Extract Month and Year from Date in R

Extract Month and Year From Date in R

This will add a new column to your data.frame with the specified format.

df$Month_Yr <- format(as.Date(df$Date), "%Y-%m")

df
#>   ID       Date Month_Yr
#> 1  1 2004-02-06  2004-02
#> 2  2 2006-03-14  2006-03
#> 3  3 2007-07-16  2007-07

# your data sample
  df <- data.frame( ID=1:3,Date = c("2004-02-06" , "2006-03-14" , "2007-07-16") )

a simple example:

dates <- "2004-02-06"

format(as.Date(dates), "%Y-%m")
> "2004-02"

side note:
the data.table approach can be quite faster in case you're working with a big dataset.

library(data.table)
setDT(df)[, Month_Yr := format(as.Date(Date), "%Y-%m") ]

Extract month and year from datetime in R

lubridate month and year will work.

as.data.frame(Order.Date) %>%
  mutate(Month = lubridate::month(Order.Date, label = FALSE),
         Year = lubridate::year(Order.Date))

  Order.Date Month Year
1 2011-10-20    10 2011
2 2011-12-25    12 2011
3 2012-04-15     4 2012
4 2012-08-23     8 2012
5 2013-09-25     9 2013

If you want month format as Jan, use month.abb and as January, use month.name

as.data.frame(Order.Date) %>%
  mutate(Month = month.abb[lubridate::month(Order.Date, label = TRUE)],
         Year = lubridate::year(Order.Date))

  Order.Date Month Year
1 2011-10-20   Oct 2011
2 2011-12-25   Dec 2011
3 2012-04-15   Apr 2012
4 2012-08-23   Aug 2012
5 2013-09-25   Sep 2013

as.data.frame(Order.Date) %>%
  mutate(Month = month.name[lubridate::month(Order.Date, label = TRUE)],
         Year = lubridate::year(Order.Date))

  Order.Date     Month Year
1 2011-10-20   October 2011
2 2011-12-25  December 2011
3 2012-04-15     April 2012
4 2012-08-23    August 2012
5 2013-09-25 September 2013

How to extract Month and Year from Date column and add two more columns for each

Looks like it is grabbing the day instead of the year. So would seem that your date isn't properly formatted when it goes through the as.Date() function

See what this

as.Date(act_weather_data$Date)

looks like on its own and format accordingly

i.e.

as.Date(act_weather_data$Date, format="%Y/%m/%d")

Then apply.the formatting as before
i.e

Year=format(as.Date(act_weather_data$Date, format="%Y/%m/%d"),"%Y")

How to extract Month from date in R

?month states:

Date-time must be a POSIXct, POSIXlt, Date, Period, chron, yearmon,
yearqtr, zoo, zooreg, timeDate, xts, its, ti, jul, timeSeries, and fts
objects.

Your object is a factor, not even a character vector (presumably because of stringsAsFactors = TRUE). You have to convert your vector to some datetime class, for instance to POSIXlt:

library(lubridate)
some_date <- c("01/02/1979", "03/04/1980")
month(as.POSIXlt(some_date, format="%d/%m/%Y"))
[1] 2 4

There's also a convenience function dmy, that can do the same (tip proposed by @Henrik):

month(dmy(some_date))
[1] 2 4

Going even further, @IShouldBuyABoat gives another hint that dd/mm/yyyy character formats are accepted without any explicit casting:

month(some_date)
[1] 2 4

For a list of formats, see ?strptime. You'll find that "standard unambiguous format" stands for

The default formats follow the rules of the ISO 8601 international
standard which expresses a day as "2001-02-28" and a time as
"14:01:02" using leading zeroes as here.

extract month date year into new columns from char column in R

You can use tidyr::separate

library(tidyverse)
data <- data.frame(date = "2021-18-02 16:08:48")

data %>% 
  separate(date, into = c("Year","Day","Month","Time"), extra = "merge")

#   Year Day Month     Time
# 1 2021  18    02 16:08:48

Converting a date 'year - month - date' to only 'year and month' in r with SQL data

Up front, your attempt of as.Date(df$Posting_Date, format="%Y %m") seems backwards: the function as.Date is for converting from a string to a Date-class, and its format= argument is to identify how to find the year/month/day components of the string, not how you want to convert it later. (Note that in R, a Date is shown as YYYY-MM-DD. Always. Telling R you want a date to be just year/month is saying that you want to convert it to a string, no longer date-like or number-like. lubridate and perhaps other packages allow you to have similar-to-Date like objects.)

For df, one can just subset the strings without parsing to Date-class:

substring(df$Posting_Date, 1, 7)
# [1] "2020-05" "2020-10" "2021-10"

If you want to do anything number-like to them, you can convert to Date-class first, and then use format(.) to convert to a string with a specific format.

as.Date(df$Posting_Date)
# [1] "2020-05-28" "2020-10-09" "2021-10-19"
format(as.Date(df$Posting_Date), format = "%Y-%m")
# [1] "2020-05" "2020-10" "2021-10"

For df2, though, since it is numeric you need to specify an origin= instead of a format=. I'm inferring that these are based off of epoch, so

as.Date(df2$Posting_Date, origin = "1970-01-01")
# [1] "2020-05-28" "2020-10-09" "2021-10-19"
format(as.Date(df2$Posting_Date, origin = "1970-01-01"), format = "%Y-%m")
# [1] "2020-05" "2020-10" "2021-10"

Note that R stores Date (and POSIXct, incidentally) as numbers internally:

dput(as.Date(df2$Posting_Date, origin = "1970-01-01"))
# structure(c(18410, 18544, 18919), class = "Date")

extract year and month from character date field R

dplyr and lubridate -

library(dplyr)
library(lubridate)

data <- data %>%
  mutate(Date = dmy_hm(Date), 
         month = month(Date), 
         year = year(Date))

#                 Date month year
#1 2020-06-02 11:23:00     6 2020
#2 2020-06-12 07:56:00     6 2020
#3 2020-06-12 07:56:00     6 2020
#4 2020-06-29 16:54:00     6 2020
#5 2020-06-03 15:09:00     6 2020
#6 2020-06-25 17:11:00     6 2020

Base R -

data$Date <- as.POSIXct(data$Date, tz = 'UTC', format = '%d/%m/%Y %H:%M')
data <- transform(data, Month = format(Date, '%m'), Year = format(Date, '%Y'))

data

data <- structure(list(Date = c("2/06/2020 11:23", "12/06/2020 7:56", 
"12/06/2020 7:56", "29/06/2020 16:54", "3/06/2020 15:09", "25/06/2020 17:11"
)), class = "data.frame", row.names = c(NA, -6L))

Format Date to Year-Month in R

lubridate only handle dates, and dates have days. However, as alistaire mentions, you can floor them by month of you want work monthly:

library(tidyverse)

df_month <-
  df %>%
  mutate(Date = floor_date(as_date(Date), "month"))

If you e.g. want to aggregate by month, just group_by() and summarize().

df_month %>%
  group_by(Date) %>%
  summarize(N = sum(N)) %>%
  ungroup()

#> # A tibble: 4 x 2
#>  Date           N
#>  <date>     <dbl>
#>1 2017-01-01    59
#>2 2018-01-01    20
#>3 2018-02-01    33
#>4 2018-03-01    45

Extract year from date

if all your dates are the same width, you can put the dates in a vector and use substring

Date
a <- c("01/01/2009", "01/01/2010" , "01/01/2011")
substring(a,7,10) #This takes string and only keeps the characters beginning in position 7 to position 10

output

[1] "2009" "2010" "2011"

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