I have a datetime field in this format "dd/MM/yyyy, HH:mm:ss".
I nedd to get the nearest up hour using spoon tranformation options.
Any suggestion?
For example:
Original date and time object: 10/08/2021, 15:51:25
Date and Time in Integer Format: 10/08/2021, 16:00:00
Thank you in advance.
You can use the Modified JavaScript value step to round time to the nearest hour using JS and then use the Select values step to format the date.
function roundToHour(date) {
p = 60 * 60 * 1000; // milliseconds in an hour
return new Date(Math.round(date.getTime() / p ) * p);
}
var new_dt = roundToHour(dt);
var new_dt_fmt = roundToHour(dt);
Related
I have column with cell format date or time (DD.MM.YYYY HH:MM:SS) and values like
03.12.2013 14:01:49
04.12.2013 10:19:27
04.12.2013 12:44:56
04.12.2013 14:20:12
04.12.2013 18:30:21
I need those values converted to unix epoch (seconds since 1970). Somehow it feels like the values are not recognized as dates, but rather as strings. I tried different formats, had little luck with dates without time.
Operations performed on date data should be automatic provided that the cells are formatted as as a user defined DD.MM.YYYY HH:MM:SS in the 'Format' > 'Cells' > 'Numbers' tab.
If you're using the standard settings, LibreOffice Calc uses 12/30/1899 as it's default date. So the first step is getting the number of days between 12/30/1899 and 1/1/1970:
=(DATE(1970,1,1) - DATE(1899,12,30)) = 25569
Number of seconds in a day:
=(60 * 60 * 24) = 86400
If, for example, in cell A2 you have the date 03.12.2013 14:01:49. I subtract the difference between Calc's default date and the Unix Epoch we just calculated, and multiply it by the number of seconds in a day:
=(A2 - 25569) * 86400
The result is a value of 1363096909 which is the Epoch time in seconds. If you need it in milliseconds, multiply the equation by 1000.
If it's something you use a lot, you can create a custom function that does this. Go to Tools > Macros > Edit Macros, and type out the following into whichever module comes up:
REM ***** BASIC *****
Function EPOCH(date_cell)
EPOCH = (date_cell - 25569)*86400
End Function
Close the macro IDE, and now you can use your EPOCH() like any other function!
This formula worked for me, where the others above did not:
= DATE( 1970, 1, 1 ) + ( A1 / 86400 )
Is there a method in matlab to convert seconds from a known date to a standard date time format?
For example, if I have a vector of values shown as seconds from 1901/01/01, how would I convert them to a dateTime? In this case a value of 28125 would correspond to 1981/01/01. Is there an efficient method for doing this?
The numbers in your example do not make sense so it is not clear if your time is in seconds or days but since you asked for seconds I will use this.
What you want to achieve can be done using datenum function. This function returns the number of (fractional) days from 1/1/0000. So first you need to find your offset, e.g.:
offsetInDays = datenum(1901,1,1);
Next, you convert the date from seconds to days:
dateInDays = YourRequiredDateInSec * 3600 * 24;
Finally, you date is given by
RequiredDate = datestr(offsetInDays + dateInDays);
I have date & time in intervals of one hour(3600 seconds) in number format, e.g 0,3600, 7200, 10800, 14400, 18000 etc.
I have starting date and time , e.g 0 corresponds to 2005/06/01 01:00 in 'yyyy/mm/dd HH:MM' format.
I am writing this data to Excel file, so I am looking for way where I can convert time given in Hour (in seconds) to Date Time (2005/06/01 01:00, 2005/06/01 02:00 etc) before writing to excel file.
I have explored 'datenum' and 'datestr' functions but they are not useful since I can not give them customised start time i.e (0 corresponds to 2005/06/01 01:00).
May be if some one can help me to point me in right direction.
tempMatrix = [NrID time_inSec ff X Y];
tempMatrix_dataCell=num2cell(tempMatrix);
col_header={'NrID','Time','ff','X','Y'};
data_for_xls_file=[col_header; tempMatrix_dataCell];
xlswrite('My_file.xls',data_for_xls_file);
time_inSec is column with values 0, 3600, 7200, 10800 etc which need to be converted.
When I use datenum it returns 7.3246e+05 so when I add 3600 to get 2005/06/01 02:00 and pass it to datestr it returns 2015/04/10 01:00.
temp_time = datenum('2005/06/01 01:00','yyyy/mm/dd HH:MM')
This works with a given start time:
startTime = datenum('2005/06/01 01:00', 'yyyy/mm/dd HH:MM'); % Define start time.
currentTime = datenum('2005/06/01 02:00', 'yyyy/mm/dd HH:MM'); % Current time.
timePassedHours = (currentTime - startTime) * 24; % Time that has passed in hours.
display(timePassedHours); % Print the output.
I have some time as string format in my data. Can anyone help me to convert this date to milliseconds in Matlab.
This is an example how date looks like '00:26:16:926', So, that is 0 hours 26 minutes 16 seconds and 926 milliseconds. After converting this time, I need to get only milliseconds such as 1576926 milliseconds for the time that I gave above. Thank you in advance.
Why don't you try using datevec instead? datevec is designed to take in various time and date strings and it parses the string and spits out useful information for you. There's no need to use regexp or split up your string in any way. Here's a quick example:
[~,~,~,hours,minutes,seconds] = datevec('00:26:16:926', 'HH:MM:SS:FFF');
out = 1000*(3600*hours + 60*minutes + seconds);
out =
1576926
In this format, the output of datevec will be a 6 element vector which outputs the year, month, day, hours, minutes and seconds respectively. The millisecond resolution will be added on to the sixth element of datevec's output, so all you have to do is convert the fourth to sixth elements into milliseconds and add them all up, which is what is done above. If you don't specify the actual day, it just defaults to January 1st of the current year... but we're not using the date anyway... we just want the time!
The beauty with datevec is that it can accept multiple strings so you're not just limited to a single input. Simply put all of your strings into a single cell array, then use datevec in the following way:
times = {'00:26:16:926','00:27:16:926', '00:28:16:926'};
[~,~,~,hours,minutes,seconds] = datevec(times, 'HH:MM:SS:FFF');
out = 1000*(3600*hours + 60*minutes + seconds);
out =
1576926
1636926
1696926
One solution could be:
timeString = '00:26:16:926';
cellfun(#(x)str2num(x),regexp(timeString,':','split'))*[3600000;60000;1000;1]
Result:
1576926
Assuming that your date string comes in that format consistently, you could use something as simple as this:
test = '00:26:16:926';
H = str2num(test(1:2)); % hours
M = str2num(test(4:5)); % minutes
S = str2num(test(7:8)); % seconds
MS = str2num(test(10:12)); % milliseconds
totalMS = MS + 1000*S + 1000*60*M + 1000*60*60*H;
Output:
1576926.00
you can convert a single string with a date or even a vector by using datevec for conversion and the dot product
a = ['00:26:16:926' ; '08:42:12:936']
datevec(a,'HH:MM:SS:FFF') * [0 0 0 3600e3 60e3 1e3]'
ans =
1576926
31332936
How can I convert the following timestamp into the an integer form, ideally milliseconds after 1970....
s = '2014-02-11-00_40_05'
I have tried using:
out = datevec(s)
However I receive an error saying 'Too many date fields in
2014-02-11-00_40_0'
Thanks
Try datenum with a format specifier:
>> datenum(s,'yyyy-mm-dd-hh_MM_ss')
ans =
7.3564e+05
Convert to epoch:
mtime = datenum(s,'yyyy-mm-dd-hh_MM_ss');
unix_time = round(8.64e7 * (mtime - datenum('1970', 'yyyy')))
Assuming 02 in your example is month (otherwise change format string in the obvious way):
datenum('2014-02-11-00_40_05','yyyy-mm-dd-HH_MM_SS')
gives you seconds after Jan-1-0000. From that it's easy to get miliseconds after 1970:
( datenum('2014-02-11-00_40_05','yyyy-mm-dd-HH_MM_SS') - ...
datenum('1970-01-01-00_00_00','yyyy-mm-dd-HH_MM_SS') ) * 1000