stability.par: Stability analysis. SHUKLA'S STABILITY VARIANCE AND KANG'S

Description

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.

Usage

stability.par(data,rep,MSerror,alpha=0.1,main=NULL,cova = FALSE,name.cov=NULL,file.cov=0)

Arguments

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

Value

dataNumeric
repConstant numeric
MSerrorConstant numeric
alphaConstant numeric
mainText
covaFALSE or TRUE
name.covText
file.covVector numeric

References

Kang, M. S. 1993. Simultaneous selection for yield and stability: Consequences for growers. Agron. J. 85:754-757

Examples

Run this code

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)