Average column in daily information at every n-th row - r

Question

I am very new on R. I have daily observations of temperature and PP for 12-year period (6574 row, 6col, some NA ). I want to calculate, for example, the average from 1st to 10thday of January-2001, then 11-20 and finally 21 to 31 and so on for every month until december for each year in the period I mentioned before.
But also I have problems because February sometimes has 28 or 29 days (leap years).
This is how i open my file is a CSV, with read.table
# READ CSV
setwd ("C:\\Users\\GVASQUEZ\\Documents\\ESTUDIO_PAMPAS\\R_sheet")
huancavelica<-read.table("huancavelica.csv",header = TRUE, sep = ",",
dec = ".", fileEncoding = "latin1", nrows = 6574 )
This is the output of my CSV file
Año Mes Dia PT101 TM102 TM103
1 1998 1 1 6.0 15.6 3.4
2 1998 1 2 8.0 14.4 3.2
3 1998 1 3 8.6 13.8 4.4
4 1998 1 4 5.6 14.6 4.6
5 1998 1 5 0.4 17.4 3.6
6 1998 1 6 3.4 17.4 4.4
7 1998 1 7 9.2 14.6 3.2
8 1998 1 8 2.2 16.8 2.8
9 1998 1 9 8.6 18.4 4.4
10 1998 1 10 6.2 15.0 3.6
. . . . . . .

Answer 1

With the data setup that you have a fairly tried and true method should work:
# add 0 in front of single digit month variable to account for 1 and 10 sorting
huancavelica$MesChar <- ifelse(nchar(huancavelica$Mes)==1,
paste0("0",huancavelica$Mes), as.character(huancavelica$Mes))
# get time of month ID
huancavelica$timeMonth <- ifelse(huancavelica$Dia < 11, 1,
ifelse(huancavelica$Dia > 20, 3, 2)
# get final ID
huancavelica$ID <- paste(huancavelica$Año, huancavelica$MesChar, huancavelica$timeMonth, sep=".")
# average stat
huancavelica$myStat <- ave(huancavelica$PT101, huancavelica$ID, FUN=mean, na.rm=T)

Answer 2

We can try
library(data.table)
setDT(df1)[, Grp := (Dia - 1)%/%10+1, by = .(Ano, Mes)
][Grp>3, Grp := 3][,lapply(.SD, mean, na.rm=TRUE), by = .(Ano, Mes, Grp)]

Answer 3

It adds a bit more complexity, but you could cut each month into thirds and get the average for each third. For example:
library(dplyr)
library(lubridate)
# Fake data
set.seed(10)
df = data.frame(date=seq(as.Date("2015-01-01"), as.Date("2015-12-31"), by="1 day"),
value=rnorm(365))
# Cut months into thirds
df = df %>%
mutate(mon_yr = paste0(month(date, label=TRUE, abbr=TRUE) , " ", year(date))) %>%
group_by(mon_yr) %>%
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")),
cutMonth = paste0(mon_yr, ", ", cutMonth)) %>%
ungroup %>%
mutate(cutMonth = factor(cutMonth, levels=unique(cutMonth)))
date value cutMonth
1 2015-01-01 0.01874617 Jan 2015, 1st third
2 2015-01-02 -0.18425254 Jan 2015, 1st third
3 2015-01-03 -1.37133055 Jan 2015, 1st third
...
363 2015-12-29 -1.3996571 Dec 2015, 3rd third
364 2015-12-30 -1.2877952 Dec 2015, 3rd third
365 2015-12-31 -0.9684155 Dec 2015, 3rd third
# Summarise to get average value for each 1/3 of a month
df.summary = df %>%
group_by(cutMonth) %>%
summarise(average.value = mean(value))
cutMonth average.value
1 Jan 2015, 1st third -0.49065685
2 Jan 2015, 2nd third 0.28178222
3 Jan 2015, 3rd third -1.03870698
4 Feb 2015, 1st third -0.45700203
5 Feb 2015, 2nd third -0.07577199
6 Feb 2015, 3rd third 0.33860882
7 Mar 2015, 1st third 0.12067388
...