I have the following problem: I take a date (as a string data type) from the user. Now, I want to know if there is a a function in actionscript that will convert it to a date format. Right now, I am just parsing the string and concatenating the pieces back together. Ie:
changeDateString = date.getFullYear().toString() + '/' + (date.getMonth()+1).toString() + '/' + date.getDate();
But for months like May, it will return "5" and not "05". I have similar problems for days like "9" or "7." Is there something in the library that will do this for me? (For the moment, I can go ahead and manually concatenate the "0" in front, but this seems like a hassle to do.)
I know this is a simple question, but I a novice.
Thanks.
Use a date formatter for that:
http://help.adobe.com/en_US/FlashPlatform/beta/reference/actionscript/3/mx/formatters/DateFormatter.html
You configure your formatter to use the format based on the types listed and use it to output your date.
var formatter:DateFormatter = new DateFormatter();
formatter.formatString = "m/d/Y";
var example:Date = new Date(2010, 0, 5, 10, 25);
trace(formatter.format(example)); // Displays: 01/05/2010
Just use the Pattern Letter/Description grid in the docs to find the right format for your needs.
Related
How can I convert a DataFrame column of strings (in dd/mm/yyyy format) to datetime dtype?
The easiest way is to use to_datetime:
df['col'] = pd.to_datetime(df['col'])
It also offers a dayfirst argument for European times (but beware this isn't strict).
Here it is in action:
In [11]: pd.to_datetime(pd.Series(['05/23/2005']))
Out[11]:
0 2005-05-23 00:00:00
dtype: datetime64[ns]
You can pass a specific format:
In [12]: pd.to_datetime(pd.Series(['05/23/2005']), format="%m/%d/%Y")
Out[12]:
0 2005-05-23
dtype: datetime64[ns]
If your date column is a string of the format '2017-01-01'
you can use pandas astype to convert it to datetime.
df['date'] = df['date'].astype('datetime64[ns]')
or use datetime64[D] if you want Day precision and not nanoseconds
print(type(df_launath['date'].iloc[0]))
yields
<class 'pandas._libs.tslib.Timestamp'>
the same as when you use pandas.to_datetime
You can try it with other formats then '%Y-%m-%d' but at least this works.
You can use the following if you want to specify tricky formats:
df['date_col'] = pd.to_datetime(df['date_col'], format='%d/%m/%Y')
More details on format here:
Python 2 https://docs.python.org/2/library/datetime.html#strftime-strptime-behavior
Python 3 https://docs.python.org/3.7/library/datetime.html#strftime-strptime-behavior
If you have a mixture of formats in your date, don't forget to set infer_datetime_format=True to make life easier.
df['date'] = pd.to_datetime(df['date'], infer_datetime_format=True)
Source: pd.to_datetime
or if you want a customized approach:
def autoconvert_datetime(value):
formats = ['%m/%d/%Y', '%m-%d-%y'] # formats to try
result_format = '%d-%m-%Y' # output format
for dt_format in formats:
try:
dt_obj = datetime.strptime(value, dt_format)
return dt_obj.strftime(result_format)
except Exception as e: # throws exception when format doesn't match
pass
return value # let it be if it doesn't match
df['date'] = df['date'].apply(autoconvert_datetime)
Try this solution:
Change '2022–12–31 00:00:00' to '2022–12–31 00:00:01'
Then run this code: pandas.to_datetime(pandas.Series(['2022–12–31 00:00:01']))
Output: 2022–12–31 00:00:01
Multiple datetime columns
If you want to convert multiple string columns to datetime, then using apply() would be useful.
df[['date1', 'date2']] = df[['date1', 'date2']].apply(pd.to_datetime)
You can pass parameters to to_datetime as kwargs.
df[['start_date', 'end_date']] = df[['start_date', 'end_date']].apply(pd.to_datetime, format="%m/%d/%Y")
Use format= to speed up
If the column contains a time component and you know the format of the datetime/time, then passing the format explicitly would significantly speed up the conversion. There's barely any difference if the column is only date, though. In my project, for a column with 5 millions rows, the difference was huge: ~2.5 min vs 6s.
It turns out explicitly specifying the format is about 25x faster. The following runtime plot shows that there's a huge gap in performance depending on whether you passed format or not.
The code used to produce the plot:
import perfplot
import random
mdYHM = range(1, 13), range(1, 29), range(2000, 2024), range(24), range(60)
perfplot.show(
kernels=[lambda x: pd.to_datetime(x), lambda x: pd.to_datetime(x, format='%m/%d/%Y %H:%M')],
labels=['pd.to_datetime(x)', "pd.to_datetime(x, format='%m/%d/%Y %H:%M')"],
n_range=[2**k for k in range(19)],
setup=lambda n: pd.Series([f"{m}/{d}/{Y} {H}:{M}"
for m,d,Y,H,M in zip(*[random.choices(e, k=n) for e in mdYHM])]),
equality_check=pd.Series.equals,
xlabel='len(df)'
)
I am working on a project that uses a different timezone than the one I currently live in:
moment.tz.add("Asia/Seoul|LMT KST JST KST KDT KDT|-8r.Q -8u -90 -90 -9u -a0|0123141414141414135353|-2um8r.Q 97XV.Q 1m1zu kKo0 2I0u OL0 1FB0 Rb0 1qN0 TX0 1tB0 TX0 1tB0 TX0 1tB0 TX0 2ap0 12FBu 11A0 1o00 11A0|23e6")
Then formatted it as such:
const kST = moment().tz('Asia/Seoul').format("HH:mm");
Now I want to be able to subtract a year from that. Looking at examples of how to do that I found something like var foo = moment(blah, "HH:mm).subtract(1, "years")
I assumed that since I already defined the moment with the 'const kST' I could simply substitute "kST" for "moment" as such:
const firstTimeConverted = kST(firstBus, "HH:mm").subtract(1, "years");
Unfortunately that doesn't work. Any thoughts how I might do this would be much appreciated.
In your code, kST isn't a function - it's a string. The format returns a string formatted as specified.
You probably want:
const firstTimeConverted = moment.tz(firstBus, "HH:mm", `Asia/Seoul`).subtract(1, "years");
This will parse the string in your firstBus variable in HH:mm format (such as "23:45"), and interpret as belonging to the Asia/Seoul time zone on the current date there. Then it will subtract a year and return the result as a moment object. If you want a string, you would then need to call the format function.
I am using intl's NumberFormat class to format my numbers.
my pubspec.yaml is something like this:
intl:
I have this very huge number when a user type on the textfield:
1,000,000,000,000,000,000,000
Without using intl, the output is this:
1e+21
And I wanted to format it so it could be readable by a normal human being.
So I did something like this:
NumberFormat myFormat = new NumberFormat("#,##0.00", "en_US");
...
String _input = _textController.text; // user's input
double _number = double.tryParse(_input);
String _formatted = '${myFormat.format(_number)}';
This worked on numbers lesser than 1e+21. However, when the user's input is equal to or greater than 1e+21, it gives a value of:
922,337,203,685,477,580,700.00 // output of 1e+21
922,337,203,685,477,580,700,000,000,000.00 // output of 1e+30
I am completely clueless why it gave a number like that.
I would appreciate any form of help.
You can use this flutter package bignum.
I have a start month (3), start year (2004), and I have an end year (2008). I want to calculate the time in words between the start and end dates. This is what I'm trying and it's not working..
# first want to piece the start dates together to make an actual date
# I don't have a day, so I'm using 01, couldn't work around not using a day
st = (start_year + "/" + start_month + "/01").to_date
ed = (end_year + "/01/01").to_date
# the above gives me the date March 1st, 2004
# now I go about using the method
distance_of_time_in_words(st, ed)
..this throws an error, "string can't me coerced into fixnum". Anyone seen this error?
You can't just concatenate strings and numbers in Ruby. You should either convert numbers to strings as mliebelt suggested or use string interpolation like that:
st = "#{start_year}/#{start_month}/01".to_date
But for your particular case I think there is no need for strings at all. You can do it like that:
st = Date.new(start_year, start_month, 1)
ed = Date.new(end_year, 1, 1)
distance_of_time_in_words(st, ed)
or even like that:
st = Date.new(start_year, start_month)
ed = Date.new(end_year)
distance_of_time_in_words(st, ed)
See Date class docs for more information.
Given that the context in which you are calling the method is one that knows the methods from ActionView::Helpers::DateHelper, you should change the following:
# first want to piece the start dates together to make an actual date
# I don't have a day, so I'm using 01, couldn't work around not using a day
st = (start_year.to_s + "/" + start_month.to_s + "/01").to_date
ed = (end_year.to_s + "/01/01").to_date
# the above gives me the date March 1st, 2004
# now I go about using the method
distance_of_time_in_words(st, ed)
=> "almost 3 years"
So I have added calls to to_s for the numbers, to ensure that the operation + is working. There may be more efficient ways to construct a date, but yours is sufficient.
I'm trying to parse a full ISO8601 datetime from JSON data in Lua.
I'm having trouble with the match pattern.
So far, this is what I have:
-- Example datetime string 2011-10-25T00:29:55.503-04:00
local datetime = "2011-10-25T00:29:55.503-04:00"
local pattern = "(%d+)%-(%d+)%-(%d+)T(%d+):(%d+):(%d+)%.(%d+)"
local xyear, xmonth, xday, xhour, xminute,
xseconds, xmillies, xoffset = datetime:match(pattern)
local convertedTimestamp = os.time({year = xyear, month = xmonth,
day = xday, hour = xhour, min = xminute, sec = xseconds})
I'm stuck at how to deal with the timezone on the pattern because there is no logical or that will handle the - or + or none.
Although I know lua doesn't support the timezone in the os.time function, at least I would know how it needed to be adjusted.
I've considered stripping off everything after the "." (milliseconds and timezone), but then i really wouldn't have a valid datetime. Milliseconds is not all that important and i wouldn't mind losing it, but the timezone changes things.
Note: Somebody may have some much better code for doing this and I'm not married to it, I just need to get something useful out of the datetime string :)
The full ISO 8601 format can't be done with a single pattern match. There is too much variation.
Some examples from the wikipedia page:
There is a "compressed" format that doesn't separate numbers: YYYYMMDD vs YYYY-MM-DD
The day can be omited: YYYY-MM-DD and YYYY-MM are both valid dates
The ordinal date is also valid: YYYY-DDD, where DDD is the day of the year (1-365/6)
When representing the time, the minutes and seconds can be ommited: hh:mm:ss, hh:mm and hh are all valid times
Moreover, time also has a compressed version: hhmmss, hhmm
And on top of that, time accepts fractions, using both the dot or the comma to denote fractions of the lower time element in the time section. 14:30,5, 1430,5, 14:30.5, or 1430.5 all represent 14 hours, 30 seconds and a half.
Finally, the timezone section is optional. When present, it can be either the letter Z, ±hh:mm, ±hh or ±hhmm.
So, there are lots of possible exceptions to take into account, if you are going to parse according to the full spec. In that case, your initial code might look like this:
function parseDateTime(str)
local Y,M,D = parseDate(str)
local h,m,s = parseTime(str)
local oh,om = parseOffset(str)
return os.time({year=Y, month=M, day=D, hour=(h+oh), min=(m+om), sec=s})
end
And then you would have to create parseDate, parseTime and parseOffset. The later should return the time offsets from UTC, while the first two would have to take into account things like compressed formats, time fractions, comma or dot separators, and the like.
parseDate will likely use the "^" character at the beginning of its pattern matches, since the date has to be at the beginning of the string. parseTime's patterns will likely start with "T". And parseOffset's will end with "$", since the time offsets, when they exist, are at the end.
A "full ISO" parseOffset function might look similar to this:
function parseOffset(str)
if str:sub(-1)=="Z" then return 0,0 end -- ends with Z, Zulu time
-- matches ±hh:mm, ±hhmm or ±hh; else returns nils
local sign, oh, om = str:match("([-+])(%d%d):?(%d?%d?)$")
sign, oh, om = sign or "+", oh or "00", om or "00"
return tonumber(sign .. oh), tonumber(sign .. om)
end
By the way, I'm assuming that your computer is working in UTC time. If that's not the case, you will have to include an additional offset on your hours/minutes to account for that.
function parseDateTime(str)
local Y,M,D = parseDate(str)
local h,m,s = parseTime(str)
local oh,om = parseOffset(str)
local loh,lom = getLocalUTCOffset()
return os.time({year=Y, month=M, day=D, hour=(h+oh-loh), min=(m+om-lom), sec=s})
end
To get your local offset you might want to look at http://lua-users.org/wiki/TimeZone .
I hope this helps. Regards!
There is also the luadate package, which supports iso8601. (You probably want the patched version)
Here is a simple parseDate function for ISO dates. Note that I'm using "now" as a fallback. This may or may not work for you. YMMV 😉.
--[[
Parse date given in any of supported forms.
Note! For unrecognised format will return now.
#param str ISO date. Formats:
Y-m-d
Y-m -- this will assume January
Y -- this will assume 1st January
]]
function parseDate(str)
local y, m, d = str:match("(%d%d%d%d)-?(%d?%d?)-?(%d?%d?)$")
-- fallback to now
if y == nil then
return os.time()
end
-- defaults
if m == '' then
m = 1
end
if d == '' then
d = 1
end
-- create time
return os.time{year=y, month=m, day=d, hour=0}
end
--[[
--Tests:
print( os.date( "%Y-%m-%d", parseDate("2019-12-28") ) )
print( os.date( "%Y-%m-%d", parseDate("2019-12") ) )
print( os.date( "%Y-%m-%d", parseDate("2019") ) )
]]