R每周平均值

我有一组网格海面温度的日常值34年（12418每日文件x 4248点）并假装计算每周值 . 我在这篇文章https://stackoverflow.com/a/15102394/709777后几乎成功了 . 但是日期和周之间存在一些分歧 . 我可以't find the point and I want to be sure I'获得正确的日期来计算每周平均值 .

我使用这段R脚本来读取每日数据并构建一个大数据框，其中包含列中单个点的所有每日值（12418行/天乘4248列/温度）

# Paths
ruta_datos_diarios<-"/home/meteo/PROJECTES/VERSUS/DATA/SST/CSV/"
ruta_files<-"/home/meteo/PROJECTES/VERSUS/SCRIPTS/CLUSTER/FILES/"
ruta_eixida<-"/home/meteo/PROJECTES/VERSUS/OUTPUT/DATA/SEMANAL/"

# List of daily files
files <- list.files(path = ruta_datos_diarios, pattern = "SST-diaria-MED")

output <- matrix(ncol=4248, nrow=length(files))
fechas <- matrix(ncol=1, nrow=length(files))

for (i in 1:length(files)){
  # read data
  datos<-read.csv(paste0(ruta_datos_diarios,files[i],sep=""),header=TRUE,na.strings = "NA")
  datos<-datos[complete.cases(datos),]

  # Extract dates from daily file names
  yyyy<-substr(files[i],16,19)
  mm<-substr(files[i],20,21)
  dd<-substr(files[i],22,23)
  dates[i,]<-paste0(yyyy,"-",mm,"-",dd,sep="")

  output[i,]<-t(datos$sst)
}

datos.df<-as.data.frame(output)

# Build a dataframe with the dates  (day, week and year)
fechas<-as.data.frame(fechas)
fechas$V1<-as.Date(fechas$V1)
fechas$Week <- week(fechas$V1)
fechas$Year <- year(fechas$V1)

# Extract day of the week (Saturday = 6)
fechas$Week_Day <- as.numeric(format(fechas$V1, format='%w'))
# Adjust end-of-week date (first saturday from the original Date)
fechas$End_of_Week <- fechas$V1 + (6 - fechas$Week_Day)

# new dataframe from End_of_Week
fechas.semana<-fechas[!duplicated(fechas$End_of_Week),]
fechas.semana<-as.data.frame(fechas.semana)

colnames(fechas)<-c("Day","Week","Year","Week_Day","End_of_Week")
colnames(fechas.semana)<-c("Day","Week","Year","Week_Day","End_of_Week")

这就是我阅读数据和日期的方式 . 为了举一个简短的例子，我在这个文件temp-sst.csv中保存了数据帧的一个子集（10个变量中有1000个，包括"Day"，"Week"，"Year"，"Week_Day"，"End_of_Week"） .

sst.dat <- read.csv("temp-dat.csv",header=TRUE)

# Join dates and SST values
sst.dat <- cbind(fechas, sst.dat)

# Build new dates data frame
fechas<-as.data.frame(sst.dat$Day)
colnames(fechas)<-c("Day")
fechas$Day<-as.Date(fechas$Day)
fechas$Week <- week(fechas$Day)
fechas$Year <- year(fechas$Day)
# Extract day of the week (Saturday = 6)
fechas$Week_Day <- as.numeric(format(fechas$Day, format='%w'))
# Adjust end-of-week date (first saturday from the original Date)
fechas$End_of_Week <- fechas$Day + (6 - fechas$Week_Day)

fechas.semana<-fechas[!duplicated(fechas$End_of_Week),]
fechas.semana<-as.data.frame(fechas.semana)

colnames(fechas)<-c("Day","Week","Year","Week_Day","End_of_Week")
colnames(fechas.semana)<-c("Day","Week","Year","Week_Day","End_of_Week")

# Weekly aggregation function from the referred post
media.semanal <- function(x, column){
  a<-aggregate(x[,column]~End_of_Week+Year, FUN=mean, data=x, na.rm=TRUE)
  colnames(a)<-c("End_of_Week","Year","SSTmean")
  return(a)
}

# Matrix to be populated by weekly function
SST.mat<-matrix(nrow=nrow(fechas.semana), ncol=length(sst.dat)-5)  # 5 son las columnas de fecha

for (j in 6:length(sst.dat)){   # comienza en 6 para evitar las columnas de fecha
b<-media.semanal(sst.dat,j)
SST.mat[,j-5]<-b$SSTmean
}

但问题来了 . 来自循环的“b”数据帧有145行，而SST.mat和fechas.semana只有144行 . 我没有找到这种分歧出现的点 .

任何帮助将不胜感激，我被困在这里 . 谢谢