I have a dataset, df with a column containing dates in yyyy format (ex: 2018). I’m trying to make a time series graph, and therefore need to convert them to a date format.
I initially tried, df$year <- as.Date(df$year) but was told I needed to specify an origin.
I then tried to convert to a character, then a date format:
df$year <- as.character(df$year)
df$year <- as.Date(df$year, format = “%Y”)
This seems to have worked, however when it changed the all the years to yyyy-mm-dd format, and set the month and day to April 5th, today. For example 2018 becomes 2018-04-05.
Does anyone have an idea for how to fix this? I would like it to start on January 1, not the day I am performing the conversion. I also tried strptime(as.character(beer_states$year), “%Y”) with the same result.
Any help would be very much appreciated. Thanks!
Add an arbitrary date and month before converting to date.
df$Date <- as.Date(paste(df$year, 1, 1), '%Y %m %d')
We can use as.yearmon
library(zoo)
df$Date <- as.Date(as.yearmon(df$year, '-01'), '%Y-%m'))
Related
Have an large data frame where there's 2 columns (POSIXct) and need to calculate length of ride.
Dates are formatted as follows:
format: "2020-10-31 19:39:43"
Can use the difftime function, correct?
Thanks
Given your data is using the correct POSIXct format you can simply subtract two dates to get the difference. No need for additional functions.
date1 <- as.POSIXct(strptime("2020-10-31 19:39:43", format = "%Y-%m-%d %H:%M:%OS"))
date2 <- as.POSIXct(strptime("2020-10-31 19:20:43", format = "%Y-%m-%d %H:%M:%OS"))
date1 - date2
Output: Time difference of 19 mins
It depends what output format you want.
For example if you want month difference between two dates, you can use the "interval" function from library "lubridate"
library(lubridate)
interval(as.Date(df$date1),as.Date(df$date2) %/% months(1))
It also works with years, weeks, days, hours
Currently I am attempting to convert dates in the YYYYMMDD format to separate columns for year, month, and day. I know that using the as.Date function I can convert YYYYMMDD to YYYY-MM-DD, and work from there, however R is misinterpreting the dates and I'm not sure what to do. The function is converting the values into dates, but not correctly.
For example: R is converting '19030106' to '2019-03-01', when it should be '1903-01-06'. I'm not sure how to fix this, but this is the code I am using.
library(lubridate)
PrecipAll$Date <- as.Date(as.character(PrecipAll$YYYYMMDD), format = "%y%m%d")
YYYYMMDD is currently numeric, and I needed to include as.character in order for it to output a date at all, but if there are better solutions please help.
Additionally, if you have any tips on separating the corrected dates into separate Year, Month, and Date columns that would be greatly appreciated.
With {lubridate}, try ymd() to parse the YYYYMMDD varaible, regradless if it is in numeric or character form. Also use {lubridate}'s year, month, and day functions to get those variables as numeric signals.
library(lubridate)
PrecipAll <- data.frame(YYYYMMDD = c(19030106, 19100207, 20001130))
mutate(.data = PrecipAll,
date = lubridate::ymd(YYYYMMDD),
year = year(date),
month_n = month(date),
day_n = day(date))
YYYYMMDD date year month_n day_n
1 19030106 1903-01-06 1903 1 6
2 19100207 1910-02-07 1910 2 7
3 20001130 2000-11-30 2000 11 30
I'm trying to convert a column of full integers into date format of abbreviated months. The column has numbers like : 01 02 04 15 13. etc. I want these numbers to show the month they correspond to. Could someone please tell me how. the code I'm trying is this:
#Changing integers to Month Abbrev.
dets_per_month$monthcollected = as.POSIXlt(dets_per_month$monthcollected, format = "%m", origin = "%m")
but I realize the column doesn't have an origin because it's not in date format.
month.abb[as.integer(dets_per_month$monthcollected)]
I would recommend the lubridate package for all things date-time related. It's a nifty package and has more utility than base R, but YMMV.
library(lubridate)
x <- rep(1:12, 2)
lubridate::month(x, label=TRUE)
I have a column of my dataframe as
date
17-Feb
17-Mar
16-Dec
16-Nov
16-Sep
17-Feb
I am trying to convert it into a date column from string. I am using the following pieces of code:
as.Date(df$Date, format="%y-%b")
and
as.POSIXct(df$Date, format="%y-%b")
Both of them give NAs
I am getting the format from this link
The starting number is year. Sorry for the confusion.
I assume from your approach that the 17 and 16 refer to the year 2017 and 2016 respectively. You need to also specify the day of month. If you don't care about it, then set it to the 1st.
A slight modification to your code will work, by appending '-01' to the date then updating your format argument to reflect this:
df = data.frame(Date = c("17-Feb", "17-Mar", "16-Dec"))
as.Date(paste0(df$Date, "-01"), format="%y-%b-%d")
I have a date frame df that simply looks like this:
month values
2012M01 99904
2012M02 99616
2012M03 99530
2012M04 99500
2012M05 99380
2012M06 99103
2013M01 98533
2013M02 97600
2013M03 96431
2013M04 95369
2013M05 94527
2013M06 93783
with month that was written in form of "M01", "M02"... and so on.
Now I want to convert this column to date format, is there a way to do it in R with lubridate?
I also want to select columns that contain one certain month from each year, like only March columns from all these years, what is the best way to do it?
The short answer is that dates require a year, month and day, so you cannot convert directly to a date format. You have 2 options.
Option 1: convert to a year-month format using zoo::as.yearmon.
library(zoo)
df$yearmon <- as.yearmon(df$month, "%YM%m")
# you can get e.g. month from that
months(df$yearmon[1])
# [1] "January"
Option 2: convert to a date by assuming that the day is always the first day of the month.
df$date <- as.Date(paste(df$month, "01", sep = "-"), "%YM%m-%d")
For selection (and I think you mean select rows, not columns), you already have everything you need. For example, to select only March 2013:
library(dplyr)
df %>% filter(month == "2013M03")
Something like this will get it:
raw <- "2012M01"
dt <- strptime(raw,format = "%YM%m")
dt will be in a Posix format. The strptime function will assign a '1' as the default day of month to make it a complete date.