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我们已经知道,与RCB或RIB设计相比,空间分析可以减少试验误差并提高农作物品种评估的准确性。在林业上,空间分析可明显提高种源、家系、亲本和无性系遗传效应估算的准确性。
在前面的博文(http://blog.sciencenet.cn/blog-1114360-735785.html)中,已经演示了空间分析模型,但那模型仅适用于规则的空间模型,而在实际的试验中,往往会遇见不规则的情况,比方缺株,或者随机抽样,或者采用GPS定位数据,等等。在后者的条件下,就属于不规则的空间分析。
分析代码如下:
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 | library(asreml) df1<- asreml.read.table(file='zhao.sp.csv', header=T, sep=',') df2<- asreml.read.table(file='zhao.sp2.csv', header=T, sep=',') # regular sp model df3b.asr<-asreml(dbh13~1,random=~ped(Tree)+units,rcov=~ar1(Row):ar1(Col), ginverse=list(Tree=df.ped.inv), data=df1) # irregular sp model df5.asr<-asreml(dbh13~1,random=~ped(Tree)+units,rcov=~aexp(Row,Col),maxit=50, ginverse=list(Tree=df.ped.inv), data=df2) df5a.asr<-asreml(dbh13~1,random=~Prov+units,rcov=~aexp(Row,Col),maxit=50, data=df2) summary(df3b.asr)$varcomp plot(variogram(df3b.asr),col="blue",main="DBH13 for 3b") |
运行结果如下:
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 | # regular sp model > summary(df3b.asr)$varcomp gamma component std.error z.ratio constraint ped(Tree)!ped 0.5024994 1.8775717 0.84901509 2.211470 Positive units!units.var 1.0590226 3.9570016 0.69795773 5.669400 Positive R!variance 1.0000000 3.7364656 0.93225456 4.007989 Positive R!Row.cor 0.8673566 0.8673566 0.04249064 20.412887 Unconstrained R!Col.cor 0.9068556 0.9068556 0.02887082 31.410798 Unconstrained # irregular sp model > summary(df5.asr)$varcomp gamma component std.error z.ratio constraint ped(Tree)!ped 0.5026788 1.8799947 0.84912301 2.214043 Positive units!units.var 1.0595114 3.9625216 0.69802763 5.676740 Positive R!variance 1.0000000 3.7399520 0.93463257 4.001521 Positive R!Row.pow 0.8671931 0.8671931 0.04242899 20.438693 Unconstrained R!Col.pow 0.9068278 0.9068278 0.02886393 31.417338 Unconstrained > summary(df5a.asr)$varcomp gamma component std.error z.ratio constraint Prov!Prov.var 0.1256704 0.4700073 0.21228122 2.214078 Positive units!units.var 1.4365297 5.3726226 0.27504822 19.533384 Positive R!variance 1.0000000 3.7400011 0.93466659 4.001428 Positive R!Row.pow 0.8671906 0.8671906 0.04242805 20.439086 Unconstrained R!Col.pow 0.9068274 0.9068274 0.02886387 31.417389 Unconstrained |
本例中的数据,df1和df2差异不大,df2仅仅是有几个缺株。从运行的结果来看,两者的结果差异不大。但是方差残差图的样式差异比较大。
图1 规则的空间分析之残差方差图
图2 不规则的空间分析之残差方差图
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