如何使用R计算栅格堆栈上的Sen斜率

Question

I have 36 layers of raster stack(annual composits of rainfall over specified region). When I tried to compute Sen's slope by using the following code:

library(raster)
library(trend)
# example data
s <- stack(system.file("external/rlogo.grd", package="raster")) 
s <- stack(s, s* 2, s*3)

func <- function(x) { unlist(sens.slope(x)) }
sen.slop <- calc(s, fun=func)

It returns the following error

Error in .local(x, values, ...) : 
 values must be numeric, integer or logical.

Is there anybody who can help me to resolve the problem?

Answer 1

sens.slope returns an object of class htest that includes numeric values, but also character values. To make a Raster, you need to select the numeric values you want. Eg :

library(raster)
library(trend)
s <- stack(system.file("external/rlogo.grd", package="raster")) 
s <- stack(s, s* 2, s*3)
func <- function(x) { unlist(sens.slope(x)[1:3]) }
sen.slop <- calc(s, fun=func)

The thing to understand is that before supplying your own function to calc you should inspect its behavior. For example, compare:

set.seed(9);
v <- runif(100) * 1:100
# original function
func <- function(x) { unlist(sens.slope(x)) }

func(v)
# estimates.Sen's slope            statistic.z                p.value           null.value.z            alternative              data.name                 method            parameter.n 
# "0.40383510858131"      "6.6084411517969" "3.88387866698504e-11"                    "0"            "two.sided"                    "x"          "Sen's slope"                  "100" 

# Yikes! character output. 

... with what is returned by this function

func <- function(x) { unlist(sens.slope(x)[1:3]) }
func(v)
# estimates.Sen's slope           statistic.z               p.value 
#          4.038351e-01          6.608441e+00          3.883879e-11 

 # better!