similar to: calculate area of outer polygon

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