[R-sig-Geo] Isolating only land areas on a global map for computing averages

[r@i@1290 m@iii@g oii @im@com]

Hi Tom and others,

Thank you for your response and suggestions! 
Yes, I loaded and used the maptools package previously to create continents on my world map, among other things. I do think that the easiest approach would be to create a raster layer for land, and then water, with the values that I have. However, my precipitation values are globally distributed - every grid cell has a precipitation value for each year (i.e. each grid cell has 138 values/layers/years). So, if I were to create a raster of only land areas, how would I have my grid cells coincide with the land areas only on that raster?
Also, would it be possible to accomplish this with the raster stack that I already have? If so, is there a way to separate all land/water areas this way using the maptools package?
Thanks, again, and I really appreciate your help!
-----Original Message-----
From: Tom Philippi <tephilippi using gmail.com>
To: rain1290 <rain1290 using aim.com>
Cc: r-sig-geo <r-sig-geo using r-project.org>
Sent: Thu, Nov 7, 2019 12:44 am
Subject: Re: [R-sig-Geo] Isolating only land areas on a global map for computing averages

The easiest approach would be to create a separate aligned raster layer for land vs water.  There are plenty of coastline polygons available out there (e.g., maptools, rworldmap, rnaturalearth packages): choose one in your raster CRS (or choose one and spTransform() it).  Then, use a grid version of your raster to extract values from that land/water SpatialPolygons object.
1: Your idea of extracting the land/water value at each grid cell centroid gives one estimate.  Instead of TRUE/FALSE, think of the numeric equivalents 1,0,  then using those as weights for averaging across your grid cells.2: A "better" estimate would be to compute the fraction of each grid cell that is land, then use those fractional [0, 1] values as weights to compute a weighted average of precipitation over land.  At 2.8deg grid cells, a lot of heavy rainfall coastal areas would have the grid cell centroid offshore and be omitted by approach #1.3: I recommend that you think hard about averaging across cells in Lat Lon to estimate average precipitation over land.  The actual area of a ~2.8 by 2.8 deg grid cell at the equator is much larger than the same at 70 deg N.  I would spend the extra hour computing the actual area (in km^2) of land in each of your 8192 grid cells, then using those in a raster as weights for whatever calculations you do on the raster stack.  [Or you can cheat, as the area of a grid cell in degrees is a function of only the latitudes, and your required weights are multiplicative.]
Your mileage may vary...
Tom
On Wed, Nov 6, 2019 at 6:18 PM rain1290--- via R-sig-Geo <r-sig-geo using r-project.org> wrote:

Hi there,
I am interested in calculating precipitation medians globally. However, I only want to isolate land areas to compute the median. I already first created a raster stack, called "RCP1pctCO2median", which contains the median values. There are 138 layers, with each layer representing one year.  This raster stack has the following attributes:
class       : RasterStack 
dimensions  : 64, 128, 8192, 138  (nrow, ncol, ncell, nlayers)
resolution  : 2.8125, 2.789327  (x, y)
extent      : -181.4062, 178.5938, -89.25846, 89.25846  (xmin, xmax, ymin, ymax)
coord. ref. : +proj=longlat +datum=WGS84 +ellps=WGS84 +towgs84=0,0,0 
names       :    layer.1,    layer.2,    layer.3,    layer.4,    layer.5,    layer.6,    layer.7,    layer.8,    layer.9,   layer.10,   layer.11,   layer.12,   layer.13,   layer.14,   layer.15, ... 
min values  : 0.42964514, 0.43375653, 0.51749371, 0.50838983, 0.45366730, 0.53099146, 0.49757186, 0.45697752, 0.41382199, 0.46082401, 0.45516687, 0.51857087, 0.41005131, 0.45956529, 0.47497867, ... 
max values  :   96.30350,  104.08584,   88.92751,   97.49373,   89.57201,   90.58570,   86.67651,   88.33519,   96.94720,  101.58247,   96.07792,   93.21948,   99.59785,   94.26218,   90.62138, ...  

Previously, I was isolating a specific region by specifying a range of longitudes and latitudes to obtain the medians for that region, like this:
expansion1<-expand.grid(103:120, 3:15)lonlataaa<-extract(RCP1pctCO2Median, expansion1)Columnaaa<-colMeans(lonlataaa)

However, with this approach, too much water can mix with land areas, and if I narrow the latitude/longitude range on land, I might miss too much land to compute the median meaningfully.
Therefore, with this data, would it be possible to use an IF/ELSE statement to tell R that if the "center point" of each grid cell happens to fall on land, then it would be considered as land (i.e. that would be TRUE - if not, then FALSE)? Even if a grid cell happens to have water mixed with land, but the center point of the grid is on land, that would be considered land. But can R even tell what is land or water in this case?
Thank you, and I would greatly appreciate any assistance!

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