Sorting data frame based on month-year time format

Question

I'm struggling with something very basic: sorting a data frame based on a time format (month-year, or, “%B-%y” in this case). My goal is to calculate various monthly statistics, starting with sum.

The part of relevant part of the data frame looks like this * (This goes well and in accordance of my goal. I'm including it here to show where the problem could originate from)*:

> tmp09
   Instrument AccountValue   monthYear   ExitTime
1         JPM         6997    april-07 2007-04-10
2         JPM         7261      mei-07 2007-05-29
3         JPM         7545     juli-07 2007-07-18
4         JPM         7614     juli-07 2007-07-19
5         JPM         7897 augustus-07 2007-08-22
10        JPM         7423 november-07 2007-11-02
11        KFT         6992      mei-07 2007-05-14
12        KFT         6944      mei-07 2007-05-21
13        KFT         7069     juli-07 2007-07-09
14        KFT         6919     juli-07 2007-07-16
# Order on the exit time, which corresponds with 'monthYear'
> tmp09.sorted <- tmp09[order(tmp09$ExitTime),]
> tmp09.sorted
   Instrument AccountValue   monthYear   ExitTime
1         JPM         6997    april-07 2007-04-10
11        KFT         6992      mei-07 2007-05-14
12        KFT         6944      mei-07 2007-05-21
2         JPM         7261      mei-07 2007-05-29
13        KFT         7069     juli-07 2007-07-09
14        KFT         6919     juli-07 2007-07-16
3         JPM         7545     juli-07 2007-07-18
4         JPM         7614     juli-07 2007-07-19
5         JPM         7897 augustus-07 2007-08-22
10        JPM         7423 november-07 2007-11-02

So far, so good, and sorting based on ExitTime works. The trouble starts when I try to calculate the totals per month, followed by an attempt to sort this output:

# Calculate the total results per month
> Tmp09Totals <- tapply(tmp09.sorted$AccountValue, tmp09.sorted$monthYear, sum)
> Tmp09Totals <- data.frame(Tmp09Totals)
> Tmp09Totals
            Tmp09Totals
april-07           6997
augustus-07        7897
juli-07           29147
mei-07            21197
november-07        7423

How can I sort this output in a chronological Way?

I've already tried (besides various attempts to convert the monthYear to another date format): order, sort, sort.list, sort_df, reshape, and calculating the sum based on tapply, lapply, sapply, aggregate. And even rewriting the rownames (by giving them a number from 1 to length (tmp09.sorted2$AccountValue) didn't work. I also tried to give each month-year a different ID based on what I've learned in another question, but R also experienced difficulties in discriminating between the various month-year values.

The correct order of this output would be april-07,mei-07,juli-07,augustus07, november-07:

apr-07  6997
mei-07  21197
jul-07  29147
aug-07  7897
nov-07  7423

Answer 1

It would be easier to have separate Month and Year factors, in the correct order, and use tapply on the union of both variables, e.g.:

## The Month factor
tmp09 <- within(tmp09,
                Month <- droplevels(factor(strftime(ExitTime, format = "%B"),
                                                    levels = month.name)))
## for @Jura25's locale, we can't use the in built English constant
## instead, we can use this solution, from ?month.name:
## format(ISOdate(2000, 1:12, 1), "%B"))
tmp09 <- within(tmp09,
                Month <- droplevels(factor(strftime(ExitTime, format = "%B"),
                                                    levels = format(ISOdate(2000, 1:12, 1), "%B"))))
##
## And the Year factor
tmp09 <- within(tmp09, Year <- factor(strftime(ExitTime, format = "%Y")))

Which gives us (in my locale):

> head(tmp09)
   Instrument AccountValue   monthYear   ExitTime    Month Year
1         JPM         6997    april-07 2007-04-10    April 2007
2         JPM         7261      mei-07 2007-05-29      May 2007
3         JPM         7545     juli-07 2007-07-18     July 2007
4         JPM         7614     juli-07 2007-07-19     July 2007
5         JPM         7897 augustus-07 2007-08-22   August 2007
10        JPM         7423 november-07 2007-11-02 November 2007

Then use tapply with both factors:

> with(tmp09, tapply(AccountValue, list(Month, Year), sum))
          2007
April     6997
May      21197
July     29147
August    7897
November  7423

or via aggregate:

> with(tmp09, aggregate(AccountValue, list(Month = Month, Year = Year), sum))
     Month Year     x
1    April 2007  6997
2      May 2007 21197
3     July 2007 29147
4   August 2007  7897
5 November 2007  7423

Answer 2

Try using the "yearmon" class in zoo as it sorts appropriately. Below we create the sample DF data frame and then we add a YearMonth column of class "yearmon" . Finally we perform our aggregation. The actual processing is just the last two lines (the other part is just to create the sample data frame).

Lines <-   "Instrument AccountValue   monthYear   ExitTime
JPM         6997    april-07 2007-04-10
JPM         7261      mei-07 2007-05-29
JPM         7545     juli-07 2007-07-18
JPM         7614     juli-07 2007-07-19
JPM         7897 augustus-07 2007-08-22
JPM         7423 november-07 2007-11-02
KFT         6992      mei-07 2007-05-14
KFT         6944      mei-07 2007-05-21
KFT         7069     juli-07 2007-07-09
KFT         6919     juli-07 2007-07-16"
library(zoo)
DF <- read.table(textConnection(Lines), header = TRUE)

DF$YearMonth <- as.yearmon(DF$ExitTime)
aggregate(AccountValue ~ YearMonth + Instrument, DF, sum)

This gives the following:

> aggregate(AccountValue ~ YearMonth + Instrument, DF, sum)
  YearMonth Instrument AccountValue
1  Apr 2007        JPM         6997
2  May 2007        JPM         7261
3  Jul 2007        JPM        15159
4  Aug 2007        JPM         7897
5  Nov 2007        JPM         7423
6  May 2007        KFT        13936
7  Jul 2007        KFT        13988

A slightly different approach and output uses read.zoo directly. It produces one column per instrument and one row per year/month. We read in the columns assigning them appropriate classes using "NULL" for the monthYear column since we won't use that one. We also specify that the time index is the 3rd column of the remaining columns and that we want the input split into columns by the 1st column. FUN=as.yearmon indicates that we want the time index to be converted from "Date" class to "yearmon" class and we aggregate everything using sum.

z <- read.zoo(textConnection(Lines),  header = TRUE, index = 3, 
     split = 1, colClasses = c("character", "numeric", "NULL", "Date"),
     FUN = as.yearmon, aggregate = sum)

The resulting zoo object looks like this:

> z
           JPM   KFT
Apr 2007  6997    NA
May 2007  7261 13936
Jul 2007 15159 13988
Aug 2007  7897    NA
Nov 2007  7423    NA

We may prefer to keep it as a zoo object to take advantage of other functionality in zoo or we can convert it to a data frame like this: data.frame(Time = time(z), coredata(z)) which makes the time a separate column or as.data.frame(z) which uses row names for the time. fortify.zoo()z) also works.