I'm working on a machine which has some code running on it which sets the time when I set the password. The time set is epoch time, but it has 13 digits in it, and when I wrote a simple program to get the epoch time and ran it on my personal computer running linux, it returns the epoch time which has 10 digits. Would anyone know what the extra three digits signify?
Thanks in advance
Probably seconds vs milliseconds.
You'd have to consult the specific documentation though.
Related
The following code calculates the epoch time of boot.
console.log(Math.floor(new Date() /1000) - os.uptime())
It should be constant, but there is a 1-second anomaly.
Anyone knows why?
The OS doesn't necessarily start exactly on the second boundary. While you're getting OS uptime down to the second, it may have came up part of the way through a second.
I'm making use of open trip planner using the jython scripting method explained here: http://docs.opentripplanner.org/en/latest/Scripting/
(specifically 'Using OTP as a library') and am using a script very similar to their example script
For testing purposes I have two csv files containing 40 locations each. The locations are inside the Netherlands and I have loaded both the dutch gtfs and map. The strange thing is that the code that calculates the public transport trip times (line 32 in the example script: res = spt.eval(colleges), using modes WALK,TRANSIT) takes longer when I specify a day other than today.
An example:
req.setDateTime(2018, 12, 8, 16, 00, 00) # today
spt.eval(my_data) # -> takes ~7 - 10 seconds
req.setDateTime(2018, 12, 7, 16, 00, 00) # yesterday
spt.eval(my_data) # -> takes ~30 - 40 seconds
When not setting req.setDateTime(), spt.eval() is even faster. Note that I ran the script on the 6th, for the 6th, as well, and it was fast then too, so it's certainly related to "today" and not specifically the 8th.
Of course my primary question is, how do I make it fast for days other than today? (my main interest is actually tomorrow)
Is it related to when the OTP instance is started or is it some internal optimization? I don't think it's related to the building of the graph because that was built a couple of days ago. I was looking into providing a day or datetime setting when initializing OTP but am unable to find that in the docs.
(I haven't tried messing with my system time yet, but that's also an option I'm not very fond of). Any ideas or comments are welcome. If necessary I will provide a reproducible sample tomorrow.
This problem was actually caused because of how I used req.setDateTime() in combination with req.setMaxTimeSec().
Basically, setMaxTimeSec() uses the date set by setDateTime() as a starting point, and defines a worstTime (aka the last possible time) to that date time + the maxTimeSec. However, if setDateTime() was not yet set when calling setMaxTimeSec(), the current date time is used instead. This will consequently cause problems when you happen to call setDateTime() AFTERWARDS. Example:
setMaxTimeSec(60*60) # Sets worst time to now + 1 hour
setDateTime(yesterday) # Sets departure time to yesterday
This example has a very long time window to search for solutions! Instead of only looking within an hour time, we are now looking in a window of 25 hours!
Anyway, a simple solution is to first call setDateTime(), and then setMaxTimeSec():
setDateTime(yesterday) # Sets departure time to yesterday
setMaxTimeSec(60*60) # Sets worst time to yesterday + 1 hour
Alternatively, if, for some reason, you can't switch these methods, you can always correct the setMaxTimeSec() with the time difference between now and your setDateTime()-value:
date = datetime.strptime('2019-01-08 21:00', '%Y-%m-%d %H:%M')
date_seconds = time.mktime(date.timetuple())
now_seconds = time.mktime(datetime.now().timetuple())
date_diff_seconds = int(round(date_seconds - now_seconds))
req.setMaxTimeSec(60*60 + date_diff_seconds)
req.setDateTime(date.year, date.month, date.day, date.hour, date.minute, 00)
I have been trying to find a nice way to format a timestamp of the current date & time with milliseconds using Python 3's native time library.
However there's no directive for milliseconds in the standard documentation https://docs.python.org/3/library/time.html#time.strftime.
There's undocumented directives though, like %s which will give the unix timestamp. Is there any other directives like this?
Code example:
>>> import time
>>> time.strftime('%Y-%m-%d %H:%M:%S %s')
'2017-08-28 09:27:04 1503912424'
Ideally I would just want to change the trailing %s to some directive for milliseconds.
I'm using Python 3.x.
I'm fully aware that it's quite simple to get milliseconds using the native datetime library, however I'm looking for a solution using the native time library solely.
If you insist on using time only:
miliSeconds = time.time()%1*1000
time() returns accurately the time since the epoch. Since you already have the the date up to a second, you don't really care this is a time delta, since the remaining fraction is what you need to add anyway to what you have already to get the accurate date. %1 retrieves the fraction and then I convert the numbers to millisecond by multiplying by 1000.
note
Note that even though the time is always returned as a floating point
number, not all systems provide time with a better precision than 1
second. While this function normally returns non-decreasing values, it
can return a lower value than a previous call if the system clock has
been set back between the two calls.
Taken from https://docs.python.org/3/library/time.html#time.time. But this means there is no way to do what you want. You may be able to do something more robust with process_time, but that would have to be elaborate.
In a Linux CentOS 5 machine, I am running process.sh using a cronjob at #reboot, every day (the machine gets shut off every night and turned on every morning).
process.sh takes the 'date' of the computer, and writes it to a log file, then exits.
When I check the logfile, the timestamp in it says "Tue Jan 1 13:14:38 GMT 2008"
When I go to the console of the computer and give it the 'date' command, it prints the correct date.
My best guess is that my cronjob is running BEFORE the computer sets its correct time.
Is there a way to fix this?
The battery that powers the CMOS memory on your motherboard may have run out. Try replacing it by a fresh one. It should look something like this.
This advice is based on the fact that the date of your log entry is "Jan 1 2008" which looks conspicuously like an epoch your motherboard may use. However, the time-of-day 13:14:38 is a little off for this; while the 13 hour shift can be explained if you are in the correct time zone, the nearly 15 minute offset seems odd. Unless your computer takes that long to boot to the point where cron executes your job. And of course, the reason why you eventually see the correct time is probably that ntp fixed the system time, as others have noted.
Was reading "Understanding Linux Kernel" book and in it says that, "number of microseconds is calculated by do_fast_gettimeoffset( )". Also it says that "to count the number of microseconds that have elapsed within the current second."
Couldnt understand what the author means by last sentence. Could anyone explain more on that?
If you want to understand the linux kernel, you should be aware that that book has been outdated for a long time and that do_fast_gettimeoffset no longer exists.
do_get_fast_time returns the number of seconds, and is always fast.
do_gettimeoffset returns the number of microseconds since the start of the second, and might be slow.