[R-sig-Geo] Antw: Re: Creating a spatialPolygonsDataFrame from a data frame

[Matteo Mattiuzzi]

If MODIS tiles are your aim why don't you use?


MODIS:::getTile()



>>> steven mosher  07.10.12 19.11 Uhr >>>
Well I'm just puttering around and got kinda tired of pulling up Google
Earth everytime I wanted to locate a MODIS Tile from  lat/lon.   so I found
the the following

gring <-"
http://landweb.nascom.nasa.gov/developers/sn_tiles/sn_gring_10deg.txt"



GRING <- read.table(gring, skip = 7,   stringsAsFactors =FALSE, nrows=648,
                    na.strings=c("-999.0000","-99.0000"))

colnames(GRING)<-c("v","h","ll_lon","ll_lat",
                   "ul_lon","ul_lat","ur_lon","ur_lat",
                   "lr_lon","lr_lat")




On Sun, Oct 7, 2012 at 9:51 AM, Matteo Mattiuzzi <
matteo.mattiuzzi at boku.ac.at> wrote:

> Hi Steve, what about this?
>
>
> DF <- structure(list(v = c(4L, 4L, 4L, 4L, 4L, 4L, 4L), h = 8:14, ll_lon =
> c(-131.0149,
> -117.7464, -104.5202, -91.3388, -78.2083, -65.15, -52.1199),
>    ll_lat = c(39.7081, 39.7342, 39.7557, 39.7728, 39.7858, 39.7937,
>    39.7994), ul_lon = c(-156.8405, -140.7952, -124.8854, -109.0855,
>    -93.3822, -77.7862, -62.229), ul_lat = c(49.8983, 49.9394,
>    49.9677, 49.9863, 49.9972, 50, 50), ur_lon = c(-140.2398,
>    -124.6153, -109.0021, -93.3968, -77.7506, -62.1191, -46.5357
>    ), ur_lat = c(50.1258, 50.1159, 50.1047, 50.0921, 50.0754,
>    50.0582, 50.0429), lr_lon = c(-117.2848, -104.2354, -91.191,
>    -78.1497, -65.0781, -52.0094, -38.9707), lr_lat = c(39.8699,
>    39.8624, 39.8554, 39.8489, 39.8411, 39.8337, 39.828)), .Names = c("v",
> "h", "ll_lon", "ll_lat", "ul_lon", "ul_lat", "ur_lon", "ur_lat",
> "lr_lon", "lr_lat"), row.names = 153:159, class = "data.frame")
>
>
>
>
> SPDF <- list()
> id <- matrix(NA,nrow(DF),ncol=2)
> colnames(id)<- c("h","v")
>       for (u in 1:nrow(DF))
> {
> SPDF[[u]] <- Polygons(list(Polygon(cbind(
> c(DF$ll_lon[u], DF$ul_lon[u], DF$ur_lon[u], DF$lr_lon[u], DF$ll_lon[u]),
> c(DF$ll_lat[u], DF$ul_lat[u], DF$ur_lat[u], DF$lr_lat[u], DF$ll_lat[u])),
> hole = FALSE)),u)
> id[u,] <- c(DF$h[u], DF$v[u]) # probably h/v information can be taken
> direcly from DF
>       }
> SPDF <- SpatialPolygons(SPDF,1:length(SPDF),proj4string=CRS("+proj=longlat
> +ellps=WGS84 +datum=WGS84 +no_defs"))
>       SPDF <- SpatialPolygonsDataFrame(SPDF,
> as.data.frame(id),match.ID=FALSE)
> plot(SPDF,col="green")
>
>
>
>
> >>> steven mosher  07.10.12 8.18 Uhr >>>
> Hi,
>
>   I have a dataframe that has a few hundred rows like this.
>
> DF <- structure(list(v = c(4L, 4L, 4L, 4L, 4L, 4L, 4L), h = 8:14, ll_lon =
> c(-131.0149,
> -117.7464, -104.5202, -91.3388, -78.2083, -65.15, -52.1199),
>     ll_lat = c(39.7081, 39.7342, 39.7557, 39.7728, 39.7858, 39.7937,
>     39.7994), ul_lon = c(-156.8405, -140.7952, -124.8854, -109.0855,
>     -93.3822, -77.7862, -62.229), ul_lat = c(49.8983, 49.9394,
>     49.9677, 49.9863, 49.9972, 50, 50), ur_lon = c(-140.2398,
>     -124.6153, -109.0021, -93.3968, -77.7506, -62.1191, -46.5357
>     ), ur_lat = c(50.1258, 50.1159, 50.1047, 50.0921, 50.0754,
>     50.0582, 50.0429), lr_lon = c(-117.2848, -104.2354, -91.191,
>     -78.1497, -65.0781, -52.0094, -38.9707), lr_lat = c(39.8699,
>     39.8624, 39.8554, 39.8489, 39.8411, 39.8337, 39.828)), .Names = c("v",
> "h", "ll_lon", "ll_lat", "ul_lon", "ul_lat", "ur_lon", "ur_lat",
> "lr_lon", "lr_lat"), row.names = 153:159, class = "data.frame")
>
>
> The columns  are  v and h, two attributes,  and then  4 points in lat and
> lon    ll_lon  is lower left lon, lr_lon is lower right,
> ul_lon is upper left and ur_lon is upper right.
>
> What I want to do is create a spatialPolygonsDataFrame from this data.
> As best as I can figure I have to take the  lon/lat coordinates and put
> them in to a matrix format, replicate the first point to close
> the ring and create a Polygon, and then Create a Polygons() object with an
> ID and then a SpatialPolygon, and then a
> SpatialPolygonDataFrame  from a list of the spatial polygons.
>
> What I'm struggling with is how to do this for the several hundred rows
> that I have.
>
> Steve
>
>     [[alternative HTML version deleted]]
>
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