无法有趣地计算R在栅格堆栈上的逐像素回归 [英] Can't Calculate pixel-wise regression in R on raster stack with fun

问题描述

我正在处理栅格，并且我有一个具有7n层的RasterStack.我想使用下面的公式来计算逐像素回归.我试图使用 raster :: calc 来做到这一点，但是我的函数失败并显示消息:

I am working with rasters and I've a RasterStack with 7n layers. I would like to calculate pixel-wise regression, using formula beneath. I was trying to do it with raster::calc, but my function failed with message :

'lm.fit(x，y，offset = offset，singular.ok = singular.ok，中的错误...):0(非NA)案件.'

'Error in lm.fit(x, y, offset = offset, singular.ok = singular.ok, ...) : 0 (non-NA) cases.'

但是所有栅格都可以，并且包含数字(不仅是NA)，我可以绘制它，我可以用公式计算一般线性回归

But all rasters are OK, and contain numbers (not only NAs), I can plot it, and I can calculate general linear regression with formula

 cr.sig=lm (raster::as.array(MK_trend.EVI.sig_Only) ~ raster::as.array(stack.pet)+raster::as.array(stack.tmp)+raster::as.array(stack.vap)+raster::as.array(stack.pre)+raster::as.array(stack.wet)+raster::as.array(stack.dtr))

但是当我堆叠层时

allData = stack(MK_trend.EVI.sig_Only,stack.dtr,stack.wet,stack.pre,stack.vap,stack.tmp,stack.pet)

并尝试计算功能

    # Regression Function, R2
lmFun=function(x){
    x1=as.vector(x);
    if (is.na(x1[1])){
        NA 
    } else {
        m = lm(x1[1] ~ x1[2]+x1[3]+x1[4]+x1[5]+x1[6]+x1[7])
        return(summary(m)$r.squared)
    }
}

我看到了错误消息.
我在R和编程方面还很陌生，所以，也许有一些愚蠢的错误?为了使处理工作正常进行，我将不胜感激.

I see the error message.
I am pretty new in R and progranning, so, maybe, there is some silly mistake? I would appreciate any hint in order to make the processing work.

推荐答案

您可以使用 calc 进行像素级(局部)回归，但是您的公式似乎暗示您需要其他东西(全局模型).

You can use calc for pixel-wise (local) regression, but your formula seems to suggest you want something else (a global model).

如果回归是像素级的，则每个单元格将具有相等数量的x和y值，并且可以使用 calc .有关示例，请参见?calc .

If the regression were pixel wise, you would have an equal number of x and y values for each cell, and you can use calc. See ?calc for examples.

相反，每个单元格具有1 y(独立)和6 x(因变量)变量值.这表明您需要一个全局模型.为此，您可以执行以下操作:

Instead you have 1 y (independent) and 6 x (dependent) variable values for each cell. This suggests you want a global model. For that, you can do something like this:

library(raster)
# example data
r <- raster(nrow=10, ncol=10)
set.seed(0) 
s <- stack(lapply(1:7, function(i) setValues(r, rnorm(ncell(r), i, 3))))  
x <- values(s)

# model
m <- lm(layer.1 ~ ., data=data.frame(x))

# prediction
p <- predict(s, m)

这需要将所有值加载到内存中.如果您不能这样做，则可以进行大量常规采样.参见 sampleRegular

This requires loading all values into memory. If you can not do that, you could take a large regular sample. See sampleRegular

并说明为什么您的方法行不通:

And to show why your approach does not work:

testFun=function(x1){
    lm(x1[1] ~ x1[2]+x1[3]+x1[4]+x1[5]+x1[6]+x1[7])
}

# first cell
v <- s[1]
v
#      layer.1  layer.2   layer.3  layer.4  layer.5  layer.6  layer.7
#[1,] 4.788863 4.345578 -0.137153 3.626695 3.829971 4.120895 1.936597

m <- testFun(v)
m
#Call:
#lm(formula = x1[1] ~ x1[2] + x1[3] + x1[4] + x1[5] + x1[6] + x1[7])

#Coefficients:
#(Intercept)        x1[2]        x1[3]        x1[4]        x1[5]        x1[6]        x1[7]  
#      4.789           NA           NA           NA           NA           NA           NA  


summary(m)$r.squared
# 0

即使我没有收到您报告的错误消息(但所有R ^ 2值均为零).

Even though I do not get the error message you report (but all R^2 values are zero).

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