I have some data files with Unix timestamps (in this case, number of milliseconds since Jan 1, 1970 00:00 UTC). I would like to convert these to human-friendly date/time strings (e.g. 31-Aug-2012 11:36:24) in Matlab. Is there an easy way to do this in Matlab, or am I better off using an external library (e.g. java.text.SimpleDateFormat)?
How about
date = datestr(unix_time/86400 + datenum(1970,1,1))
if unix_time is given in seconds, unix_time/86400 will give the number of days since Jan. 1st 1970. Add to that the offset used by Matlab's datenum (datenum(0000,1,1) == 1), and you have the amount of days since Jan. 1st, 0000. This can be easily converted to human-readable form by Matlab's datestr.
If you have milliseconds, just use
date = datestr(unix_time/86400/1000 + datenum(1970,1,1))
Wrapped in functions, these would be
function dn = unixtime_to_datenum( unix_time )
dn = unix_time/86400 + 719529; %# == datenum(1970,1,1)
end
function dn = unixtime_in_ms_to_datenum( unix_time_ms )
dn = unix_time_ms/86400000 + 719529; %# == datenum(1970,1,1)
end
datestr( unixtime_to_datenum( unix_time ) )
Newer versions of MATLAB (verified in R2015a) have a datetime type that is useful for working with and formatting dates and times. You can convert UNIX timestamps into a MATLAB datetime with
dt = datetime( unix_time, 'ConvertFrom', 'posixtime' );
and then use datestr as before for formatting as a string
datestr( 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 way to convert time= 08/10/2014 23:34:02 to Epoch seconds (array of numbers) in MATLAB?
So you want the Unix standard which can be calculated as follows:
InputDate=datenum('20141008 233402','yyyymmdd HHMMSS');
UnixOrigin=datenum('19700101 000000','yyyymmdd HHMMSS');
EpochSecond=round((InputDate-UnixOrigin)*86400);
>> 1412811242
EDIT for the OP's date format:
MYSTRING = '08/10/2014 23:34:02';
InputDate = datenum(MYSTRING,'dd/mm/yyyy HH:MM:SS');
UnixOrigin=datenum('19700101 000000','yyyymmdd HHMMSS'); %//This can stay the same, unless you want to change it for consistency.
EpochSecond=round((InputDate-UnixOrigin)*86400);
>>1412811242
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
I am looking for helping doing time conversions from UTC time to string using MATLAB.
I am trying to extract time from a data file collected at the end of October 2010.
The data file says it is reporting in UTC time and the field is an integer string value in milliseconds that is around 3.02e11. I would like to convert this to a string but am have some trouble.
I figured out that the units are most definitely in milliseconds so I convert this to fractions of days to be compatible with datenum format.
If the data was collected at the end of October (say, October 31, 2010) then I can guess what kind of number I might get. I thought that January 1, 2001 would be a good epoch and calculated what sort of number (in days) I might get:
suspectedDate = datenum('October 31, 2010')
suspectedEpoch = datenum('January 1, 2001')
suspectedTimeInDays = suspectedDate - suspectedEpoch
Which comes out as 3590.
However, my actual time, in days, comes out with the following code
actualTime = 3.02e11
actualTimeInDays = 3.02e11/1000/24/3600
as 3495.4.
This is troubling as the difference is only 94.6 -- not a full year. This would mean either the documentation for the file is wrong or the epoch is close to April 1-5, 2001:
calculatedEpoch = suspectedDate - actualTimeInDays
calculatedEpochStr = datestr(calculatedEpoch)
Alternately, if the epoch is January 1, 2001 then the actual date in the file is from the end of July.
ifEpochIsJanuaryDate = suspectedEpoch + actualTimeInDays
ifEpochIsJanuaryDateStr = datestr(ifEpochIsJanuaryDate)
Is this a known UTC format and can anyone give suggestions on how to get an October date from 3.02e11 magnitude number?
Unix time today is about 13e11, and is measured in ms since 1970.
If your time is ~3e11, then it's probably since year 2000.
>> time_unix = 1339116554872; % example time
>> time_reference = datenum('1970', 'yyyy');
>> time_matlab = time_reference + time_unix / 8.64e7;
>> time_matlab_string = datestr(time_matlab, 'yyyymmdd HH:MM:SS.FFF')
time_matlab_string =
20120608 00:49:14.872
Notes:
1) change 1970 into 2000 if your time is since 2000;
2) See the definition of matlab's time.
3) 8.64e7 is number of milliseconds in a day.
4) Matlab does not apply any time-zone shifts, so the result is the same UTC time.
5) Example for backward transformation:
>> matlab_time = now;
>> unix_time = round(8.64e7 * (matlab_time - datenum('1970', 'yyyy')))
unix_time =
1339118367664
You can't just make up your own epoch. Also datenum returns things in days. So the closeness you got with doing your math was just a coincidence.
Turns out that
>> datenum('Jan-1-0000')
ans =
1
and
>> datenum('Jan-1-0001')
ans =
367
So Matlab should be returning things in days since Jan. 1, 0000. (Not a typo)
However, I'd look carefully at this 3.02e11 number and find out exactly what it means. I'm pretty sure it's not standard Unix UTC, which should be seconds since January 1, 1970. It's way too big. It's close to GMT: Mon, 1 Jan 11540 08:53:20 UTC.
I have a dataset for which I have extracted the date at which an event occurred. The date is in the format of MMDDYY although MatLab does not show leading zeros so often it's MDDYY.
Is there a method to find the mean or median (I could use either) date? median works fine when there is an odd number of days but for even numbers I believe it is averaging the two middle ones which doesn't produce sensible values. I've been trying to convert the dates to a MatLab format with regexp and put it back together but I haven't gotten it to work. Thanks
dates=[32381 41081 40581 32381 32981 41081 40981 40581];
You can use datenum to convert dates to a serial date number (1 at 01/01/0000, 2 at 02/01/0000, 367 at 01/01/0001, etc.):
strDate='27112011';
numDate = datenum(strDate,'ddmmyyyy')
Any arithmetic operation can then be performed on these date numbers, like taking a mean or median:
mean(numDates)
median(numDates)
The only problem here, is that you don't have your dates in a string type, but as numbers. Luckily datenum also accepts numeric input, but you'll have to give the day, month and year separated in a vector:
numDate = datenum([year month day])
or as rows in a matrix if you have multiple timestamps.
So for your specified example data:
dates=[32381 41081 40581 32381 32981 41081 40981 40581];
years = mod(dates,100);
dates = (dates-years)./100;
days = mod(dates,100);
months = (dates-days)./100;
years = years + 1900; % set the years to the 20th century
numDates = datenum([years(:) months(:) days(:)]);
fprintf('The mean date is %s\n', datestr(mean(numDates)));
fprintf('The median date is %s\n', datestr(median(numDates)));
In this example I converted the resulting mean and median back to a readable date format using datestr, which takes the serial date number as input.
Try this:
dates=[32381 41081 40581 32381 32981 41081 40981 40581];
d=zeros(1,length(dates));
for i=1:length(dates)
d(i)=datenum(num2str(dates(i)),'ddmmyy');
end
m=mean(d);
m_str=datestr(m,'dd.mm.yy')
I hope this info to be useful, regards
Store the dates as YYMMDD, rather than as MMDDYY. This has the useful side effect that the numeric order of the dates is also the chronological order.
Here is the pseudo-code for a function that you could write.
foreach date:
year = date % 100
date = (date - year) / 100
day = date % 100
date = (date - day) / 100
month = date
newdate = year * 100 * 100 + month * 100 + day
end for
Once you have the dates in YYMMDD format, then find the median (numerically), and this is also the median chronologically.
You see above how to present dates as numbers.
I will add no your issue of finding median of the list. The default matlab median function will average the two middle values when there are an even number of values.
But you can do it yourself! Try this:
dates; % is your array of dates in numeric form
sdates = sort(dates);
mediandate = sdates(round((length(sdates)+1)/2));