I have dataframe df with x,y,and monthly.year data for each x,y point.
I am trying to get the seasonal aggregate.
I need to calculate seasonal means i.e.
For winter mean of (December,January,February); for Spring mean of (March,April,May), for Summer mean of (June,July,August) and for autumn mean of (September,October,November).
The data looks similar to:
set.seed(1)
df <- data.frame(x=1:3,y=1:3, matrix(rnorm(72),nrow=3) )
names(df)[3:26] <- paste(month.abb,rep(2009:2010,each=12),sep=".")
x y Jan.2009 Feb.2009 ... Dec.2010
1 1 1 -0.6264538 1.5952808 ... 2.1726117
2 2 2 0.1836433 0.3295078 ... 0.4755095
3 3 3 -0.8356286 -0.8204684 ... -0.7099464
I could not think of going any further except melting the data and making new data frame as
ddt.m<-melt(df,id=c("x","y"))
I want result like x,y,mean of season of each year. Please suggest me how I may be able to do that.
Here is one possible approach:
melt the data, as you suggested...
...and, use colsplit to split up the "variable" into the "Mon" and "Year" columns.
library(reshape2)
ddt.m <- melt(df, id = c("x", "y"))
ddt.m <- cbind(ddt.m, colsplit(ddt.m$variable, "\\.", c("Mon", "Year")))
Use factor and levels to get your seasons
(which I've left in the "Mon" column. Oops.)
ddt.m$Mon <- factor(ddt.m$Mon)
levels(ddt.m$Mon) <- list(Winter = month.abb[c(12, 1, 2)],
Spring = month.abb[c(3:5)],
Summer = month.abb[c(6:8)],
Autumn = month.abb[c(9:11)])
head(ddt.m)
# x y variable value Mon Year
# 1 1214842 991964.4 Jan.2009 -1.332933 Winter 2009
# 2 1220442 991964.4 Jan.2009 -1.345808 Winter 2009
# 3 1226042 991964.4 Jan.2009 -1.314435 Winter 2009
# 4 1231642 991964.4 Jan.2009 -1.236600 Winter 2009
# 5 1237242 991964.4 Jan.2009 -1.261989 Winter 2009
# 6 1242842 991964.4 Jan.2009 -1.306614 Winter 2009
Use dcast to aggregate the data
dfSeasonMean <- dcast(ddt.m, x + y ~ Mon + Year,
value.var="value", fun.aggregate=mean)
head(dfSeasonMean)
# x y Winter_2009 Winter_2010 Spring_2009 Spring_2010 Summer_2009
# 1 1214842 991964.4 -1.439480 -1.006512 -0.02509008 0.2823048 1.392440
# 2 1220442 964154.4 -1.457407 -1.039266 -0.04337596 0.2315217 1.422541
# 3 1220442 973424.4 -1.456991 -1.035115 -0.04117584 0.2423561 1.414473
# 4 1220442 982694.4 -1.456479 -1.029627 -0.03799926 0.2544062 1.405813
# 5 1220442 991964.4 -1.456234 -1.027081 -0.03815661 0.2610397 1.400743
# 6 1226042 945614.4 -1.463465 -1.031665 -0.04288670 0.2236609 1.434002
# Summer_2010 Autumn_2009 Autumn_2010
# 1 1.256840 0.06469363 -0.03823892
# 2 1.263593 0.04521096 -0.04485553
# 3 1.258328 0.04860321 -0.04477636
# 4 1.252779 0.05337575 -0.04729598
# 5 1.247251 0.05742809 -0.05152524
# 6 1.272742 0.04692731 -0.04915314
Convert each year/month name to a zoo yearmon object and add 1/12 to push it to the next month. After adding one month the 4 seasons correspond to calendar quarters so convert to a zoo yearqtr object. Then regress the data against the year quarters with no intercept and the coefficients will be the desired means:
library(zoo)
df0 <- df[-(1:2)]
Y <- format(as.yearqtr(as.yearmon(names(df0), "%b.%Y") + 1/12)) # "2009 Q1" "2009 Q1" ...
cbind(df[1:2], t(coef(lm(t(df0) ~ Y + 0))))
giving the following. Note that the seasons are labelled by the calendar quarter in which the season ends:
x y Y2009 Q1 Y2009 Q2 Y2009 Q3 Y2009 Q4 Y2010 Q1 Y2010 Q2 Y2010 Q3 Y2010 Q4 Y2011 Q1
1 1 1 0.4844135 -0.1464000 0.51947463 0.1692510 0.1050269 -0.04095933 -0.07437911 0.2082204 2.1726117
2 2 2 0.2565755 0.0118020 0.21742535 -0.2123555 -0.5336322 0.60078430 0.96374641 0.2276805 0.4755095
3 3 3 -0.8280485 0.6968518 -0.04217937 0.2166059 0.1438897 0.15929437 -0.54387973 0.3439283 -0.7099464
For each period of interest, you could follow an approach like this:
spring_months <- paste0(c("Mar","Apr","May"),".2009"
spring_mean <- rowMeans(df[,spring_months], na.rm=T)
winter_months <- c("Dec.2009","Jan.2010","Feb.2010")
winter_mean <- rowMeans(df[,winter_months], na.rm=T)
Then you can just take those variables and make a data frame with df$x and df$y:
data.frame(x=df$x, y=df$y, spring_mean = spring_mean, winter_mean = winter_mean)
来源：https://stackoverflow.com/questions/22124315/seasonal-aggregate-of-monthly-data