As you can see from the dataframe below, RBloomberg returns NAs for weekend dates.
I want to remove the entire row if it falls on a weekend. How would I do this?
I don't want to use the na.omit as this might remove weekday rows if/when I get an NA there in the data for a legitimate reason.
ticker date yld_ytm_mid
1 R206 2011-05-11 6.946
2 R206 2011-05-12 6.969
3 R206 2011-05-13 7.071
4 R206 2011-05-14 NA
5 R206 2011-05-15 NA
6 R201 2011-05-11 7.201
7 R201 2011-05-12 7.213
8 R201 2011-05-13 7.323
9 R201 2011-05-14 NA
10 R201 2011-05-15 NA
11 R157 2011-05-11 7.611
12 R157 2011-05-12 7.622
13 R157 2011-05-13 7.718
14 R157 2011-05-14 NA
15 R157 2011-05-15 NA
16 R203 2011-05-11 8.165
17 R203 2011-05-12 8.170
18 R203 2011-05-13 8.279
19 R203 2011-05-14 NA
20 R203 2011-05-15 NA
21 R204 2011-05-11 8.303
22 R204 2011-05-12 8.296
23 R204 2011-05-13 8.386
24 R204 2011-05-14 NA
25 R204 2011-05-15 NA
26 R207 2011-05-11 8.361
27 R207 2011-05-12 8.371
28 R207 2011-05-13 8.479
29 R207 2011-05-14 NA
30 R207 2011-05-15 NA
31 R208 2011-05-11 8.392
32 R208 2011-05-12 8.393
33 R208 2011-05-13 8.514
34 R208 2011-05-14 NA
35 R208 2011-05-15 NA
36 R186 2011-05-11 8.546
37 R186 2011-05-12 8.571
38 R186 2011-05-13 8.664
39 R186 2011-05-14 NA
40 R186 2011-05-15 NA
41 R213 2011-05-11 8.783
42 R213 2011-05-12 8.802
43 R213 2011-05-13 8.898
44 R213 2011-05-14 NA
45 R213 2011-05-15 NA
46 R209 2011-05-11 8.785
47 R209 2011-05-12 8.807
48 R209 2011-05-13 8.898
49 R209 2011-05-14 NA
50 R209 2011-05-15 NA
51 R214 2011-05-11 8.841
52 R214 2011-05-12 8.861
53 R214 2011-05-13 8.958
54 R214 2011-05-14 NA
55 R214 2011-05-15 NA
For completeness' sake, I would add to blindjesse's answer that typing ?weekdays reveals that R has base functions weekdays(), months() and quarters() that work on both the posix and date types, and are I believe vectorized, so this would work as well:
!(weekdays(as.Date(date)) %in% c('Saturday','Sunday'))
Convert the date column to a POSIXlt ,eg
date <- as.POSIXlt(date,format="%Y-%m-%d")
Then you can access the day of the week using
date$wday
and subset the frame appropriately
The answer by blindJesse is correct and useful as it falls back to base R functions.
Many packages have additional helper wrappers. Here is one from timeDate which requires conversion to its type:
R> isWeekend( as.timeDate( seq( as.Date("2011-01-01"),
+ to=as.Date("2011-01-07"), by=1 ) ) )
2011-01-01 2011-01-02 2011-01-03 2011-01-04 2011-01-05 2011-01-06 2011-01-07
TRUE TRUE FALSE FALSE FALSE FALSE FALSE
R>
and here is another approach using a function from RcppBDT:
R> sapply(seq(as.Date("2011-01-01"),to=as.Date("2011-01-07"), by=1),getDayOfWeek)
[1] 6 0 1 2 3 4 5
R>
R> sapply(seq(as.Date("2011-01-01"),to=as.Date("2011-01-07"), by=1),getDayOfWeek)
+ %%6 == 0
[1] TRUE TRUE FALSE FALSE FALSE FALSE FALSE
R>
The lubridate package also has wday()
and there are undoubtedly more he;per functions in other packages.
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