lubridate | 易学教程

Using lubridate and ggplot2 effectively for date axis

阅读更多关于 Using lubridate and ggplot2 effectively for date axis

问题 Consider this example: library(ggplot2) library(lubridate) set.seed(4) date <- seq(from = as.POSIXct("2012-01-01"), to = as.POSIXct("2014-12-31"), by = "days") value <- c(rnorm(274, 50, 1), rnorm(274, 55, 1), rnorm(274, 55, 2), rnorm(274, 60, 2)) df <- data.frame(date, value) head(df) # date value # 1 2012-01-01 50.21675 # 2 2012-01-02 49.45751 # 3 2012-01-03 50.89114 # 4 2012-01-04 50.59598 # 5 2012-01-05 51.63562 # 6 2012-01-06 50.68928 ggplot(df, aes(x=yday(date), y=value, color=factor

Is there a fast parser for date

阅读更多关于 Is there a fast parser for date

问题 For datetimes fasttime provides very fast parsing to POSIXct library('fasttime') library('lubridate') library('microbenchmark') # parse character to POSIXct Sys.setenv(TZ='UTC') test <- rep('2011-04-02 11:01:00',1e4) microbenchmark( test1 <- fastPOSIXct(test), test2 <- fast_strptime(test,format='%Y-%m-%d %H:%M:%S'), test3 <- as.POSIXct(test, format='%Y-%m-%d %H:%M:%S'), test4 <- ymd_hms(test), times=100) Unit: microseconds expr min lq mean median uq max test1 <- fastPOSIXct(test) 663.123 692

dplyr: grouping and summarizing/mutating data with rolling time windows

阅读更多关于 dplyr: grouping and summarizing/mutating data with rolling time windows

问题 I have irregular timeseries data representing a certain type of transaction for users. Each line of data is timestamped and represents a transaction at that time. By the irregular nature of the data some users might have 100 rows in a day and other users might have 0 or 1 transaction in a day. The data might look something like this: data.frame( id = c(1, 1, 1, 1, 1, 2, 2, 3, 4), date = c("2015-01-01", "2015-01-01", "2015-01-05", "2015-01-25", "2015-02-15", "2015-05-05", "2015-01-01", "2015

Time difference in years with lubridate?

阅读更多关于 Time difference in years with lubridate?

问题 I would like to use lubridate to calculate age in years given their date of birth and today's date. Right now I have this: library(lubridate) today<-mdy(08312015) dob<-mdy(09071982) today-dob which gives me their age in days. 回答1: This is the lubridate approach I would take: interval(dob, today) / years(1) Yields the answer of 32 years. Note that the function will complain that it cannot express the remainder of the fraction of the year. This is because year is not a fixed concept, i.e. 366

Why are lubridate functions so slow when compared with as.POSIXct?

阅读更多关于 Why are lubridate functions so slow when compared with as.POSIXct?

问题 As the title goes. Why is the lubridate function so much slower? library(lubridate) library(microbenchmark) Dates <- sample(c(dates = format(seq(ISOdate(2010,1,1), by='day', length=365), format='%d-%m-%Y')), 50000, replace = TRUE) microbenchmark(as.POSIXct(Dates, format = "%d-%b-%Y %H:%M:%S", tz = "GMT"), times = 100) microbenchmark(dmy(Dates, tz ="GMT"), times = 100) Unit: milliseconds expr min lq median uq max 1 as.POSIXct(Dates, format = "%d-%b-%Y %H:%M:%S", tz = "GMT") 103.1902 104.3247

Add/subtract 6 months (bond time) in R using lubridate

阅读更多关于 Add/subtract 6 months (bond time) in R using lubridate

I am looking to add and subtract six months reliably with lubridate . For example, adding six months to 12/31/2014 should result in 6/30/2015 , and adding to 2/28/2014 should result in 8/31/2014 The issue with as.Date("2014-12-31") + months(6) , is that it yields an NA . Alternatively, the second result is 8/28/2014 because it doesn't just add 6 months to the month and then know where the day should end up dependent upon the month. Is there any way to quickly correct this? At the moment, I am building a function to basically use a switch and consider each month, but this is very long and I am

Create a Vector of All Days Between Two Dates

阅读更多关于 Create a Vector of All Days Between Two Dates

Is there an easy way in R for me to itemize all valid days that occurred between two specified dates? For instance, I'd like the following inputs: itemizeDates(startDate="12-30-11", endDate="1-4-12") To produce the following dates: "12-30-11" "12-31-11", "1-1-12", "1-2-12", "1-3-12", "1-4-12" I'm flexible on classes and formatting of the dates, I just need an implementation of the concept. You're looking for seq > seq(as.Date("2011-12-30"), as.Date("2012-01-04"), by="days") [1] "2011-12-30" "2011-12-31" "2012-01-01" "2012-01-02" "2012-01-03" [6] "2012-01-04" Or, you can use : > as.Date(as.Date

extract hours and seconds from POSIXct for plotting purposes in R

阅读更多关于 extract hours and seconds from POSIXct for plotting purposes in R

Suppose I have the following data.frame foo start.time duration 1 2012-02-06 15:47:00 1 2 2012-02-06 15:02:00 2 3 2012-02-22 10:08:00 3 4 2012-02-22 09:32:00 4 5 2012-03-21 13:47:00 5 And class(foo$start.time) returns [1] "POSIXct" "POSIXt" I'd like to create a plot of foo$duration v. foo$start.time . In my scenario, I'm only interested in the time of day rather than the actual day of the year. How does one go about extracting the time of day as hours:seconds from POSIXct class of vector? This is a good question, and highlights some of the difficulty in dealing with dates in R. The lubridate

Is there a more elegant way to convert two-digit years to four-digit years with lubridate?

阅读更多关于 Is there a more elegant way to convert two-digit years to four-digit years with lubridate?

If a date vector has two-digit years, mdy() turns years between 00 and 68 into 21st Century years and years between 69 and 99 into 20th Century years. For example: library(lubridate) mdy(c("1/2/54","1/2/68","1/2/69","1/2/99","1/2/04")) gives the following output: Multiple format matches with 5 successes: %m/%d/%y, %m/%d/%Y. Using date format %m/%d/%y. [1] "2054-01-02 UTC" "2068-01-02 UTC" "1969-01-02 UTC" "1999-01-02 UTC" "2004-01-02 UTC" I can fix this after the fact by subtracting 100 from the incorrect dates to turn 2054 and 2068 into 1954 and 1968. But is there a more elegant and less

Assigning Dates to Fiscal Year

阅读更多关于 Assigning Dates to Fiscal Year

I'm trying to come up with some code that will look at a date and then assign it to a fiscal year. I'm totally stuck. I have a variable that contains dates in POSIXct format: df$Date #2015-05-01 CST #2015-04-30 CST #2014-09-01 CST What I need to be able to do is take those dates and return a fiscal year, which runs from May 1 - April 30. For example, Fiscal Year 2016 runs 2015-05-01 through 2016-04-30. Results would look something like this: df$Date df$FiscalYear #2015-05-01 CST #FY2016 #2015-04-30 CST #FY2015 #2014-09-01 CST #FY2015 Is there any easy way to do this? Here are some alternatives