How to cut yearly time-based data into 36 parts with R? - r

Question

I have a df like the following with 30 years until 2015. I want to cut every month into three data like 1-10, 11-20, and 21-31 and average all ten (less then ten) data. Thus, each month has three data. How can I do it?
1993-01-29 28.92189
1993-02-01 29.12760
1993-02-02 29.18927
1993-02-03 29.49786
1993-02-04 29.62128
1993-02-05 29.60068
1993-02-08 29.60068
1993-02-09 29.39498
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2015-08-18 209.92999
2015-08-19 208.28000
2015-08-20 204.01000
2015-08-21 197.63001
2015-08-24 189.55000
2015-08-25 187.23000
2015-08-26 194.67999
2015-08-27 199.16000
2015-08-28 199.24000

Answer 1

tryCatch is for eliminate data start date problem. I will provide more info when i have time.
library(xts)
dates<-seq(as.Date("1993-01-29"),as.Date("2015-08-25"),"days")
sample<-rnorm(length(dates))
tmpxts<-split.xts(xts(x = sample,order.by = dates),f = "months")
mxts<-lapply(tmpxts,function(x) {
tmp<-data.frame(val=tryCatch(c(mean(x[1:10]),mean(x[11:20]),mean(x[21:length(x)])),
error=function(e) matrix(mean(x),1)))
row.names(tmp)<-tryCatch(index(x[c(1,11,21)]),error=function(e) index(x[1]))
tmp
})
do.call(rbind,mxts)

Answer 2

This is a base solution that builds cuts from an increasing sequence the cycles through years, months and your cuts at 1st, 11th and 21th of the month, The default for the base cut functions are to include the breaks as the "right-side" of intervals, but your specification required cuts at 1,11,and 21 (to leave 10, and 20 in the lower intervals) so I used right=TRUE:
tapply(dat$V2, cut.Date(dat$V1,
breaks=as.Date(
apply( expand.grid( c(1,11,21), 1:12, 1993:2015), 1,
function( x) paste(rev(x), collapse="-")) ), right=TRUE), FUN=mean)
1993-01-01 1993-01-11 1993-01-21 1993-02-01 1993-02-11 1993-02-21 1993-03-01
NA NA 29.02475 29.48412 NA NA NA
snipped many empty intervals
And the bottom of results included:
2015-07-21 2015-08-01 2015-08-11 2015-08-21 2015-09-01 2015-09-11 2015-09-21
NA NA 204.96250 193.97200 NA NA NA
2015-10-01 2015-10-11 2015-10-21 2015-11-01 2015-11-11 2015-11-21 2015-12-01
NA NA NA NA NA NA NA
2015-12-11
NA

Answer 3

The code below cuts each month separately into thirds, based on the number of days in each month.
library(dplyr)
library(lubridate)
library(ggplot2)
# Fake data
df = data.frame(date=seq.Date(as.Date("2013-01-01"),
as.Date("2013-03-31"), by="day"))
set.seed(394)
df$value = rnorm(nrow(df), sqrt(1:nrow(df)), 2)
# Cut months into thirds
df = df %>%
# Create a new column to group by Year-Month
mutate(yr_mon = paste0(year(date) , "_", month(date, label=TRUE, abbr=TRUE))) %>%
group_by(yr_mon) %>%
# Cut each month into thirds
mutate(cutMonth = cut(day(date),
breaks=c(0, round(1/3*n()), round(2/3*n()), n()),
labels=c("1st third","2nd third","3rd third")),
# Add yr_mon to cutMonth so that we have a unique group label for
# each third of each month
cutMonth = paste0(yr_mon, "\n", cutMonth)) %>%
ungroup() %>%
# Turn cutMonth into a factor with correct date ordering
mutate(cutMonth = factor(cutMonth, levels=unique(cutMonth)))
And here is the result:
# Show number of observations in each group
as.data.frame(table(df$cutMonth))
Var1 Freq
1 2013_Jan\n1st third 10
2 2013_Jan\n2nd third 11
3 2013_Jan\n3rd third 10
4 2013_Feb\n1st third 9
5 2013_Feb\n2nd third 10
6 2013_Feb\n3rd third 9
7 2013_Mar\n1st third 10
8 2013_Mar\n2nd third 11
9 2013_Mar\n3rd third 10
# Plot means by group (just to visualize the result of the date grouping operations)
ggplot(df, aes(cutMonth, value)) +
stat_summary(fun.y=mean, geom='point', size=4, colour="red") +
coord_cartesian(ylim=c(-0.2,10.2)) +
theme(axis.text.x = element_text(size=14))