我有一个(x,y,z)值的网格,我想要一个函数,当它被赋予超出网格的(x,y)点时,它可以近似z值。
我试图使用Akima包(代码块3)来解决这个问题,但我似乎无法使用interp函数来处理线性超出网格所需的linear = FALSE选项。
数据:
# Grid data
x <- seq(0,1,length.out=6)
y <- seq(0,1,length.out=6)
z <- outer(x,y,function(x,y){sqrt(x^2+y^3)})
可视化数据(对于问题不是必需的):
## Visualize the data - Not important for question ##
jet.colors <- colorRampPalette( c("Royal Blue", "Lime Green") )
nbcol <- 100
color <- jet.colors(nbcol)
nrz <- nrow(z)
ncz <- ncol(z)
zfacet <- z[-1, -1] + z[-1, -ncz] + z[-nrz, -1] + z[-nrz, -ncz]
facetcol <- cut(zfacet, nbcol)
pmat <- persp(x, y, z, d = 1,r = 4,
ticktype="detailed",
col = color[facetcol],
main = "Title",
xlab="x value",
ylab = "y value",
zlab= "z value",
scale=FALSE,
expand=0.6,
theta=-40,
phi=25)
## End visualization ##
我尝试使用Akima包解决问题
library(akima)
# Vectorize the grid:
zz <- as.vector(z)
# create all combinations of x and y
xy <- expand.grid(x,y)
# What we want:
sqrt(0.7^2 + 0.7^3) # c(0.7, 0.7) = 0.9126883
sqrt(0.7^2 + 1.2^3) # c(0.7, 1.2) = 1.489295
# We get a result for the first point inside the grid,
# but not for the second one outside the grid.
# This is expected behaviour when linear=TRUE:
interp(xy[,1], xy[,2], zz, xo = c(0.7), yo= c(0.7, 1.2), linear=TRUE)
# = (0.929506, NA)
# When LINEAR = FALSE we get z= 0, 0!!
interp(xy[,1], xy[,2], zz, xo = c(0.7), yo= c(0.7, 1.2), linear=FALSE, extrap = TRUE)
# = (0, 0)
# Dropping extrap=TRUE we see that both are actually NA in this case
interp(xy[,1], xy[,2], zz, xo = c(0.7), yo= c(0.7, 1.2), linear=FALSE)
# = (NA, NA)
# What is going on?
使用R 3.1.3和akima_0.5-11。
答案 0 :(得分:1)
直接使用interp.old()获得结果:
interp.old(xy[,1], xy[,2], zz, xo = c(0.7), yo= c(0.7, 1.2), ncp=2, extrap=TRUE)
# = (0.9211469, 1.472434)
我不确定这是否是interp()函数中的错误。 interp()是一个包装器,如果linear = TRUE则调用interp.old(),如果是linear = FALSE,则调用interp.new()。 interp.new()函数似乎失败了。
请注意,不推荐使用ncp参数。从手册:
ncp - 已弃用,请使用参数线性。现在只用于interp.old()。含义是:在每个数据点计算偏导数时使用的附加点数。 ncp必须为0(不使用偏导数),或者至少为2但小于数据点数(且小于25)