Advertisements I am running into issues with subsetting after subtracting dates using lubridate. I have a dataframe: customerid <- c("A1", "A1", "A2", "A2", "A3", "A3", "A3", "A4") orderdate <- c("2018-09-14", "2018-09-14", "2018-09-15", "2018-09-15", "2020-08-21", "2020-08-21","2020-08-21", "2018-08-10") returndate <- c("2018-09-15", "2018-09-18", "2018-09-20", "2019-09-15", "2021-08-20", "2020-07-21","2020-09-21", "2018-08-15") orderid <- c("1", "2", "3", "4", "5", "6", "7", "8")… Read More Running into issues with subsetting after subtracting dates?

Advertisements I am trying to make a graph of a time series, but I’m having trouble achieving my desired result. I have data from a time series from May 2021 to December 2022. Original data is by day, but for my visualization, months will suffice. I used lubridate::floor_date() and summarised the observations. The resulting data… Read More Plotting dates on the x-axis and customizing labels

Advertisements When I’m trying to store date in a DF it is getting represented as a number instead of a data format. Here is my example. library(lubridate) df1 = data.frame(task = c(‘do somthing’, ‘do something else’, ‘do something more’) ,Start_date = c(NA, NA, NA)) Start_date = ymd("2023-12-01") df1$Date_Start[1] = Start_date Start_date #this returns a "2023-12-01"… Read More R – Problem Storing lubridate dates in a dataframe

Advertisements I want to add months to a date by conditition. That means if my ISO column is ET I want to add 92 month. When I only run this code, i recieve the dates I want. ymd(paste(comb_extract_all$hv007,comb_extract_all$hv006,"01", sep = "-")) %m+% months(92) first lines of my output: [1] "2011-04-01" "2011-04-01" "2011-04-01" "2011-04-01" "2011-04-01" "2011-04-01"… Read More add months to a date ifelse

Advertisements I’m trying to find an elegant way to create a vector of dates that follows a pattern like: x <- c(ymd("2000/01/01"), ymd("2000/01/31"), ymd("2000/02/01"), ymd("2000/02/28")) …and so on. So far I’ve just been doing this: library(lubridate) start <- ymd("2000/01/01") x <- c(start, rollback(start + month(1)), start + months(1), rollback(start + months(2)), start + months(2), rollback(start… Read More Is there an easy way to create a vector following the pattern: 1st day of the month, last day of the month, 1st day of the month etc.?

Advertisements I have 2 intervals of dates, and I want to see if any of the dates in interval_A are within interval_B. I am ideally looking for a dplyr solution. The data library(lubridate) interval_A <- new("Interval", .Data = c(20822400, 10454400, 42508800, 18662400, 12355200, 16243200, 10195200, 14774400, 37324800, 31276800, 27734400, 62985600, 15724800, 32054400, 21427200), start =… Read More Check if any dates within an Interval are within any of the dates in another Interval

Advertisements I am looking to determine the difference in days by groups across two columns and two rows. Essentially subtract from the End Day by the subsequent Start Day in the subsequent row and record the difference as new column in the data frame and start over when a new group (ID) has been identified.… Read More How to determine difference in days between two dates across two columns and two rows by group?

Advertisements I have a date column whose date values are something like this: date = c("1/6/2022", "1/6/2022", "1/19/2022", "1/20/2022") When I try to convert it to new date column using lubridate::mdy I get: library(lubridate) date_new = lubridate::mdy(date) print(date_new) [1] "2022-01-06" "2022-01-06" "2022-01-19" "2022-01-20" # Desired output (mm-dd-yyyy or mm/dd/yyyy): [1] "01-06-2022" "01-06-2022" "01-19-2022" "01-20-2022" How… Read More lubridate mdy format issue

Advertisements Lubridate makes working with time easier, but sometimes it is hard for me to understand. What I would like to do is to divide a period by integer. For example: If I run 5 km in 24m:45s what would I run 1 km in? # period gives me a time object > as.period(ms("24:30")) [1]… Read More R: how to divide period by integer in lubridate?

Advertisements I calculate some durations in R as shown below. Depending of the value of the duration, the unit can be minutes, hours, days, etc.: library(lubridate) start <- "2022-08-27 10:06:58" end <- "2022-08-27 10:10:24" start <- as_datetime(start) end <- as_datetime(end) end – start # Time difference of 3.433333 mins start – Sys.time() # Time difference… Read More How to control the unit of a time difference in R?