Get the number of days from the date column [duplicate] - r

Question

This question already has answers here:
Find out the number of days of a month in R
(15 answers)
Closed 3 years ago.
I have column - Month, and there are few months, I need to find out the number of day in that month.
I need to find the number of days in a month considering the leap year and that the month can go to June 2019, or current month. The logic should be applied on the column.
MONTH no_of_days
Jan,2017
Feb,2017
Mar,2017
Apr,2017
May,2017
Jun,2017
Jul,2017
Aug,2017
Sep,2017
Oct,2017
Nov,2017
Dec,2017
structure(list(MONTH = structure(c(9L, 7L, 15L, 1L, 17L, 13L,
11L, 3L, 23L, 21L, 19L, 5L, 10L, 8L, 16L, 2L, 18L, 14L, 12L,
4L, 24L, 22L, 20L, 6L), .Label = c("Apr,2017", "Apr,2018", "Aug,2017",
"Aug,2018", "Dec,2017", "Dec,2018", "Feb,2017", "Feb,2018", "Jan,2017",
"Jan,2018", "Jul,2017", "Jul,2018", "Jun,2017", "Jun,2018", "Mar,2017",
"Mar,2018", "May,2017", "May,2018", "Nov,2017", "Nov,2018", "Oct,2017",
"Oct,2018", "Sep,2017", "Sep,2018"), class = "factor")), class = "data.frame", row.names = c(NA,
-24L))

Answer 1

You need to first get the column MONTH to Date and then use any of the function shown here to get number of days in a month.
as.Date(paste0(1, df$MONTH), "%d%b,%Y")
#[1] "2017-01-01" "2017-02-01" "2017-03-01" "2017-04-01" "2017-05-01" "2017-06-01"
# "2017-07-01" "2017-08-01" "2017-09-01" "2017-10-01" "2017-11-01" "2017-12-01"
Hmisc::monthDays(as.Date(paste0(1, df$MONTH), "%d%b,%Y"))
#[1] 31 28 31 30 31 30 31 31 30 31 30 31
data
df <- structure(list(MONTH = structure(c(5L, 4L, 8L, 1L, 9L, 7L, 6L,
2L, 12L, 11L, 10L, 3L), .Label = c("Apr,2017", "Aug,2017", "Dec,2017",
"Feb,2017", "Jan,2017", "Jul,2017", "Jun,2017", "Mar,2017", "May,2017",
"Nov,2017", "Oct,2017", "Sep,2017"), class = "factor")), class =
"data.frame", row.names = c(NA, -12L))

Answer 2

You'll need to wrangle your MONTH column into a date object, then you can use lubridate::days_in_month():
if (!require(lubridate)) install.packages('lubridate')
if (!require(dplyr)) install.packages('dplyr')
library(lubridate)
library(dplyr)
dat %>%
mutate(MONTH = paste("01", MONTH, sep = ",") %>% dmy(),
no_of_days = days_in_month(MONTH))
# A tibble: 12 x 2
MONTH no_of_days
<date> <int>
1 2017-01-01 31
2 2017-02-01 28
3 2017-03-01 31
4 2017-04-01 30
5 2017-05-01 31
6 2017-06-01 30
7 2017-07-01 31
8 2017-08-01 31
9 2017-09-01 30
10 2017-10-01 31
11 2017-11-01 30
12 2017-12-01 31

Answer 3

The package lubridate has a function called days_in_month(). So you can first convert the MONTH column into dates and then use days_in_month:
library(lubridate)
library(zoo)
df <- structure(list(MONTH = structure(c(9L, 7L, 15L, 1L, 17L, 13L, 11L, 3L, 23L, 21L, 19L, 5L, 10L, 8L, 16L, 2L, 18L, 14L, 12L, 4L, 24L, 22L, 20L, 6L), .Label = c("Apr,2017", "Apr,2018", "Aug,2017", "Aug,2018", "Dec,2017", "Dec,2018", "Feb,2017", "Feb,2018", "Jan,2017", "Jan,2018", "Jul,2017", "Jul,2018", "Jun,2017", "Jun,2018", "Mar,2017", "Mar,2018", "May,2017", "May,2018", "Nov,2017", "Nov,2018", "Oct,2017", "Oct,2018", "Sep,2017", "Sep,2018"), class = "factor")), class = "data.frame", row.names = c(NA, -24L))
df$no_of_days <- days_in_month(as.Date(as.yearmon(df$MONTH, "%b,%Y")))