R help - Nov 2010 - Generating kml lines output

Dear,

I would like to generate kml file with lines (<LineString>) of
different style (definitely color, but width would also be nice to
see). However with kmlLine from maptools package I am able only to
output the single first Lines object (first row of
SpatialLinesDataFrame) 

I guess the option would be to have single Lines object (instead of
list of them), but then also the style wouldn't vary among them... My
code so far:  

library(maptools)
library(sp)

out <- structure(list(I = structure(c(1L, 7L, 3L, 4L, 5L, 1L, 2L, 9L, 
								6L, 3L,
10L, 8L), .Label = c("0.34", "0.35", "0.43",
"0.47", "0.48", "0.52",
"0.64", "0.72", "0.77", "0.92"), class =
"factor"), BF structure(c(1L, 9L, 4L, 6L, 7L, 2L, 3L, 11L, 8L, 5L,
12L, 10L), .Label
= c("3.16", "3.2", "3.44", "4.71",
"4.75", "5.54", "5.76", "6.77",
"11.33", "16.45", "21.24", "75.16" ),
class = "factor"), from structure(c(4L, 3L, 3L, 5L, 5L, 5L, 1L, 2L,
4L, 4L, 4L, 1L), .Label c("Bissau", "Caio",
"CotedIvoire", "Gambia", "Ghana"), class
"factor"), to = structure(c(2L, 4L, 1L, 5L, 1L, 2L, 2L, 6L, 3L, 4L,
6L,
2L), .Label = c("BurkinaFaso", "Caio",
"CotedIvoire", "Ghana",
"Nigeria", "Senegal"), class = "factor"), x =
c(-16.57534, -4.027751,
-4.027751, -0.1963, -0.1963, -0.1963, -15.5982, -16.2027, -16.57534,
-16.57534, -16.57534, -15.5982), y = c(13.4526, 5.336318, 5.336318,
5.555, 5.555, 5.555, 11.86398, 11.9277, 13.4526, 13.4526, 13.4526,
11.86398 ), xend = c(-16.2027, -0.1963, -1.5352, 7.489, -1.5352,
-16.2027, -16.2027, -17.3333, -4.027751, -0.1963, -17.3333, -16.2027
				), yend = c(11.9277, 5.555, 12.357,
9.058, 12.357, 11.9277, 11.9277, 14.75, 5.336318, 5.555, 14.75,
11.9277)), .Names = c("I", "BF", "from",
"to", "x", "y", "xend",
"yend"), row.names = c(NA, -12L), class = "data.frame")


N <- dim(out)[1]
RowAsMatrix <- function(lines, row) {	
	rbind(c(lines[row,]$x, lines[row,]$y), c(lines[row,]$xend,
lines[row,]$yend) ) }

# list of coordinate matrixes to list of Line objects
lines_list <- list()
for(i in 1 : N ) {	
	lines_list[[i]] <- Line( RowAsMatrix(out[5:8], i) )
}

# list of Line objects to Lines object
lines_list_list <- list()
for (i in 1 : N) {
	lines_list_list[[i]] <- Lines(lines_list[[i]], ID paste( out$BF[i] ) ) }

Lines object to SpatialLines object
s_l <- SpatialLines(lines_list_list)

SpatialLines object to SpatialLinesDataFrame object
df = data.frame(z = c(1 : N), row.names = sapply(slot(s_l, "lines"),
function(x) slot(x, "ID"))) sl_df <- SpatialLinesDataFrame(s_l,
data df)

kmlLine(sl_df, kmlfile = "output.kml")



-- 
while(!succeed) { try(); }

On Fri, Nov 26, 2010 at 5:03 PM, fbielejec <fbielejec at gmail.com>
wrote:
> Dear,
>
> I would like to generate kml file with lines (<LineString>) of
> different style (definitely color, but width would also be nice to
> see). However with kmlLine from maptools package I am able only to
> output the single first Lines object (first row of
> SpatialLinesDataFrame)
>
> I guess the option would be to have single Lines object (instead of
> list of them), but then also the style wouldn't vary among them... My
> code so far:
 If you cant get what you want from package code, and you know a bit
about the KML format you are trying to produce, then try using the R
'brew' package, which lets your write a template which can have loops,
variable substitutions, and other bits of R code in it.

 For example, here creating a simple set of KML points from a data
frame called 'cpts':

<?xml version="1.0" encoding="UTF-8"?>
<kml xmlns="http://earth.google.com/kml/2.1">
<Document>
<% for(i in 1:nrow(cpts)){ %>
<Placemark>
<Name><%=paste("pt-",i,sep="")%></Name>
<Point>
<coordinates><%=cpts[i,]$Xt%>,<%=cpts[i,]$Yt%>,0</coordinates>
</Point>
</Placemark>
<% } %>
</Document>
</kml>


results in a KML with lots of these:

<Placemark>
<Name>pt-4</Name>
<Point>
<coordinates>-43.93769,-19.92065,0</coordinates>
</Point>
</Placemark>

 - its a lot easier than writing lots of cat() and other formatting
nonsense in R code!

 But if you can get package code to do it, that's the way to go. But
for outside-the-box solutions, install.packages("brew")!

Barry