通过两个参数的功能

Question

I have some data going from 1971 until 2099. Organize as follow:

YEAR;MONTH;DAY;RES1;RES2
1971;1;1;1206.1;627
1971;1;2;1303.4;654.3
1971;1;3;1248.9;662
1971;1;4;1188.8;666.8
1971;1;5;1055.2;667.8
1971;1;6;987.1;663.3
1971;1;7;939.2;655.1
1971;1;8;883.2;644.4
1971;1;9;844.1;632.6
1971;1;10;813.2;620.7
1971;1;11;786.4;609
1971;1;12;765.9;598.2
1971;1;13;990.2;650.1
1971;1;14;1374.4;698.9
1971;1;15;1335.9;718
1971;1;16;1193.2;721.6
1971;1;17;1043.5;719.5
1971;1;18;995.7;710.9
1971;1;19;937.2;696.2
1971;1;20;877;678.2
1971;1;21;880.2;676.5
1971;1;22;1227.2;715.3
1971;1;23;1275.7;731.1
1971;1;24;1029.2;730.7
1971;1;25;934.2;724.9
1971;1;26;923.6;714.8
1971;1;27;887.6;700.1
1971;1;28;840.2;682.6
1971;1;29;791.7;664.3
1971;1;30;746.7;646.4
1971;1;31;706.8;629.3

Using this data I need to calculate several average values such as the monthly average value. In order to calculate the monthly average value I used the summaryBy function of the DoBy package. The following code provide me the Monthly average value:

indREF=which(data$YEAR > 1974 & data$YEAR < 2005)
indEND=which(data$YEAR > 2069)
dataREF=data[indREF,]
dataEND=data[indEND,]
MoyRef=c(summaryBy(dataREF[,"MONTH"]~MONTH, dataREF, FUN = function(x) {return(mean(x,na.rm=TRUE))})[,1])
MoyEnd=c(summaryBy(dataEND[,"MONTH"]~MONTH, dataEND, FUN = function(x) {return(mean(x,na.rm=TRUE))})[,1])

for ( i in 4:dim(data)[2])
{
  MoyRef=cbind(MoyRef,summaryBy(dataREF[,i]~MONTH, dataREF, FUN = function(x) {return(mean(x,na.rm=TRUE))})[,2])
  MoyEnd=cbind(MoyEnd,summaryBy(dataEND[,i]~MONTH, dataEND, FUN = function(x) {return(mean(x,na.rm=TRUE))})[,2])
}

But now, considering the fact that the data is going from 1971 until 2099 using a daily time step, I would like to calculate the daily average value of the data such as the output looks like the following:

MONTH;DAY;AVERAGE_RES1;AVERAGE_RES2
01;01;VALUE1;VALUE2
01;02;VALUE3;VALUE4
...
12;31;VALUEx;VALUEx

Does anyone has any idea about how to achieve that?

Answer 1

Unfortunately, sample data is not suitable for testing, as it contains only january and only one year, so not much to compute means. However, this should do the job:

aggregate(data[c("RES1", "RES2")], by = list(data$MONTH, data$DAY), FUN = "mean")

Answer 2

I think you should try this with dplyr package like this

library(dplyr)
df %>% group_by(MONTH,DAY) %>% summarise_each_(funs(mean),c("RES1","RES2"))

Answer 3

A dplyr answer has been posted and probably a data.table answer will follow soon. I still stand by my "R without packages" answer using aggregate() . Even though dplyr and data.table obviously have their justification, I like the idea of sqldf: You learn SQL syntax once and then can use it for the rest of your life, whilst other languages and packages come and go, SQL, like basic R, are here to stay. Thus:

library(sqldf)
sqldf("SELECT DAY, MONTH, AVG(RES1), AVG(RES2) FROM data GROUP BY MONTH, DAY")