stability.par {agricolae} | R Documentation |
This procedure calculates the stability variations as well as the statistics of selection for the yield and the stability. The averages of the genotype through the different environment repetitions are required for the calculations. The mean square error must be calculated from the joint variance analysis.
stability.par(data,rep,MSerror,alpha=0.1,main=NULL,cova = FALSE,name.cov=NULL,file.cov=0)
data |
matrix of averages, by rows the genotypes and columns the environment |
rep |
Number of repetitions |
MSerror |
Mean Square Error |
alpha |
Label significant |
main |
Title |
cova |
Covariable |
name.cov |
Name covariable |
file.cov |
Data covariable |
data |
Numeric |
rep |
Constant numeric |
MSerror |
Constant numeric |
alpha |
Constant numeric |
main |
Text |
cova |
FALSE or TRUE |
name.cov |
Text |
file.cov |
Vector numeric |
Felipe de Mendiburu
Kang, M. S. 1993. Simultaneous selection for yield and stability: Consequences for growers. Agron. J. 85:754-757
library(agricolae) # example 1 # Experimental data, # replication rep= 4 # Mean square error, MSerror = 1.8 # 12 environment # 13 genotype = 1,2,3,.., 13 # yield averages of 13 genotypes in localities V1 <- c(10.2, 8.8, 8.8, 9.3, 9.6, 7.2, 8.4, 9.6, 7.9, 10, 9.3, 8.0, 10.1) V2 <- c(7, 7.8, 7.0, 6.9, 7, 8.3, 7.4, 6.5, 6.8, 7.9, 7.3, 6.8, 8.1) V3 <- c(5.3, 4.4, 5.3, 4.4, 5.6, 4.6, 6.2, 6.0, 6.5, 5.3, 5.7, 4.4, 4.2) V4 <- c(7.8, 5.9, 7.3, 5.9, 7.8, 6.3, 7.9, 7.5, 7.6, 5.4, 5.6, 7.8, 6.5) V5 <- c(9, 9.2, 8.8, 10.6, 8.3, 9.3, 9.6, 8.8, 7.9, 9.1, 7.7, 9.5, 9.4) V6 <- c(6.9, 7.7, 7.9, 7.9, 7, 8.9, 9.4, 7.9, 6.5, 7.2, 5.4, 6.2, 7.2) V7 <- c(4.9, 2.5, 3.4, 2.5, 3,2.5, 3.6, 5.6,3.8, 3.9, 3.0, 3.0, 2.5) V8 <- c(6.4, 6.4, 8.1, 7.2, 7.5, 6.6, 7.7, 7.6, 7.8, 7.5, 6.0, 7.2, 6.8) V9 <- c(8.4, 6.1, 6.8, 6.1, 8.2, 6.9, 6.9, 9.1, 9.2, 7.7, 6.7, 7.8, 6.5) V10 <-c(8.7, 9.4, 8.8, 7.9, 7.8, 7.8, 11.4, 9.9, 8.6, 8.5, 8.0, 8.3, 9.1) V11 <-c(5.4, 5.2, 5.6, 4.6, 4.8, 5.7, 6.6, 6.8, 5.2, 4.8, 4.9, 5.4, 4.5) V12 <-c(8.6, 8.0, 9.2, 8.1, 8.3, 8.9, 8.6, 9.6, 9.5, 7.7, 7.6, 8.3, 6.6) data<-data.frame(V1,V2,V3,V4,V5,V6,V7,V8,V9,V10,V11,V12) rownames(data)<-LETTERS[1:13] stability.par(data, rep=4, MSerror=1.8, alpha=0.1, main="Genotype") #example 2 covariable. precipitation precipitation<- c(1000,1100,1200,1300,1400,1500,1600,1700,1800,1900,2000,2100) stability.par(data, rep=4, MSerror=1.8, alpha=0.1, main="Genotype", cova=TRUE, name.cov="Precipitation", file.cov=precipitation)