[英]R weekly averaging
我有一组34年的网格化海面温度的每日值(12418每日文件x 4248点),并假装计算每周值。 我几乎成功地遵循了这篇文章https://stackoverflow.com/a/15102394/709777 。 但是日期和星期之间有些分歧。 我找不到要点,我想确保我能找到正确的日期来计算每周平均值。
我使用这段R脚本读取每日数据并构建一个大数据框,该数据框包含列中单个点的所有每日值(12418行/天乘以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$End_of_Week),]
fechas.semana<-as.data.frame(fechas.semana)
colnames(fechas)<-c("Day","Week","Year","Week_Day","End_of_Week")
colnames(fechas.semana)<-c("Day","Week","Year","Week_Day","End_of_Week")
这就是我读取数据和日期的方式。 为了简短起见,我已在此文件temp-sst.csv中保存了数据帧的子集(1000个观察点,共10个变量,包括“日”,“周”,“年”,“周日”,“周日结束” )。
sst.dat <- read.csv("temp-dat.csv",header=TRUE)
# Join dates and SST values
sst.dat <- cbind(fechas, sst.dat)
# Build new dates data frame
fechas<-as.data.frame(sst.dat$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$SSTmean
}
但是问题来了。 循环中的“ b”数据帧有145行,而SST.mat和fechas.semana只有144行。我还没有发现这种分歧的出处。
任何帮助将不胜感激,我被困在这里。 谢谢
您有一个重复的b$End_of_Week
。
首先,我注意到所设置的成员资格没有区别:
setdiff(as.character(b$End_of_Week),as.character(fechas.semana$End_of_Week))
人物(0)
然后我意识到那一定是因为重复,并像这样确认了它:
table(table(as.character(b$End_of_Week))>1)
143 1 FALSE TRUE
看着桌上的骗局是1983-01-01
。
看来根本原因在于,您可以通过汇总End_of_Week + Year
,其中Year
是不必要的,因为End_of_Week
有当年一样好,如果你只通过汇总End_of_Week
你144,而不是145。
# Weekly aggregation function from the referred post
media.semanal <- function(x, column){
a<-aggregate(x[,column]~End_of_Week, FUN=mean, data=x, na.rm=TRUE)
colnames(a)<-c("End_of_Week","SSTmean")
return(a)
}
# Matrix to be populated by weekly function
SST.mat<-matrix(nrow=nrow(fechas.semana), ncol=length(sst.dat)-5) # 5 son las columnas de fecha
for (j in 6:length(sst.dat)){ # comienza en 6 para evitar las columnas de fecha
b<-media.semanal(sst.dat,j)
SST.mat[,j-5]<-b$SSTmean
}
dim(b)
144 2
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