5th of month working day
Here is a second example that does the same thing except (that is it
returns the last value on or prior to the 5th); however, in the prior
solution the date was shown as the 5th and in this one it is shown as
the date of the last filled in value. Not sure which you would
prefer. See ?na.locf for more info on moving values forward into NAs
and also read the three vignettes that come with zoo.
# same as your example except I have removed the 5ths of the month and
added 4/4/2006.
Lines <- "
04/04/2006 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
06/05/2006 1325.76"
library(zoo)
z <- read.zoo(textConnection(Lines), format = "%d/%m/%Y")
# z.na is same as z but with missing days added using NAs
# Its formed by merging z with a zoo-width series containing all days.
rng <- range(time(z))
z.na <- merge(z, zoo(, seq(rng[1], rng[2], by = "day")))
# form a series that has NAs wherever z.na does but has 1, 2, 3, ...
# instead of z.na's data values and then use na.locf to fill in NAs
idx <- na.locf(seq_along(z.na) + (0 * z.na))
# pick off elements of z.na corresponding to 5th of month
z.na[idx[format(time(z.na), "%d") == "05"]]
Here is the final result:
2006-04-04 2006-05-04
1311.56 1312.25
On Mon, Jan 18, 2010 at 10:16 AM, Gabor Grothendieck
<ggrothendieck at gmail.com> wrote:
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"]