【发布时间】:2012-08-12 17:19:43
【问题描述】:
【问题讨论】:
标签: r
【问题讨论】:
标签: r
这里有一些信息来创建你自己的库或例程
常量:
day_of_month
the day of the month
e.g. if input mm-dd-yyy then dd
月:
march = 1
april = 2
may = 3
...
年份
yy[yy] (last to digits from yyyy)
*subtract 1 if month jan or feb
e.g. if input date is 02-01-2012 (mm-dd-yyyy)
year = (12-1) = 11
世纪
[yy]yy (first two digits from yyyy)
e.g. if input year is 2012 then 20 = century
* year 2000, 1900, ... are 20-1, 19-1 respectively
算法
step1: floor(century / 4)
step2: year
step3: floor(year/4)
step4: floor(month*2.6 -0.2) #this is the leap year correction
step5: day_of_month
step6: add step1...step5
step7: divide by 7 # modulo 7 in codespeak
step8: the remainder is the day of the week
解释结果:
Sun = 0, Mon = 1, Tues = 3, etc..
不是图书馆,而是公共服务的叮当声……
“阅读:你知道的更多”
【讨论】:
weekdays(as.Date('16-08-2012','%d-%m-%Y'))
[1] "Thursday"
【讨论】:
weekdays() 在base R中。+1
lubridate 包非常适合这类东西。
> wday(as.Date('16-08-2012','%d-%m-%Y'))
[1] 5
> wday(as.Date('16-08-2012','%d-%m-%Y'), label=TRUE)
[1] Thurs
Levels: Sun < Mon < Tues < Wed < Thurs < Fri < Sat
> wday(as.Date('16-08-2012','%d-%m-%Y'), label=TRUE, abbr = FALSE)
[1] Thursday
Levels: Sunday < Monday < Tuesday < Wednesday < Thursday < Friday < Saturday
【讨论】: