Time Difference
in [Matlab]
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From: Karan on 2 Sep 2007 01:40 hi lorenzo Mailed u a copy as well but here it goes 733283 733283 733283 733283 733283 733283.0001 733283.0001 733283.0001 733283.0001 733283.0001 733283.0001 733283.0001 733283.0001 733283.0002 733283.0002 733283.0002 733283.0002 733283.0002 733283.0002 733283.0002 733283.0002 733283.0002 733283.0003 733283.0003 733283.0003 733283.0003 733283.0003 733283.0003 733283.0003 733283.0003 733283.0003 733283.0004 733283.0004 733283.0004 733283.0004 733283.0004 733283.0004 733283.0004 733283.0004 733283.0005 733283.0005 733283.0005 733283.0005 733283.0005 733283.0005 733283.0005 733283.0005 733283.0005 733283.0006 733283.0006 733283.0006 733283.0006 733283.0006 733283.0006 733283.0006 733283.0006 733283.0006 733283.0007 733283.0007 733283.0007 733283.0007 733283.0007 733283.0007 733283.0007 733283.0007 733283.0008 733283.0008 733283.0008 733283.0008 733283.0008 733283.0008 733283.0008 733283.0008 733283.0008 733283.0009 733283.0009 733283.0009 "Lorenzo " <lorenz4matlab(a)gmail.com> wrote in message <fbc9hn$67j$1(a)fred.mathworks.com>... > Can you post (part of) your initial data? >
From: Walter Roberson on 2 Sep 2007 01:57 In article <fbc1rh$6n$1(a)fred.mathworks.com>, Karan <ksumbaly(a)hotmail.com> wrote: >Have a huge vector of serial date time and i wish to >compute the difference between each element of the vector >with the next element. Basically the pseudo command is >something like this >find(diff(serial_Time_vector)>1800 seconds) >I wish to find the index numbers in the vector where the >difference is greater than 30 minutes. Whats the quickest >way to do this? thirtyserialmins = 30./60./24. %fraction of a day find(diff(serial_Time_vector) > thirtyserialmins) As long as you don't change the pivot date/time somewhere along the vector ;-) -- "It is important to remember that when it comes to law, computers never make copies, only human beings make copies. Computers are given commands, not permission. Only people can be given permission." -- Brad Templeton
From: Lorenzo on 2 Sep 2007 11:07 "Karan " <ksumbaly(a)hotmail.com> wrote in message <fbdico$a9b$1(a)fred.mathworks.com>... > hi lorenzo > > Mailed u a copy as well > > but here it goes > > 733283 > 733283 <snip> > 733283.0009 > > > "Lorenzo " <lorenz4matlab(a)gmail.com> wrote in message > <fbc9hn$67j$1(a)fred.mathworks.com>... > > Can you post (part of) your initial data? > > > Ok... from the Matlab documentation and these numbers, I gather that this is an array of values in "Serial Date Number" format, which "represents the whole and fractional number of days from a specific date and time, where datenum('Jan-1-0000 00:00:00') returns the number 1". (from the help page for datenum). So, basically, these numbers are the *days* from "beginning of time". Let's call Time the vector you posted. It can be transformed in seconds using this code: secsInADay = 24*3600; % seconds in a day Time_days = Time - Time(1); % first element is the zero time Time_sec = Time_days * secsInADay; It turns out that the file you emailed me represents 24 hrs sampled at 1 second. Is this correct? If it is, then the code above is correct and your original problem would be solved by find( diff(Time_sec) > 1800 ) Lorenzo
From: Karan on 11 Sep 2007 06:45 Thanks Lorenzo I actually took an easy way out. Its not foolproof but hey it works start_time = datevec(startTime); start_time = start_time(:,6)+start_time(:,5) *60+start_time(:,4)... *3600+start_time(:,3)*24*3600; "Lorenzo " <lorenz4matlab(a)gmail.com> wrote in message <fbejiv$gef$1(a)fred.mathworks.com>... > "Karan " <ksumbaly(a)hotmail.com> wrote in message > <fbdico$a9b$1(a)fred.mathworks.com>... > > hi lorenzo > > > > Mailed u a copy as well > > > > but here it goes > > > > 733283 > > 733283 > <snip> > > 733283.0009 > > > > > > "Lorenzo " <lorenz4matlab(a)gmail.com> wrote in message > > <fbc9hn$67j$1(a)fred.mathworks.com>... > > > Can you post (part of) your initial data? > > > > > > > Ok... from the Matlab documentation and these numbers, I gather that this is > an array of values in "Serial Date Number" format, which "represents the > whole and fractional number of days from a specific date and time, where > datenum('Jan-1-0000 00:00:00') returns the number 1". (from the help page > for datenum). > So, basically, these numbers are the *days* from "beginning of time". > Let's call Time the vector you posted. It can be transformed in seconds using > this code: > > secsInADay = 24*3600; % seconds in a day > Time_days = Time - Time(1); % first element is the zero time > Time_sec = Time_days * secsInADay; > > It turns out that the file you emailed me represents 24 hrs sampled at 1 > second. Is this correct? If it is, then the code above is correct and your > original problem would be solved by > > find( diff(Time_sec) > 1800 ) > > Lorenzo
From: us on 11 Sep 2007 07:37
Karan: <SNIP where's the time warp... one of the solutions % create some sorted dates ds=[ datestr(now:now+5) datestr(now+21:now+25) ]; % the engine % ...use <datenum> dn=datenum(ds); ix=find(diff(dn)>10); % the result, eg, ds(ix-1:ix,:) us |