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バッチ正規分布テストを実行しようとしています。

私のデータは次のようになります。

"Date","Department","Discipline","Employee ID","SumOfBillable Hrs"
"10/09/2012","D","B",50084.00,8.00
"10/09/2012","D","C",51870.00,10.00
"10/09/2012","D","E",50216.00,10.00
"10/09/2012","D","E",53422.00,9.00
"10/09/2012","D","E",53765.00,10.00
"14/01/2013","E","Y",53146.00,9.00
"14/01/2013","E","Y",53202.00,9.00
"14/01/2013","E","Y",54470.00,9.00
"14/01/2013","SITE","0",54525.00,9.00
"14/02/2013","D","C",51870.00,10.00
"14/02/2013","D","E",50029.00,8.50
"14/02/2013","D","E",50216.00,9.00
"14/02/2013","D","E",53422.00,4.00

各 の下の時間の分布を確認したいEmployee_ID。

これを行うバッチ方法はありますか？私は80歳以上IDsです。したがって、それぞれを個別に取得IDし、それを説明する統計をプロット/作成するのはかなり面倒です。

ありがとう

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1 に答える 1

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このようなものから始めることができます。何か違うものが必要な場合は、具体的に何をしたいのかについて、より多くの情報を提供する必要があります。

data <- read.table(header=T, sep=",", 
 text='"Date","Department","Discipline","Employee ID","SumOfBillable Hrs"
"10/09/2012","D","B",50084.00,8.00
"10/09/2012","D","C",51870.00,10.00
"10/09/2012","D","E",50216.00,10.00
"10/09/2012","D","E",53422.00,9.00
"10/09/2012","D","E",53765.00,10.00
"14/01/2013","E","Y",53146.00,9.00
"14/01/2013","E","Y",53202.00,9.00
"14/01/2013","E","Y",54470.00,9.00
"14/01/2013","SITE","0",54525.00,9.00
"14/02/2013","D","C",51870.00,10.00
"14/02/2013","D","E",50029.00,8.50
"14/02/2013","D","E",50216.00,9.00
"14/02/2013","D","E",53422.00,4.00')



# Means:
aggregate(SumOfBillable.Hrs ~ Employee.ID, data=data, FUN=mean)

# Standard Deviations:
aggregate(SumOfBillable.Hrs ~ Employee.ID, data=data, FUN=sd)

# Or a Shapiro normality test: (only works if you have more than 3 observations per Employee.ID
aggregate(SumOfBillable.Hrs ~ Employee.ID, data=data, FUN=shapiro.test)
于 2013-02-27T08:49:02.660 に答える