Disclaimer: you might need to install
a font/typeface which supports
Japanese if you see messed up
characters.
I'm trying to replicate what I've been doing so far with setlocale and strftime:
setlocale(LC_ALL, 'ja_JP.utf8');
$time = mktime();
echo strftime('%x', $time), '<br>';
Output:
2010年01月06日
Using Zend_Date - but I haven't been able to reproduce the same formatting with the japanese symbols for year, month and day.
Attempt #1:
$locale = new Zend_Locale('ja_JP');
$date = new Zend_Date( strtotime('yesterday'), null, $locale);
//echo $date->toString('YYYY abcdefghijklmnopqrstuvwxy M dE');
echo $date->get('YYYY MMM DD');
Output:
2010 1月 004
Attempt #2:
echo $date->get(Zend_Date::DATE_FULL);
Output:
2010年1月5日火曜日
My first attempt I can't seem to find a working constant to produce the YEAR and day symbols. The latter uses a standardized format but I need to customize it so there's a 0 preceding the month, and I want to be more in control.
In the future I may want to make it flexible so for example, en_US dates won't have those letters coming after the year/month/day but it would only apply to languages such as Japanese and others, where it's more common, or if I misunderstood and it isn't really common then please inform me.
Thanks in advance.
Seems what I needed was the DATE_LONG constant, which internally points to 'FFFF' - I'm trying to learn the inner workings of how the Date class corresponds with the Locale class to generate the whole string including the symbols now.
Update: I kept trying to find where it actually used date units instead of date formats, found the right data I need:
<dateFormatLength type="long">
<dateFormat>
<pattern>y年M月d日</pattern>
</dateFormat>
</dateFormatLength>
So it parses this and replaces the y, M, d, returns the formatted date.
Related
I've looked for help on the internet for the following, but I could not find a satisfying answer: for an assignment, I need to plot the time series of a certain variable (the term spread in percentages), with years on the x-axis.
However, we use daily data. Does anybody know a convenient way in which this can be done? The 'date' variable that I've got is formulated in the following way: 20111017 represents the 17th of October 2011.
I tried to extract the first 4 numbers of the variable 'date', by using the substr(date, 1, 4) command, but the message 'type mismatch' popped up. Also, I'm not quite sure if it gives the right information if I only use the years to plot daily data (over the years). It now gives the following graph, which doesn't look that nice.
Answering the question in your title.
The date() function expects a string. If your variable with value 20111017 is in a numeric format you can convert it like this: tostring datenum , gen(datestr).
Then when using the date() function you must provide a mask that tells Stata what format the date string is in. Below is a reproducible example you can run to see how this works.
* Example generated by -dataex-. For more info, type help dataex
clear
input float datenum
20111016
end
* Convert numberic varaible to string
tostring datenum , gen(datestr)
* Convert string to date
gen date = date(datestr, "YMD")
* Display date as date
format date %td
If this does not help you, try to provide a reproducible example.
This adds some details to the helpful answer by #TheIceBear.
As he indicates, one way to get a Stata daily date from your run-together date variable is convert it to a string first. But tostring is just one way to do that and not essential. (I have nothing against tostring, as its original author, but it is better suited to other tasks.)
Here I use daily() not date(): the results are identical, but it's a good idea to use daily(): date() is all too often misunderstood as a generic date function, whereas all it does is produce daily dates (or missings).
To get a numeric year variable, just divide by 10000 and round down. You could convert to a string, extract the first 4 characters, and then convert to numeric, but that's more operations.
clear
set obs 1
gen long date = 20111017
format date %8.0f
gen ddate = daily(strofreal(date, "%8.0f"), "YMD")
format %td ddate
gen year = floor(date/10000)
list
+-----------------------------+
| date ddate year |
|-----------------------------|
1. | 20111017 17oct2011 2011 |
+-----------------------------+
* date is in %td format
gen date1 = real(string(mofd(daily(date, "DMY")), "%tmCYN"))
* type mismatch error
tostring date, gen(dt)
gen date1 = real(string(mofd(daily(dt, "DMY")), "%tmCYN"))
* the code runs but generates no results
tostring date, gen(dt)
gen date2=date(dt, "YMD")
* the code runs but generates no results
If a date variable has a display format %td it must be numeric and stored as some kind of integer. The display format is, and is only, an instruction to Stata on how to display such integers. Confusions about conversion often seem to hinge on a misunderstanding about what format means, as format is an overloaded word in computing, referring variously to file format (as in graphics file format, .png or jpg or whatever); data layout (as in wide or long layout, structure or format); variable or storage type; and (here) display format. There could well be yet other meanings.
A date displayed as 30jan2015 is stored as an integer, namely
. display mdy(1, 30, 2015)
20118
and a glance at help data types shows that your variable date could be stored as an int, float, long or double. All would work, although int is least demanding of memory. You would need (e.g.) to run describe date to find out which type is being used in your case, but nothing to come in this answer depends on knowing that type. Note that finding out what Stata is doing and thinking can be illuminated by running display with simple, single examples.
Your question is ambiguous.
Want to change display format? If you wish merely to see your dates in a display format exemplified by 20150130 then consulting help datetime display formats shows that the display format is as tested here with display, which can be abbreviated all the way down to di
. di %tdCCYYNNDD 20118
20150130
so
format date %tdCCYYNNDD
is what you need. That instructs Stata to change the display format, but the numbers stored remain precisely as they were.
Want such dates as variables held as integers? If you want the dates to be held as integers like 20150130 then you could convert it to string using the display format above, and then to a real value. A minimal sandbox dataset shows this:
. clear
. set obs 1
Number of observations (_N) was 0, now 1.
. gen date = 20118
. gen wanted = real(strofreal(date, "%tdCCYYNNDD"))
. format wanted %8.0f
. l
+------------------+
| date wanted |
|------------------|
1. | 20118 20150130 |
+------------------+
A display format such as %8.0f is needed to see such values directly.
Another method is to generate a large integer directly. You need to be explicit about a suitable storage type and (as just mentioned) need to set an appropriate format, but it can be got to work:
. gen long also = 10000 * year(date) + 100 * month(date) + day(date)
. format also %8.0f
Want such dates as variables held as strings? This is the previous solution, but leave off the real(). The default display format will work fine.
. gen WANTED = strofreal(date, "%tdCCYYNNDD")
. l
+-----------------------------+
| date wanted WANTED |
|-----------------------------|
1. | 20118 20150130 20150130 |
+-----------------------------+
I have not used tostring here but as its original author I have no bias against it. The principles needed here are better illustrated using the underlying function strofreal(). The older name string() will still work.
Turning to your code,
tostring date, gen(dt)
will just put integers like 20118 in string form, so "20118", but there is no way that Stata can understand that alone to be a daily date. You could have run tostring with a format argument, which would have been equivalent to the code above. The advantage of tostring would only be if you had several such variables you wished to convert at once, as tostring would loop over such variables for you.
I can't follow why you thought that conversion to a monthly date or use of a monthly date display format was needed or helpful, as at best you'd lose the information on day of the month. Thus at best Stata can only map a monthly date back to the first day of that month, and at worst a monthly date (here 660) could not be understood as anything you want.
. di mofd(20118)
660
. di %td mofd(20118)
22oct1961
. di %td dofm(mofd(20118))
01jan2015
There is no shortcut to understanding how Stata thinks about dates that doesn't involve reading the needed parts of help datetime and help datetime display formats.
Yet more explanation and examples can be found at https://www.stata-journal.com/article.html?article=dm0067
My collegue (who has left the company) has written a bunch of scripts, including batch and Perl scripts, and I'm getting rid of the regional settings dependencies.
In the last Perl script, he's written the following piece of code:
my $format = "%d.%m.%Y %H:%M";
my $today_converted = Time::Piece->strptime($today, $format) - ONE_HOUR - ONE_HOUR - ONE_HOUR - ONE_HOUR - ONE_HOUR;
(the idea is to get five hours before midnight of that particular date)
The value of $today seems to be "03/04/2017" (which stands for the third of April (European dateformat)), which seems not to be understood by Time::Piece implementation:
Error parsing time at C:/Perl64/lib/Time/Piece.pm line 481.
Which format can I use which is understood by Time::Piece Perl implementation?
In the format you have dots . as the date delimiter, but in the data you have slashes /. That's why it doesn't parse. It needs an exact match.
I think it's worth clarifying that strptime() will parse most date and time formats - that's the point of the method. But you need to define the format of the date string that you are parsing. That's what the second parameter to strptime() (in this case, your $format variable) is for.
The letters used in the format are taken from a standard list of definitions which used by every implementation of strptime() (and its inverse, strftime()). See man strptime on your system for a complete list of the available options.
In your case, the format is %d.%m.%Y %H:%M - which means that it will parse timestamps which have the day, month and year separated by dots, followed by a space and the hours and minutes separated by a colon. If you want to parse timestamps in a different format, then you will need to change the definition of $format.
perl code:
#!/bin/env perl
use DateTime;
print DateTime->now;
OUTPUT:
2013-01-28T06:02:33
what's mean of 'T' letter in the output string ?
ISO8601 and RFC3339 both use "T" to join the date and the time, and DateTime's default stringifier apparently adopted that convention. If you want another format, you can use one of the DateTime::Format::* modules or ->strftime.
my $now = DateTime->now( time_zone => 'local' );
say $now->strftime('%Y-%m-%d %H:%M:%S');
It stands for "Time". You can read more at:
http://en.wikipedia.org/wiki/ISO_8601#Combined_date_and_time_representations
It's the separator between date and time in ISO 8601. It's always a T in this format.
The character [T] shall be used as time designator to indicate the start of the representation of the time of day
component in these expressions. The hyphen [-] and the colon [:] shall be used, in accordance with 4.4.4, as
separators within the date and time of day expressions, respectively, when required.
NOTE By mutual agreement of the partners in information interchange, the character [T] may be omitted in
applications where there is no risk of confusing a date and time of day representation with others defined in this
International Standard.
--Data elements and interchange
formats — Information interchange —
Representation of dates and times, ISO 8601:2004(E)
Cygwin user here (though if there's a suitable solution I will carry it over to K/Ubuntu, which I also use).
I have a Welcome message in my .bashrc that looks like the following:
SAD=(`date +%A-%B-%d-%Y`)
DUB=(`date -u +%k:%M`)
printf "Today's Date is: ${SAD}.\n"
printf "Dublin time is now ${DUB}. (24-Hour Clock)\n"
After numerous attempts to use whitespaces in the variable SAD above, I gave in and used hyphens. But I am, understandably, not satisfied with this band-aid solution. The problem, as I recall, was that every time I tried using quoted space, \s or some similar escape tag, along with the variables listed in the appropriate section of the GNU manpage for date, the variable for Year was either ignored or returned an error. What I do'nt want to have to do is resort to the full string as returned by date, but rather to keep the order in which I have things in the code above.
As I write this, it occurs to me that setting the IFS around this code for the Welcome message may work, provided I return it to defaults afterwards (the above appears at lines 13-17 of a 68-line .bashrc). However, I can't recall how to do that, nor am I sure that it would work.
Generally-speaking, .bashrc files are in valid BASH syntax, aren't they? Mine certainly resemble the scripts I've either written myself or tested from other sources. All I'm missing, I suppose, is the code for setting and unsetting the field separators around this message block.
No doubt anyone who comes up with a solution will be doing a favor not only to me, but to any other relative newbies on the Net taking their first (or thirteenth through seventeenth) steps in the world of shell scripting.
BZT
Putting
SAD=$(date "+%A %B %d %Y")
DUB=$(date -u +%k:%M)
echo "Today's Date is: ${SAD}."
echo "Dublin time is now ${DUB}. (24-Hour Clock)"
in my .bash_profile prints
Today's Date is: Thursday February 18 2010.
Dublin time is now 12:55. (24-Hour Clock)
I think that's what you want.
the problem is your array declaration.
SAD=(date +%A-%B-%d-%Y) just means you are putting the string "date" as element 0, and "+%A-%B-%d-%Y" as element 1. see for yourself
$ SAD=(date +%A-%B-%d-%Y) #<-- this is an array declaration
$ echo ${SAD[0]}
date
$ echo ${SAD[1]}
+%A-%B-%d-%Y
if you want the value of "date" command to be in a variable, use $(..), eg
$ SAD=$(date +%A-%B-%d-%Y)
$ echo ${SAD}
Thursday-February-18-2010
To get spaces, you need to quote the argument to date so that it's a single string. You're also erroneously declaring SAD and DUB as arrays, when what you really meant to do was evaluate them. Try this:
[/tmp]> $(date "+%A %B %d, %Y")
Thursday February 18, 2010
date +%A\ %B\ %d\ %Y
I found the combination that works:
SAD=$(date "+%A %B %d %Y")
echo $SAD
Thursday February 18 2010
Yet another instance when:
It pays to ask a question
It helps to know where to put your double quotes.
date obviously does not know from quoted space, but Bash does, so
it's a matter of "whispering in the right ear."
Thank you ghostdog74.
BZT