python (65.1k questions)

javascript (44.2k questions)

reactjs (22.7k questions)

java (20.8k questions)

c# (17.4k questions)

html (16.3k questions)

r (13.7k questions)

android (12.9k questions)

Questions - lubridate

floor_date not working in mutate and ifelse

I am trying to write a generalize aggregation function where the user specifies the aggregation level or they can aggregate the data over all study dates. The floor_date only converts the first date. ...
test-img

Desi Middleton

r

dplyr

lubridate

Votes: 0

Answers: 1

Latest Answer

The problem does not stem from floor_date but from your use of ifelse. As per its manual: ifelse(test, yes, no) ifelse returns a value with the same shape as test which is filled with elements selecte...
test-img

Valkyr

Expand date in time series using years

I have the following time series location date value North 199001 a North 199203 b North 199402 c North 199506 d South 19...
test-img

L'Android

r

dplyr

time-series

lubridate

Votes: 0

Answers: 1

Latest Answer

A dplyr / tidyr solution: substract the first four digits of your date column, convert to as.numeric, group_by location, complete the years per location and fill the values: Code library(dplyr) librar...
test-img

Gnueghoidune

Parsing one column with 5 didgits integers as date format while reading excel file

For an excel file (download from here): df <- openxlsx::read.xlsx('sample_data.xlsx', sheet='Sheet1', colNames=TRUE) df Output: date value 1 43861 5.70 2 43890 -13.89 3 43921 -49.68 4 43951...
test-img

ah bon

r

datetime-format

lubridate

zoo

openxlsx

Votes: 0

Answers: 1

Latest Answer

origin has to be inside the as.Date call. df %>% mutate(date = as.Date(date, origin = "1899-12-30")) #> date value #> 1 2020-01-31 5.70 #> 2 2020-02-29 -13.89 #> ...
test-img

Allan Cameron

How to divide a date column in two columns only

I have a data frame containing 1 date column: > head(df$date) [1] "1952-02-03" "1958-02-08" "1958-02-08" "1958-02-08" "1965-02-07" "1966-03-03...
test-img

Recology

r

date

data.table

lubridate

Votes: 0

Answers: 1

Latest Answer

Just use format to get the parts of the date you want: df$year <- format(df$date, "%Y") df$month_day <- format(df$date, "%m-%d") df # date year month_day # 1 1952...
test-img

SamR

Posts

Questions

Blogs

Jobs