I am trying to get the current system date in Common Lisp through following function
(defun current-date-string ()
"Returns current date as a string."
(multiple-value-bind (sec min hr day mon yr dow dst-p tz)
(get-decoded-time)
(declare (ignore sec min hr dow dst-p tz))
(format nil "~A-~A-~A" yr mon day)))
Unfortunately I am getting the current date in this format "2014-1-2". However actually I need this format "2014-01-02". Is any way we can change the format? I tried replacing nil with yyyy-mm-dd but no luck. However my machine clock shows the date format is "2014-01-02".
What you need is
(format nil "~4,'0d-~2,'0d-~2,'0d" yr mon day)
~2,'0d means:
d: decimal output (instead of your generic a)
2: 1st argument: width
'0: 2nd argument: pad char 0
I suggest that you read up on Formatted Output; format is a very powerful tool.
simple-date-time is what you need, it contains some convenient function to format date and time.
Related
The format of diary is
Mar 18, 2019
This is my diary
I want to append day of the week to it.
the color of Monday does not fit to the date,
Upon checking the documentation, there is no such a weekday format
The file’s entries are lines beginning with any of the forms
specified by the variable ‘diary-date-forms’, which by default
uses the forms of ‘diary-american-date-forms’:
MONTH/DAY
MONTH/DAY/YEAR
MONTHNAME DAY
MONTHNAME DAY, YEAR
DAYNAME
Is it possible to get the weekday inserted?
Try adding this to your Emacs config file:
(require 'calendar)
(setq diary-date-forms
(cons '(backup monthname " *" day ", *" year ", *" dayname "\\W")
diary-american-date-forms))
What this does is to make sure calendar is loaded so that diary-american-date-forms is available, and then tack on an extra date form to it. The 'backup' is necessary because the first part of the new format is the same as one of the existing ones.
What is the command for displaying the time and date in qbasic? Could the syntax for the commands be given as well? And an explanation if possible?
You can use DATE$ and TIME$
These can also set the date and time as well.
The command for printing the time(current system time) is time$
The time$ is actually a function, in this case, no parameter is needed.
And the code is...
PRINT TIME$
The time is printed in hh: mm: ss format(hour: minutes: seconds).
And therefore the output would be something like this:
14:55:28
For printing the current system date, we use date$ function which is also a string function
The code is:
PRINT DATE$
The date is printed in mm-dd-yyyy format or month-day-year(American date format).
Hence the output will be:
02-17-2018
Hope it helps...
The QB date/time functions are:
DATE$ returns the date in a string in the form MM-DD-YYYY
TIME$ returns the time in a string in the form HH:MM:SS
When used as a command the date$ and time$ can be assigned to set the system date and time, for example DATE$ = "12-10-1990" or TIME$ = "12:10:10"
If the year is a leap year then the 29th day of February could be set. Otherwise if it is not a leap year then a syntax error will occur trying to set the date in February to the 29th.
I have a problem in Stata with the format of the dates. I believe it is a very simple question but I can't see how to fix it.
I have a csv file (file.csv) that looks like
v1 v2
01/01/2000 1.1
01/02/2000 1.2
01/03/2000 1.3
...
01/12/2000 1.12
01/02/2001 1.1
...
01/12/2001 1.12
The form of v1 is dd/mm/yyyy.
I import the file in Stata using import delimited ...file.csv
v1 is a string variable, v2 is a float.
I want to transform v1 in a monthly date that Stata can read.
My attempts:
1)
gen Time = date(v1, "DMY")
format Time %tm
which gives me
Time
3177m7
3180m2
3182m7
...
that looks wrong.
2) In alternative
gen v1_1=v1
replace v1_1 = substr(v1_1,4,length(v1_1))
gen Time_1 = date(v1_1, "MY")
format Time_1 %tm
which gives exactly the same result.
And if I type
tsset Time, format(%tm)
it tells me that there are gaps but there are no gaps in the data.
Could you help me to understand what I'm doing wrong?
Stata has wonderful documentation on dates and times, which you should read from beginning to end if you plan on using time-related variables. Reading this documentation will not only solve your current problem, but will potentially prevent costly errors in the future. The section related to your question is titled "SIF-to-SIF conversion." SIF means "Stata internal form."
To explain your current issue:
Stata stores dates as numbers; you interpret them as "dates" when you assign a format. Consider the following:
set obs 1
gen dt = date("01/01/2003", "DMY")
list dt
// 15706
So that date is assigned the value 15706. Let's format it to look like a day:
format dt %td
list
// 01jan2003
Now let's format it to be a month:
format dt %tm
list
// 3268m11
Notice that dt is just a number that you can format and use like a day or month. To get a "month number" from a "day number", do the following:
gen mt = mofd(dt) // mofd = month of day
format mt %tm
list
// dt mt
// 3268m11 2003m1
The variable mt now equals 516. January 2003 is 516 months from January 1960. Stata's "epoch time" is January 1, 1960 00:00:00.000. Date variables are stored as days since the epoch time, and datetime variables are stored as miliseconds since the epoch time. A month variable can be stored as months since the epoch time (that's how the %tm formatting determines which month to show).
Hi I have a date conversion problem in SAS,
I imported an excel file which has the following dates.,
2012-01-09
2011-01-31
2010-06-28
2005-06-10
2012-09-19
2012-09-19
2007-06-12
2012-09-20
2004-11-01
2007-03-27
2008-06-23
2006-04-20
2012-09-20
2010-07-14
after I imported the dates have changed like this
40917
40574
40357
38513
41171
41171
39245
41172
38292
39168
39622
38827
41172
40373
I have used the input function to convert the dates but it gives a strange result.,
the code I used.,
want_date=input(have_date, anydtdte12.);
informat want_date date9.; format have_date date9.;run;
I get very stange and out of the World dates., any idea how can I convert these?
You can encourage SAS to convert the data as date during the import, although this isn't necessarily a panacea.
proc import file=whatever out=whatever dbms=excel replace;
dbdsopts=(dbSasType=( datevar=date ) );
run;
where datevar is your date column name. This tells SAS to expect this to be a date and to try to convert it.
See So Your Data Are in Excel for more information, or the documentation.
From : http://www2.sas.com/proceedings/sugi29/068-29.pdf
Times are counted internally in SAS as seconds since midnight and
date/time combinations are calculated as the number of seconds since
midnight 1 January 1960.
Excel also uses simple numerical values for dates and times
internally. For the date values the difference with the SAS date is
only the anchor point. Excel uses 1 January 1900 as day one.
So add a constant.
EXAMPLES:
SAS_date = Excel_date - 21916;
SAS_time = Excel_time * 86400;
SAS_date_time = (Excel_date_time - 21916) * 86400;
As Justin wrote you need to correct for the different zero date (SAS vs. Excel).
Then you just need to apply a format (if you want to get a date variable to do calculations):
want_date = have_date-21916;
format want_date date9.;
Or convert it to a string:
want_date = put(have_date-21916, date9.);
In either case you can choose the date format you prefer.
When trying to format durations, Emacs returns unexpected values for the hour. Here, I print a zero-length duration.
(format-time-string "%H:%M:%S" (seconds-to-time 0))
returns 19:00:00. Why isn't it 00:00:00?
This is because of time zones. From the format-time-string documentation:
The third, optional, argument UNIVERSAL, if non-nil, means describe TIME
as Universal Time; nil means describe TIME in the local time zone.
Emacs is treating that 0 as 00:00:00 UTC, but defaulting to convert that into local time. To not convert into local time, pass a non-nil value as the UNIVERSAL argument:
(format-time-string "%H:%M:%S" (seconds-to-time 0) t)
returns 00:00:00