I currently have date data in the following format: DDMMYYYY
I want to convert the dates to proper date function (I suspect I will need to for visualizing the temporal data) using the following:
data$DATE<-as.Date(as.character(data$DATE), "%d%m%Y")
which would be fine, however days with single digits cause me to get NA results because there is no 0 infront.
Example:
17042018 = 2018-05-10
5022018 = NA
What is a work around? Should I just paste a 0 in instances where characters is less than 8?
I am quite new to R, but if you could send me in the right direction it would be much appreciated!
Regards,
G
Pad the string with zeros till length 8, then convert:
a <- '5102017'
a <- sprintf('%08d', as.numeric(a))
as.Date(a, "%d%m%Y")
Or, in your example:
data$DATE <- as.Date(sprintf('%08d', as.numeric(data$DATE)), "%d%m%Y")
You may have to transform the initial data$DATE, depending on what you are strarting from
Related
As the title suggests, I am trying to use either lubridate or ANYTIME (or similar) to convert a time from 24 hour into 12 hour.. To make life easier I don't need the whole time converted.
What I mean is I have a column of dates in this format:
2021-02-15 16:30:33
I can use inbound$Hour <- hour(inbound$Timestamp) to grab just the hour from the Timestamp which is great.. except that it is still in 24hr time. (this creates an integer column for the hour number)
I have tried several mutates such as inbound <- inbound %>% mutate(Hour = ifelse(Hour > 12, sum(Hour - 12),Hour)
This technically works.. but I get some really wonky values (I get a -294 in several rows for example)..
is there an easier way to get the 12hr time converted?
Per recommendation below I tried to use a base FORMAT as follows:
inbound$Time <- format(inbound$Timestamp, "%H:%M:%S")
inbound$Time <- format(inbound$Time, "%I:%M:%S")
and on the second format I am getting an error
Error in format.default(inbound$Time, "%I:%M:%S") :
invalid 'trim' argument
I did notice the first format converts to a class CHARACTER column.. not sure if that is causing issues with the 2nd format or not..
I then also tried:
`inbound$time <- format(strptime(inbound$Timestamp, "%H:%M:%S"), "%I:%M %p")`
Which runs without error.. but it creates a full column of NA's
Final edit::::: I made the mistake of mis-reading/applying the solution and that caused errors.. when using the inbound$Time <- format(inbound$Time, "%I:%M:%S") or as.numeric(format(inbound$Timestamp, "%I")) from the comments... both worked and solved the issue I was having.
To be clear... From 2021-02-15 16:30:33 you want just 04:30:33 as a result?
No need for lubridate or anytime. Assuming that is a Posixct
a <- as.POSIXct("2021-02-15 16:30:33")
a
# [1] "2021-02-15 16:30:33 UTC"
b <- format(a, "%H:%M:%S")
b
#[1] "16:30:33"
c <- format(a, "%I:%M:%S")
c
#[1] "04:30:33"
When I read in data from a CSV file, the column names return in this format:
X2017.04, X2017.05, X2017.06
I'm looking to format it as (or something similar to):
April-2017, May-2017, June-2017
Currently, I've tried for loops to iterate through the entire data set and reformat everything using as.Date() or as.yearmon and some of them have worked, kinda.
as.yearmon returned 1997.333333333 and similar looking floats. The as.Date code I tried returned blank values.
I'm relatively new/novice level in R and could use some help.
Thank you!
Using as.yearmon you can try :
names(df) <- zoo::as.yearmon(names(df), 'X%Y.%m')
Or in base R pasting an arbitrary date :
names(df) <- format(as.Date(paste0(names(df), '.01'), 'X%Y.%m.%d'), '%b-%Y')
As an example :
x <- c('X2017.04', 'X2017.05', 'X2017.06')
format(as.Date(paste0(x, '.01'), 'X%Y.%m.%d'), '%b-%Y')
#[1] "Apr-2017" "May-2017" "Jun-2017"
We can use
library(lubridate)
format(ymd(sub('^X', "", x,), truncated = 2), '%b-%Y')
#[1] "Apr-2017" "May-2017" "Jun-2017"
data
x <- c('X2017.04', 'X2017.05', 'X2017.06')
I have a problem with the as.date function.
I have a list of normal date shows in the excel, but when I import it in R, it becomes numbers, like 33584. I understand that it counts since a specific day. I want to set up my date in the form of "dd-mm-yy".
The original data is:
how the "date" variable looks like in r
I've tried:
as.date <- function(x, origin = getOption(date.origin)){
origin <- ifelse(is.null(origin), "1900-01-01", origin)
as.Date(date, origin)
}
and also simply
as.Date(43324, origin = "1900-01-01")
but none of them works. it shows the error: do not know how to convert '.' to class “Date”
Thank you guys!
The janitor package has a pair of functions designed to deal with reading Excel dates in R. See the following links for usage examples:
https://www.rdocumentation.org/packages/janitor/versions/2.0.1/topics/excel_numeric_to_date
https://www.rdocumentation.org/packages/janitor/versions/2.0.1/topics/convert_to_date
janitor::excel_numeric_to_date(43324)
[1] "2018-08-12"
I've come across excel sheets read in with readxl::read_xls() that read date columns in as strings like "43488" (especially when there is a cell somewhere else that has a non-date value). I use
xldate<- function(x) {
xn <- as.numeric(x)
x <- as.Date(xn, origin="1899-12-30")
}
d <- data.frame(date=c("43488"))
d$actual_date <- xldate(d$date)
print(d$actual_date)
# [1] "2019-01-23"
Dates are notoriously annoying. I would highly recommend the lubridate package for dealing with them. https://lubridate.tidyverse.org/
Use as_date() from lubridate to read numeric dates if you need to.
You can use format() to put it in dd-mm-yy.
library(lubridate)
date_vector <- as_date(c(33584, 33585), origin = lubridate::origin)
formatted_date_vector <- format(date_vector, "%d-%m-%y")
Hi this question has been bugging me for some time.
So I am trying to convert the so-called dates in my R project into actual dates. Right now the dates are arranged in a numerical manner, ie after 2/28/2020 it's not 3/1/2020 but 2/3/2020.
I've tried the
as.Date(3/14/2020, origin = "14-03-2020")
and also
df <- data.frame(Date = c("10/9/2009 0:00:00", "10/15/2009 0:00:00"))
as.Date(df$Date, "%m/%d/%Y %H:%M:%S")
and
strDates <- c("01/28/2020", "05/03/2020")%>%
dates <- as.Date(strDates, "%m/%d/%Y")
i just plugged in two dates to test out if it works or not because there are about around 40 dates. However, my output is as follows:
Error in as.Date.default(., 3/14/2020, origin = "14-03-2020") : do not know how to convert '.' to class “Date”
for the first one and then
the second one is:
data frame not found
the third one is
Error in as.Date(strDates, "%m/%d/%Y") : object 'strDates' not found
Issues with your code:
as.Date(3/14/2020, origin = "14-03-2020")
First, R will replace 3/14/2020 with 0.000106082, since that's what 3 divided by 14 divided by 2020 equals. You need to identify it as a string using single or double quotes, as in: as.Date("3/14/2020", origin = "14-03-2020").
But that is still broken. When converting to Date, if you provide a character (string) input, then you may need to provide format=, since it needs to know which numbers in the string correspond to year, month, date, etc. If you provide a numeric (or integer) input, then you do need to provide origin=, so that it knows what "day 0" is. For unix, epoch is what you need, so origin="1970-01-01". If you're using dates from Excel, you need origin="1899-12-30" (see https://stackoverflow.com/a/43230524).
Your next error is because you are mixing magrittr ops with ... base R.
strDates <- c("01/28/2020", "05/03/2020")%>%
dates <- as.Date(strDates, "%m/%d/%Y")
The issue here has nothing to do with dates. The use of %>% on line 1 is taking the output of line 1 (in R, assignment to a variable invisibly returns the assigned numbers, which is why chaining assignment works, a <- b <- 2) and injecting it as the first argument in the next function call. With this your code was eventually interpreted as
strDates <- c("01/28/2020", "05/03/2020")%>%
{ dates <- as.Date(., strDates, "%m/%d/%Y") }
which is obviously not what you intended or need. I suspect that this is just an artifact of getting frustrated and was mid-stage converting from a %>% pipe to something else, and you forgot to clean up the %>%s. This could be
dates <- c("01/28/2020", "05/03/2020") %>%
as.Date("%m/%d/%Y")
dates
# [1] "2020-01-28" "2020-05-03"
Your data.frame code seems to work fine, though you do not assign the new Date-assigned values back to the frame. Try this slight adaptation:
df <- data.frame(Date = c("10/9/2009 0:00:00", "10/15/2009 0:00:00"))
df$Date <- as.Date(df$Date, "%m/%d/%Y %H:%M:%S")
df
# Date
# 1 2009-10-09
# 2 2009-10-15
str(df)
# 'data.frame': 2 obs. of 1 variable:
# $ Date: Date, format: "2009-10-09" "2009-10-15"
I'm trying to convert the following date/time string into a zoo object:
2004:071:15:23:41.87250
2004:103:15:24:15.35931
year:doy:hour:minute:second
The date/time string is stored in a dataframe without headers. What's the best way to go about this in R?
Cheers!
Edit based on answer by Gavin:
# read in time series from CSV file; each entry as described above
timeSeriesDates <- read.csv("timeseriesdates.csv", header = FALSE, sep = ",")
# convert to format that can be used as a zoo object
timeSeriesDatesZ <- as.POSIXct(timeSeriesDates$V1, format = "%Y:%j:%H:%M:%S")
Read the data in to R in the usual way. You will have something like the following:
dats <- data.frame(times = c("2004:071:15:23:41.87250", "2004:103:15:24:15.35931"))
dats
These can be converted to one of the POSIXt classes via:
dats <- transform(dats, as.POSIXct(times, format = "%Y:%j:%H:%M:%S"))
or
data$times <- as.POSIXct(dats$times, format = "%Y:%j:%H:%M:%S"))
which can then be used in a zoo object. See ?strftime for details on the placeholders used in the format argument; essentially %j is the day of the year placeholder.
To do the zoo bit, we would do, using some dummy data for the actual time series
ts <- rnorm(2) ## dummy data
require(zoo) ## load zoo
tsZoo <- zoo(ts, dats$times)
the last line gives:
> tsZoo
2004:071:15:23:41.87250 2004:103:15:24:15.35931
0.3503648 -0.2336064
One thing to note with fractional seconds is that i) the exact fraction you have may not be recordable using floating point arithmetic. Also, R may not show the full fractional seconds given the value of an option in R; digits.secs. See ?options for more on this particular option and how to change it.
Here's a commented example for the first string:
R> s <- "2004:103:15:24:15.35931"
R> # split on the ":" and convert the result to a numeric vector
R> n <- as.numeric(strsplit(s, ":")[[1]])
R> # Use the year, hour, minute, second to create a POSIXct object
R> # for the first of the year; then add the number of days (as seconds)
R> ISOdatetime(n[1], 1, 1, n[3], n[4], n[5])+n[2]*60*60*24
[1] "2004-04-13 16:24:15 CDT"