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
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"
I have a data of from 2016-2019
Here is a sample of my data
print(myData)
Date Freq
2016-08-08 14
2016-08-09 20
2016-08-10 34
2016-08-11 32
2016-08-12 19
2016-08-15 35
2016-08-16 32
I want to create a line plot but I would like to see something like this that way I can see the trend on a quarterly basis.
Date Freq
2016 Q1 300
2016 Q2 313
2016 Q3 313
2016 Q4 432
2017 Q1 313
2017 Q2 131
How can I do this in R?
We can use as.yearqtr from zoo
library(zoo)
myData %>%
mutate(Date = as.yearqtr(Date))
If the column is not already a Date class, convert to Date class first
myData <- myData %>%
mutate(Date = as.yearqtr(as.Date(Date)))
lubridate has a function quarter:
quarter(myData$Date)
so for your data you can simply write:
myData <- myData %>% mutate(quarter = quarter(Date)
Here's my data which has 10 years in one column and 365 day of another year in second column
dat <- data.frame(year = rep(1980:1989, each = 365), doy= rep(1:365, times = 10))
I am assuming all years are non-leap years i.e. they have 365 days.
I want to create another column month which is basically month of the year the day belongs to.
library(dplyr)
dat %>%
mutate(month = as.integer(ceiling(day/31)))
However, this solution is wrong since it assigns wrong months to days. I am looking for a dplyr
solution possibly.
We can convert it to to datetime class by using the appropriate format (i.e. %Y %j) and then extract the month with format
dat$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 have a column with date formatted as MM-DD-YYYY, in the Date format.
I want to add 2 columns one which only contains YYYY and the other only contains MM.
How do I do this?
Once again base R gives you all you need, and you should not do this with sub-strings.
Here we first create a data.frame with a proper Date column. If your date is in text format, parse it first with as.Date() or my anytime::anydate() (which does not need formats).
Then given the date creating year and month is simple:
R> df <- data.frame(date=Sys.Date()+seq(1,by=30,len=10))
R> df[, "year"] <- format(df[,"date"], "%Y")
R> df[, "month"] <- format(df[,"date"], "%m")
R> df
date year month
1 2017-12-29 2017 12
2 2018-01-28 2018 01
3 2018-02-27 2018 02
4 2018-03-29 2018 03
5 2018-04-28 2018 04
6 2018-05-28 2018 05
7 2018-06-27 2018 06
8 2018-07-27 2018 07
9 2018-08-26 2018 08
10 2018-09-25 2018 09
R>
If you want year or month as integers, you can wrap as as.integer() around the format.
A base R option would be to remove the substring with sub and then read with read.table
df1[c('month', 'year')] <- read.table(text=sub("-\\d{2}-", ",", df1$date), sep=",")
Or using tidyverse
library(tidyverse)
separate(df1, date, into = c('month', 'day', 'year') %>%
select(-day)
Note: it may be better to convert to datetime class instead of using the string formatting.
df1 %>%
mutate(date =mdy(date), month = month(date), year = year(date))
data
df1 <- data.frame(date = c("05-21-2017", "06-25-2015"))