mvl_dtmov is my data attribute which I would like to output in dd/MM/YYYY format. Does anyone know how to define it?
DEF STREAM st1.
OUTPUT STREAM st1 TO c:\temp\movimentos.csv.
ASSIGN SESSION:DATE-FORMAT = "mdy".
FOR EACH multipos.mp_mvlj NO-LOCK:
PUT STREAM st1 UNFORMATTED multipos.mp_mvlj.mvl_dtmov ";" SKIP.
END.
OUTPUT STREAM st1 CLOSE.
Why do you set date-format to "mdy" if you want "dmy" ?
Use dmy and set year-offset to 0 to force 4 digit year
SESSION:DATE-FORMAT = "dmy".
SESSION:YEAR-OFFSET = 0.
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 have a numeric variable (year of birth) in SPSS and i would like to take the last for digits out of it. Most values are like 1988, 2001, 1948 etc. But about 250 respondents entered their year of birth like 30-2-1947, or 2-9-1984 etc. That means not all values have the same length. By taking the last 4 digits into a new variable I could create an age category for all the respondents.
How can I do that?
I tried by converting the variable to a string and using substr to get a part of the value, but I always had to choose a starting point. I want to start from the last digit and then move backwards.
Instead of using SUBSTR() you can try using RIGHT() to grab the last four digits.
* convert yob to string variable.
ALTER TYPE yob (A10).
EXE .
* use RIGHT to extract the last 4 digits and convert to numeric.
COMPUTE n_yob = NUMBER(RIGHT(yob, 4, F4)) .
EXE .
You can now use n_yob to calcuate age (ex: COMPUTE age = 2022-n_yob .). You can also use ALTER_TYPE again if you want to convert yob back to it's original type.
This code is printing float numbers in the file with this format f,ffffff (with comma) and the numbers are in a row, but I need to print it like this f.ffffff (with dot) and after each number skip a line, so each number has its own line. Any ideas on how do I do it?
CODE EDITED
module writeFiles =
let (w:float[]) = [|-1.3231725; 1.052134922; 1.23082055; 1.457748868; -0.3481141253; -0.06886428466; -1.473392229; 0.1103078722; -1.047231857; -2.641890652; -1.335060286; -0.9839854216; 0.1844535984; 3.087001584; -0.008467130841; 1.175365466; 1.637297522; 5.557832631; -0.2906445452; -0.4052301538; 1.766454088; -2.604325471; -1.807107036; -2.471407376; -2.204730614;|]
let write secfilePath=
for j in 0 .. 24 do
let z = w.[j].ToString()
File.AppendAllText(secfilePath, z)
//File.AppendAllLines(secfilePath, z)
done
There is couple things that could be done better in your code.
You're opening the file over and over again every time you add a number
z does not need to be mutable
You can pass format pattern and/or culture to ToString call
You can iterate over filterMod.y instead of for loop and array indexer access
I would probably go with something more like
module writeFiles =
let write secfilePath=
let data = filterMod.y
|> Array.map (fun x -> x.ToString(CultureInfo.InvariantCulture))
File.AppendAllLines(secfilePath, data)
It prepares an array of strings, where every number of filterMod.y gets formatted using CultureInfo.InvariantCulture, which will make it use . as decimal separator. And later on it uses AppendAllLines to write the whole array to the file at once, where every element will be written in a separate line.
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") ) )
]]
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.