot text as italic AND bold. I tried font=c(2,3)
in text(), but it doesn't work. It seems like the latter value is used.
Thanks in advance!
Sincerely,
Tord
-----------------------------------------------------------------------
Tord Sn?ll
Avd. f v?xtekologi, Evolutionsbiologiskt centrum, Uppsala universitet
Dept. of Plant Ecology, Evolutionary Biology
2002 Jan 04
4
line up a matrix
Dear all,
I try to rearrange my ref. database (now in Excel!! :( ) for importing it
into a reference manager program (RIS format).
My file basically look like this [3,4]-matrix:
rbind(c("a", "b", "c", "d"), c("e", "f", "g", "h"), c("i", "j", "k", "l"))
[,1] [,2] [,3]
2003 Jan 02
1
replace NA with factor class
Dear all,
I have a tree data matrix. For some trees I lack info about tree species,
but I want to set them to be spruce. For some reason the tree species names
on the remaining (non-NA) rows are changed into numbers (that I do not
recognise).
I guess that ifelse is not the correct function to use, but I have not
found any better one in my searches.
Thanks in advance!
Sincerely,
Tord
>
2003 Oct 20
4
warning from return() in 1.8 but not in 1.7.0 (PR#4687)
To whom it may concern,
I get the following message when I run my function:
Warning message:
multi-argument returns are deprecated in: return(call.fn, repl, time, from,
to, last.year, occup.m, ant.occ.m,
> version
platform i386-pc-mingw32
arch i386
os mingw32
system i386, mingw32
status
major 1
minor 8.0
2003 May 22
7
extract half a matrix
Dear all,
I'm new to matrix operations in R. I couln't find a solution to the
following problem among earlier help mails or in An introd to R, I guess
because the question is really basic.
I want to extract all above the diagonal, i.e. from
1 2 3 4
1 0 26 49 49
2 26 0 44 40
3 49 44 0 21
4 49 40 21 0
I want
26
49
44
49
40
21
Thanks in advance!
Sincerely,
Tord
2003 Nov 23
4
remove 0 rows from a data frame
Dear all,
As part of a larger function, I am randomly removing rows from a data
frame. The number of removed rows is determmined by a Poisson distribution
with a low mean. Sometimes, the random number is 0, and that's when the
problem starts:
My data frame:
> temp
occ x y dbh age
801 0 2977.196 3090.225 6 36.0
802 0 2951.892 3083.769 8 40.6
803 0 2919.111
2003 Jan 10
0
Thanks: Re: count levels per factor level
Dear Lockwood,
As you can see, I'm a beginner...
But thank you very much!
Sincerely,
Tord
Quoting "J.R. Lockwood" <lockwood at rand.org>:
> how about
>
> buskartant$buskartant <- sapply( group.list, function(x)
> sum(!is.na(unique(x))) )
>
> OR
>
> buskartant$buskartant <- sapply( group.list, function(x)
> length(unique(x[!is.na(x)])) )
2003 Dec 07
2
par(las = 1) not possible in polymap(), library(splancs)?
Dear all,
I want my PhD thesis which I hand in tomorrow to look even nicer:
Does polymap in the splancs library not allow horizontal plotting of
y-labels? I have tried
polymap(studyarea, xlab = "x (m)", ylab = "y (m)", las = 1)
but it doesn't change the labels?
Mayby some function in library(spatstat) support las?
Thanks!
Sincerely,
Tord
2000 Jun 22
1
NAs introduced when changing data type
Hi all,
I am trying to change data type in the Data Editor but when I do, all
numbers in the column changes to NA and back in the Console I get the
warning message "NAs introduced by coercion".
What's wrong? The table consists of 95 columns and 470 rows.
Tord
-----------------------------------------------------------------------
******New e-mail address: Tord.Snall at
2003 Feb 05
1
simplify a data frame
Dear all,
For the past three hours I have tried simplify a data frame. I would be
really happy if someone could help solving this, I'm sure simple, problem.
I want to "aggregate" the data frame:
ObjektID BalteNummer Baltessegment
S.13 S.13.1 S.13.1.2
S.13 S.13.1 S.13.1.3
S.13 S.13.2 S.13.2.1
S.13 S.13.2 S.13.2.2
S.13 S.13.2 S.13.2.3
S.13 S.13.3 S.13.3.6
S.13 S.13.3 S.13.3.7
2003 Jan 02
1
aggregate: "sum" not meaningful for factors
Dear all,
I try to summarise my data per category using aggregate, but for some
reason I get the error message "sum" not meaningful for factors even though
my vector is numeric. The data set is shown below.
Could someone please give a hint.
Thanks in advance!
Sincerely,
Tord
> names(test)
[1] "ObjektID" "tallstubbyta"
> is.factor(test$ObjektID);
2003 Apr 27
0
Thanks for: no response of the tab key
Dear G?tz Wiegand and Corey Moffet,
Thanks for answering while I went to get a cup of coffee. This list is
really excellent!
The answer, for you other beginners, is
in R a tab is represented "\t"
Tord
>
>Dear all,
>I want to produce a tab separated text file but R 1.6.2 does not respond to
>the tab key on my keyboard.
>
>When I paste:
2018 May 13
0
BUG: 'bibentry' methods change default bibstyle
Hi,
Some 'bibentry' methods have the side effect of changing the default bibstyle.
Background:
tools::bibstyle() defines and registers bibtex styles for formatting 'bibentry' objects.
It optionally (and by default) sets a default style. It is also used to get a particular style.
The last feature of bibstyle() is used in 'bibentry' methods defined in packages
2000 Dec 10
1
basic plot() question
Dear all,
I try to plot a logstic regression model as follows:
Valkror<- read.table("Valkror.txt", header=T)
np.bark<- glm(Npinc~bark, family=binomial, data=Valkror)
plot.formula(np.bark$fitted.values~ Valkror$bark)
The above looks nice but I want to connect the points with a line so I try:
plot.formula(np.bark$fitted.values~ Valkror$bark, type="b")
but it gives a
2003 Jan 16
1
problem with as.data.frame.table
Dear all,
I think that what I want is an as.data.frame.table-object, but see error
message below.
I have a data frame with one tree per row, diaclass tells if it is a small,
mid or large tree
> cpy.tradart[1:5, ]
ObjektID diaclass
1 AX.Grb.1 bigdia
2 AX.Grb.1 middia
3 AX.Grb.1 middia
4 AX.Grb.1 smalldia
5 AX.Grb.1 middia
>
I want a data frame telling no of trees per diameter
2001 May 27
2
library for mixed GLM?
Dear all,
I am taking a course in GLM given by a devoted SAS user. He has given us a
homework where a mixed GLM with a logistic link and binomially distributed
observations should be fitted.
I know of the library nlme for mixed effect models but as I understand it,
one cannot choose between different links and distributions in the
functions provided there.
I have so far managed very well with