When I do this code:
library(lubridate)
df$date <- format(as.Date(df$date, "%m/%d/%y") , "%Y")
Some of the dates that are meant to be in the 1900s, eg: 1960, turn to 2060. I'm not sure how to fix this. The date range I want is 1951 - 2014, and I have around 8000 observations.
It seems that you have 2-digit years. From ?strptime
Year without century (00–99). On input, values 00 to 68 are prefixed by 20 and 69 to 99 by 19 – that is the behaviour specified by the 2004 and 2008 POSIX standards, but they do also say ‘it is expected that in a future version the default century inferred from a 2-digit year will change’.
So all 2 digit years from 00-68 are prefixed with 20, hence 60 turns to 2060 and not 1960.
There could be various ways to handle this. One way would be to subtract 100 years from dates whose year is more than 2014 (since we know the range of years).
For example,
df <- data.frame(date = c('1/12/60', '1/12/78' ,'1/1/91', '1/1/54'))
df$date <- as.Date(df$date, "%m/%d/%y")
df
# date
#1 2060-01-12
#2 1978-01-12
#3 1991-01-01
#4 2054-01-01
inds <- as.numeric(format(df$date, "%Y")) > 2014
df$date[inds] <- df$date[inds] - lubridate::years(100)
df
# date
#1 1960-01-12
#2 1978-01-12
#3 1991-01-01
#4 1954-01-01
We can also do this with chron as the cut-off date is 1961 by default
as.Date(chron::dates(c('01/12/60', '01/12/78' ,'01/01/91', '01/01/54')))
#[1] "1960-01-12" "1978-01-12" "1991-01-01" "1954-01-01"
Related
I have a dataframe with dates from April 2020 to today, right now they are labelled 1 to 492 with 1 being the first date I have data on. I also have a list of dates in the format I want. How can I tell R that date 1 is april 12 2020, date 2 is april 13, 2020, and so on for each date? I'm ok either replacing the values in the column or creating a new column called real_date next to it.
Update:
Sorry I didn't describe this very well. I ended up making a look-up table with the date number and real date, and I used the inner_join function to add the real date to my dataframe.
library(tidyverse)
library(lubridate)
#Creating a sample data.frame
df <-
tibble(
dates = seq.Date(dmy("01/04/20"),today(),by = "1 day")
)
df %>%
#Format date, where: %B = month as string, %d numeric day and %y numeric year
mutate(
new_date = format(dates,"%B %d %Y")
)
*Abril is April in portuguese.
If I have understood the question correctly, you have a dataframe which has numbers from 1 to 492, now you want to change them to dates where number 1 is 12th April 2020, number 2 is 13th April 2020 and so on.
You can use as.Date to convert these numbers to date and pass the origin as 11th April.
df <- data.frame(date = 1:492)
df$real_date <- as.Date(df$date, origin = '2020-04-11')
head(df)
# date real_date
#1 1 2020-04-12
#2 2 2020-04-13
#3 3 2020-04-14
#4 4 2020-04-15
#5 5 2020-04-16
#6 6 2020-04-17
Just create a sequence of dates
data.frame(date = seq(as.Date('2020-04-12'), length.out = 492,
by = '1 day'), code = 1:492)
I have a dataframe with monthly data, one column containing the year and one column containing the month. I'd like to combine them into one column with Date format, going from this:
Year Month Data
2020 1 54
2020 2 58
2020 3 78
2020 4 59
To this:
Date Data
2020-01 54
2020-02 58
2020-03 78
2020-04 59
I think you can't represent a Date format in R without showing the day. If you want a character column, like in your example, you can do:
> x <- data.frame(Year = c(2020,2020,2020,2020), Month = c(1,2,3,4), Data = c(54,58,78,59))
> x$Month <- ifelse(nchar(x$Month == 1), paste0(0, x$Month), x$Month) # add 0 behind.
> x$Date <- paste(x$Year, x$Month, sep = '-')
> x
Year Month Data Date
1 2020 01 54 2020-01
2 2020 02 58 2020-02
3 2020 03 78 2020-03
4 2020 04 59 2020-04
> class(x$Date)
[1] "character"
If you want a Date type column you will have to add:
x$Date <- paste0(x$Date, '-01')
x$Date <- as.Date(x$Date, format = '%Y-%m-%d')
x
class(x$Date)
Maybe the simplest way would be to arbitrarily set a day (e.g. 01) to all your dates ? Therefore date intervals would be preserved.
data<-data.frame(Year=c(2020,2020,2020,2020), Month=c(1,2,3,4), Data=c(54,58,78,59))
data$Date<-gsub(" ","",paste(data$Year,"-",data$Month,"-","01"))
data$Date<-as.Date(data$Date,format="%Y-%m-%d")
You can use sprintf -
sprintf('%d-%02d', data$Year, data$Month)
#[1] "2020-01" "2020-02" "2020-03" "2020-04"
Here's my data which has 10 years in one column and 365 day of another year in second column
dat <- data.frame(year = rep(1980:1989, each = 365), doy= rep(1:365, times = 10))
I am assuming all years are non-leap years i.e. they have 365 days.
I want to create another column month which is basically month of the year the day belongs to.
library(dplyr)
dat %>%
mutate(month = as.integer(ceiling(day/31)))
However, this solution is wrong since it assigns wrong months to days. I am looking for a dplyr
solution possibly.
We can convert it to to datetime class by using the appropriate format (i.e. %Y %j) and then extract the month with format
dat$month <- with(dat, format(strptime(paste(year, doy), format = "%Y %j"), '%m'))
Or use $mon to extract the month and add 1
dat$month <- with(dat, strptime(paste(year, doy), format = "%Y %j")$mon + 1)
tail(dat$month)
#[1] 12 12 12 12 12 12
This should give you an integer value for the months:
dat$month.num <- month(as.Date(paste(dat$year, dat$doy), '%Y %j'))
If you want the month names:
dat$month.names <- month.name[month(as.Date(paste(dat$year, dat$doy), '%Y %j'))]
The result (only showing a few rows):
> dat[29:33,]
year doy month.num month.names
29 1980 29 1 January
30 1980 30 1 January
31 1980 31 1 January
32 1980 32 2 February
33 1980 33 2 February
I want to know how to find out which part of string is month and which part of string is day while parsing dates.
The problem is 01-06-2017 can be 1 June or it can be 6 January. How to parse it correctly. In India we write dates as Day Month Year mostly, in west it is Month Day Year mostly, when I have mixed data how do I impute which is the month and which is the day
because the data is not clean enough, it sometimes have dates in mdy and sometimes in dmy format and if the number is less than 12, it is difficult to know if it is a day or a month
11/1/11 can be 11 Jan 2011 or 1 November 2011
Example
I am using lubridate package and I have dates in this format
library(lubridate)
fundates2=c("1Apr2017","12-30-2017","1/6/17")
fun3=dmy(fundates2)
## Warning: 1 failed to parse.
fun3
## [1] "2017-04-01" NA "2017-06-01"
fun4=mdy(fundates2)
## Warning: 1 failed to parse.
fun4
## [1] NA "2017-12-30" "2017-01-06"
Well, you have yo know from your context which one is the correct.
To check which one your date is you can simply add 1 day to it:
In fun3:
fun3 + 1
[1] "2017-04-02" NA "2017-06-02"
You can see that the month is the 06.
In fun4:
fun4 + 1
[1] NA "2017-12-31" "2017-01-07"
You can see the month is 01
I have a time series tt.txt of daily data from 1st May 1998 to 31 October 2012 in one column as this:
v1
296.172
303.24
303.891
304.603
304.207
303.22
303.137
303.343
304.203
305.029
305.099
304.681
304.32
304.471
305.022
304.938
304.298
304.120
Each number in the text file represents the maximum temperature in kelvin for the corresponding day. I want to put the data in 3 columns as follows by adding year, jday, and the value of the data:
year jday MAX_TEMP
1 1959 325 11.7
2 1959 326 15.6
3 1959 327 14.4
If you have a vector with dates, we can convert it to 'year' and 'jday' by
v1 <- c('May 1998 05', 'October 2012 10')
v2 <- format(as.Date(v1, '%b %Y %d'), '%Y %j')
df1 <- read.table(text=v2, header=FALSE, col.names=c('year', 'jday'))
df1
# year jday
#1 1998 125
#2 2012 284
To convert back from '%Y %j' to 'Date' class
df1$date <- as.Date(do.call(paste, df1[1:2]), '%Y %j')
Update
We can read the dataset with read.table. Create a sequence of dates using seq if we know the start and end dates, cbind with the original dataset after changing the format of 'date' to 'year' and 'julian day'.
dat <- read.table('tt.txt', header=TRUE)
date <- seq(as.Date('1998-05-01'), as.Date('2012-10-31'), by='day')
dat2 <- cbind(read.table(text=format(date, '%Y %j'),
col.names=c('year', 'jday')),MAX_TEMP=dat[1])
You can use yday
as.POSIXlt("8 Jun 15", format = "%d %b %y")$yday