5th of month working day

Since I do not understand Gabor's solution I add my suggestion to
solve the problem ...

 > require(timeSeries)
 >
 > # Compose for 2006 a calendar with the first 5 days in each month:
 > years = rep(2006, times = 60)
 > months = rep(1:12, each = 5)
 > days = rep(1:5, times = 12)
 > tD = timeCalendar(years, months, days)
 > tD
GMT
 [1] [2006-01-01] [2006-01-02] [2006-01-03] [2006-01-04] [2006-01-05]
 [6] [2006-02-01] [2006-02-02] [2006-02-03] [2006-02-04] [2006-02-05]
[11] [2006-03-01] [2006-03-02] [2006-03-03] [2006-03-04] [2006-03-05]
[16] [2006-04-01] [2006-04-02] [2006-04-03] [2006-04-04] [2006-04-05]
[21] [2006-05-01] [2006-05-02] [2006-05-03] [2006-05-04] [2006-05-05]
[26] [2006-06-01] [2006-06-02] [2006-06-03] [2006-06-04] [2006-06-05]
[31] [2006-07-01] [2006-07-02] [2006-07-03] [2006-07-04] [2006-07-05]
[36] [2006-08-01] [2006-08-02] [2006-08-03] [2006-08-04] [2006-08-05]
[41] [2006-09-01] [2006-09-02] [2006-09-03] [2006-09-04] [2006-09-05]
[46] [2006-10-01] [2006-10-02] [2006-10-03] [2006-10-04] [2006-10-05]
[51] [2006-11-01] [2006-11-02] [2006-11-03] [2006-11-04] [2006-11-05]
[56] [2006-12-01] [2006-12-02] [2006-12-03] [2006-12-04] [2006-12-05]
 >
 > # Then extract the Business days (not weekdays!) according
 > #   to a given holiday Calendar, here I used the NYSE holiday 
calendar for 2006
 > #   Note with Rmetrics you can create your own business calendars!
 > tM = matrix(as.integer(isBizday(tD, holidayNYSE(2006))), byrow = 
TRUE, ncol = 5)
 > rownames(tM) = paste(200600+1:12)
 > colnames(tM) = paste(1:5)
 > tM
       1 2 3 4 5
200601 0 0 1 1 1
200602 1 1 1 0 0
200603 1 1 1 0 0
200604 0 0 1 1 1
200605 1 1 1 1 1
200606 1 1 0 0 1
200607 0 0 1 0 1
200608 1 1 1 1 0
200609 1 0 0 0 1
200610 0 1 1 1 1
200611 1 1 1 0 0
200612 1 0 0 1 1
 >
 > # Then isolate the 5th day of each month if this was a business day
 > #     otherwise the most recent business day before the 5th working
 > #     day for each month - this is what you want, or?
 > # Take care, there may be again holidays in between previous working
 > #     days!! Here they are handled properly.
 > tW = t(apply(tM, 1, cumsum))[,5:1]
 > tW
       5 4 3 2 1
200601 3 2 1 0 0
200602 3 3 3 2 1
200603 3 3 3 2 1
200604 3 2 1 0 0
200605 5 4 3 2 1
200606 3 2 2 2 1
200607 2 1 1 0 0
200608 4 4 3 2 1
200609 2 1 1 1 1
200610 4 3 2 1 0
200611 3 3 3 2 1
200612 3 2 1 1 1
 > tIndex = which(t(tW) == 1)
 >
 >
 > # After having the Index, you can get the timeDate objects for 2006
 > tD[tIndex]
GMT
 [1] [2006-01-03] [2006-02-05] [2006-03-05] [2006-04-03] [2006-05-05]
 [6] [2006-06-05] [2006-07-02] [2006-07-03] [2006-08-05] [2006-09-02]
[11] [2006-09-03] [2006-09-04] [2006-09-05] [2006-10-04] [2006-11-05]
[16] [2006-12-03] [2006-12-04] [2006-12-05]
 >
 > # and finally index your time series with the timeDate objects.
 > # Isn't it powerful to use timeDate and timeSeries objects?

Exercise: write a small function to extract the n-th business day for
each month of a timeDate calendar object given a specific holiday Calendar

enjoy Rmetrics!

Diethelm 

PS: I found this example really nice to show what timeDate and timeSeries
methods can do for you, I will add this example to the FAQ's in the next
edition of our timeSeries FAQ e-book: http://www.rmetrics.org/node/8


-----------------------

On Mon, Jan 18, 2010 at 9:54 AM, Research <risk2009 at ath.forthnet.gr> wrote:

Hello,

I have a daily data zoo object with  prices such as:

05/04/2006      1311.56
06/04/2006      1309.04
07/04/2006      1295.5
10/04/2006      1296.6
11/04/2006      1286.57
12/04/2006      1288.12
13/04/2006      1289.12
14/04/2006      1289.12
17/04/2006      1285.33
18/04/2006      1307.65
19/04/2006      1309.93
20/04/2006      1311.46
21/04/2006      1311.28
24/04/2006      1308.11
25/04/2006      1301.74
26/04/2006      1305.41
27/04/2006      1309.72
28/04/2006      1310.61
01/05/2006      1305.19
02/05/2006      1313.21
03/05/2006      1307.85
04/05/2006      1312.25
05/05/2006      1325.76



How can I isolate the 5th day of each month (if this was a
working/trading day) otherwise the most recent (before the 5th) working
day for each month?

Your sample data always has the 5th of the month filled in but
assuming that that is not the case for the real data, merge your
series with a zero width series having every date and use na.locf to
move values up into subsequent NAs.  Then just pick off the 5th of
each month.

Lines <- "05/04/2006      1311.56
06/04/2006      1309.04
07/04/2006      1295.5
10/04/2006      1296.6
11/04/2006      1286.57
12/04/2006      1288.12
13/04/2006      1289.12
14/04/2006      1289.12
17/04/2006      1285.33
18/04/2006      1307.65
19/04/2006      1309.93
20/04/2006      1311.46
21/04/2006      1311.28
24/04/2006      1308.11
25/04/2006      1301.74
26/04/2006      1305.41
27/04/2006      1309.72
28/04/2006      1310.61
01/05/2006      1305.19
02/05/2006      1313.21
03/05/2006      1307.85
04/05/2006      1312.25
05/05/2006      1325.76"
library(zoo)
z <- read.zoo(textConnection(Lines), format = "%d/%m/%Y")
rng <- range(time(z))
zz <- na.locf(merge(z, zoo(, seq(rng[1], rng[2], by = "day"))))
zz[format(time(zz), "%d") == "05"]

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