I have a CSV file with dates on the format "07/11/2022 09:21:00" - which is %d/%m/%y %H:%M:%OS. I want to read them as dates but cannot make it work.
My data:
closed$closed_at: field where the dates are stored.
I tried the following:
<- as.Date(as.character("closed$closed_at"), format = "%d/%m/%y")
<- as.Date(closed$closed_at, format="%d%m%Y")
<- as.POSIXct(closed$closed_at, format="%d-%m-%Y %H:%M:%OS") - although I am aware that it does not make sense.
All of the above brings back the result: "NA"
Related
row[0] is a date formatted such that is shows as 8/16/22 in google sheets.
And for the code,
var date = Utilities.formatDate(new Date(row[0]), SpreadsheetApp.getActiveSpreadsheet().getSpreadsheetTimeZone(), "m/d/yy");
SpreadsheetApp.getUi().alert(date);
I expected the output of 8/16/22 but instead get the following:
My end goal is to have this date be pasted into a google doc template using the following line:
body.replaceText('{{date}}', date);
And here is the output of just SpreadsheetApp.getUi().alert(new Date(row[0]));
As written in official documentation:
m is minute, use M instead.
I have a vector of dates in either dmY formats and Ymd format.
These are all dates in the last century.
From each, I need to extract just the year (Y).
I use the following code
library(lubridate)
sampleDates <- c(20100517,17052010)
result <- year(parse_date_time(x, guess_formats(as.character(x), c("Ymd","dmY"))))
result
517 2010
However, I expect something like
result
2010 2010
Here is a base R solution to your problem that takes a particular difficulty into account with your date format. Let's say you have the date 20112020, i.e. November 20th in the year 2020. For your function, it is not easy to distinguish which part of the string is the year - is it 2011 or 2020? The following code takes this difficulty into account, though let me mention that there surely must be simpler solutions.
Code
NonID <- grepl("^2", sampleDates) & (substr(sampleDates, 5, 5) == "2")
ID <- !NonID
dates_normal <- sampleDates[!NonID]
dates_special <- sampleDates[NonID]
normal_years <- as.numeric(c(substr(dates_normal, nchar(dates_normal) - 3, nchar(dates_normal)), substr(dates_normal, 1, 4)))
normal_years <- normal_years[normal_years > 1999]
special_years <- as.numeric(substr(dates_special, nchar(dates_special) - 3, nchar(dates_special)))
all_years <- c(normal_years, special_years)
all_years
> all_years
[1] 2010 2010
Explanation
First, we divide the date vector into those dates which exhibit the indistinguishability (dates_normal) and those which do not (dates_special). Then, for the normal dates, we use the substr() function to extract the first four and last four digits of the string and keep only those values which exceed 2000. For the special dates, we only keep the last four digits because the year can't be possibly included in the first four digits for this date format.
Hi I am a newbie and have a problem I have been trying to solve for weeks. I have a table imported from excel with dates in text format (because dates go back to 1700s) Most are in the format "mmmyyyy", so it is relatively easy to add "1" to the date, convert to date format, and sort in correct date order. The problem I have is that some of the dates in the table are simply "yyyy", and some are empty. I cannot find an expression that works to convert these last two to eg 1 Jan yyyy and 1 Jan 1000 within the same expression. Is this possible, or would I need to do this in two queries? Sorry if this question is very basic - I cannot find an answer anywhere.
TIA
You can do something like:
Public Function ConvertDate(Byval Expression As Variant) As Date
Dim Result As Date
If IsNull(Expression) Then
Result = DateSerial(1000, 1, 1)
ElseIf Len(Expression) = 4 Then
Result = DateSerial(Expression, 1, 1)
Else
Result = DateValue(Right(Expression, 4) & "/" & Left(Expression, 3) & "/1")
End If
ConvertDate = Result
End Function
I am trying to maintain a table using some panel data. I have all the data outputting fine, but I am having difficulty getting the correct dates to display. The method I am using is the following:
gen ymdny = date(date,"MDY"); /*<- date var from panel dataset that i import*/
sort name ymdny;
summ ymdny;
local lastdate : disp %tdM-D r(max);
local lastdate2 : disp %tdM-D (r(max)-1);
local lastw : disp %tdM-D (r(max)-7);
This would work fine if the data were daily, but the dataset I have is actually business daily (ie. missing for the weekends and bank national holidays). It seems silly but I have not been able to figure out a workaround that does the job. Ideally - there is a function that i can use to print the corresponding date to a particular value.
For example:
gen resbal_1d = round(l1.resbal,0.1);
gen dateOf = dateOf(resbal_1d); /* <- pseudocode example of what I would like */
I'm not sure what you're asking for but my guess is that you want to see a human readable form date as the output, given a numerical input. (This is your last sentence.) So simply try something like:
display %td 10
The format is important as the following shows (see help format):
display %tq 10
Same numerical input, different format, different output.
Two other examples from the manual:
* string to integer
display date("5-12-1998", "MDY")
* string to date format
display %td date("5-12-1998", "MDY")
As for your example code, I don't get what you're aiming for. In effect, you can summarize the date variable because in Stata, dates are just integers. It's legal but couldn't say if it's good form. Below a simple example.
clear all
set more off
set obs 10
gen date = _n // create the data
format date %td // give date format
list
summarize date
local onedate = r(max)
display %td `onedate'
Some references:
[U] 24 Working with dates and time
help datetime
help datetime business calendars
http://www.stata.com/support/faqs/data-management/creating-date-variables/
http://www.ats.ucla.edu/stat/stata/modules/dates.htm
(Maybe you can explain with more detail and context what it is you want.)
Edit
Your comment
I do not see how this helps with the date output. For example,
displaying r(max) - 1 on a monday will still display the sunday date.
does not explain, at all, the problems you're having with Stata's business calendars.
I'm adding what is basically an example taken from the help file I already referenced. I do this with the hope of convincing you that (re)-reading the help files is worthwhile.
*clear all
set more off
* import string dates
infile str10 sdate float x using http://www.stata-press.com/data/r13/bcal_simple
list
*----- Regular dates -----
* create elapsed dates - Stata's way of managing dates
generate rdate = date(sdate, "MD20Y")
format rdate %td
drop sdate x
list
* compute previous and next dates
generate tomorrow1 = rdate + 1
format tomorrow1 %td
generate yesterday1 = rdate - 1
format yesterday1 %td
list
*----- Business dates -----
* convert regular date to business dates
generate bdate = bofd("simple", rdate)
format bdate %tbsimple
* compute previous and next dates
generate tomorrow2 = bdate + 1
format tomorrow2 %tbsimple
generate yesterday2 = bdate - 1
format yesterday2 %tbsimple
order yesterday1 rdate tomorrow1 yesterday2 bdate tomorrow2
list
/*
The stbcal-file for simple, the calendar shown below,
November 2011
Su Mo Tu We Th Fr Sa
---------------------------
1 2 3 4 X
X 7 8 9 10 11 X
X 14 15 16 17 18 X
X 21 22 23 X X X
X 28 29 30
---------------------------
*/
Notice that if you add or substract 1 from a regular date, then business days are not taken into account. If you do the same with a business calendar date, you get what you want. Business calendars are defined by .stbcal files; the example uses a built-in calendar called simple. You maybe need to make your own .stbcal file but it is not difficult. Again, the details are in the help files.
I am using ColdFusion 9.0.1 and some database that I cannot change.
I am accessing a database that stores a date as an eight digit numeric with zero decimal places like this:
YYYYMMDD
I need to be able to read the date, add and subtract days from a date, and create new dates. I am looking for a ColdFusion solution to efficiently (not much code) to convert the date to our standard format, which is
MM/DD/YYYY
And then convert it back into the database's format for saving.
I need to code this in such a way that non-ColdFusion programmers can easily read this and use it, copy and modify it for other functions (such as adding a day to a date). So, I am not looking for the most least amount of code, but efficient and readable code.
Can you suggest anything that would make this code block more flexible, readable, or more efficient (less code)?
<cfscript>
// FORMAT DB DATE FOR BROWSER
DateFromDB = "20111116";
DatedToBrowser = createBrowserDate(DateFromDB);
writeOutput(DatedToBrowser);
function createBrowserDate(ThisDate) {
ThisYear = left(ThisDate, 4);
ThisMonth = mid(ThisDate, 4, 2);
ThisDay = right(ThisDate, 2);
NewDate = createDate(ThisYear, ThisMonth, ThisDay);
NewDate = dateFormat(NewDate, "MM/DD/YYYY");
return NewDate;
}
// FORMAT BROWSER DATE FOR DB
DateFromBrowser = "11/16/2011";
DateToDB = createDBDate(DateFromBrowser);
writeDump(DateToDB);
function createDBDate(ThisDate) {
ThisYear = year(ThisDate);
ThisMonth = month(ThisDate);
ThisDay = day(ThisDate);
NewDate = "#ThisYear##ThisMonth##ThisDay#";
return NewDate;
}
</cfscript>
First find who ever did the database and kick them in the nads...
Personally I'd Convert with sql so my code only dealt with date objects.
Select Convert(DateTime, Convert(VarChar(8),DateTimeInventedByIdjitColumn))
From SomeTable
As stated by our peers, store dates as dates.
'08/06/2011' could be 8th of june of the 6th of August depending on locale.
20111643 is a valid integer..
Not using a proper date type is just a massive collection of features and bugs that at best are waiting to happen.
You can actually rewrite each function into 1 line of code.
function createBrowserDate(ThisDate) {
return mid(ThisDate,4,2) & "/" & right(ThisDate,2) & "/" & left(ThisDate,4);
}
and
function createDBDate(ThisDate) {
return dateFormat( ThisDate, "YYYYMMDD" );
}
Don't keep dates as strings - keep dates as dates and format them when you need to.
If you can't correct the database to use actual date columns (which you should if you can), then you can use these two functions to convert to/from YYYYMMDD and a date object:
function parseYMD( YYYYMMDD )
{
if ( ! refind('^\d{8}$' , Arguments.YYYYMMDD ) )
throw "Invalid Format. Expected YYYYMMDD";
return parseDateTime
( Arguments.YYYYMMDD.replaceAll('(?<=^\d{4})|(?=\d{2}$)','-') );
}
function formatYMD( DateObj )
{
return DateFormat( DateObj , 'yyyymmdd' );
}
By using date objects it means that any level of developer can work with them, without needing to care about formatting, via built-in functions like DateAdd, DateCompare, and so on.
I'm not a regular expression fan since it's not that readable to me.
Since you're using CF9, I'd typed the argument and specify the returntype of the functions to be even more readable for the next person picking up your code.
First, right after I read the date from DB, I'd parse it to a Date object using parseDBDate()
Date function parseDBDate(required String dbDate)
{
var yyyy = left(dbDate, 4);
var mm = mid(dbDate, 4, 2);
var dd = right(dbDate, 2);
return createDate(yyyy , mm, dd);
}
Once you have the date object, you can use all those built-in Date functoin like DateAdd() or DateDiff().
Call browserDateFormat() right before you need to display it.
String function browserDateFormat(required Date date)
{
return dateFormat(date, "MM/DD/YYYY");
}
Call dBDateFormat() inside <cfqueryparam value=""> when it's time to persist to DB
String function dBDateFormat(required Date date)
{
return dateFormat(date, "YYYYMMDD");
}
One liner :)
myDateString = "20110203";
myCfDate = createObject("java","java.text.SimpleDateFormat").init("yyyyMMdd").parse(myDateString,createObject("java","java.text.ParsePosition").init(0*0));
If you want to parse different patterns, change "yyyyMMdd" to any other supported pattern.
http://download.oracle.com/javase/1.5.0/docs/api/java/text/SimpleDateFormat.html
The ParsePosition is used to say where to start parsing the string.
0*0 is shorthand for JavaCast("int",0) - in the Adobe cf engine, 0 is a string, until you apply math to it, then it becomes a Double, which the ParsePosition constructor supports. Technically, it constructs with an int, but cf is smart enough to downgrade a Double to an int.
http://download.oracle.com/javase/1.5.0/docs/api/java/text/ParsePosition.html