timedelta

Loop forever and provide delta time

半腔热情 提交于 2019-12-05 02:55:40
问题 I'm writing an HTML5 Game Development Javascript framework and I want to provide the user the difference in time between the last tick and the current one. setInterval(tick, 16.6666666); function tick() { update(); draw(); } That's what I have, but I want to have: while (true) { /* Calculate delta time */ tick(dt); } function tick(dt) { update(dt); draw(); } I tried that, using date.getTime(); to calculate delta time, but Firefox said the script crashed. Obviously, an infinite loop will crash

Displaying a timedelta object in a django template

点点圈 提交于 2019-12-05 00:46:21
I'm having trouble getting my django template to display a timedelta object consistently. I tried using the time filter in my template, but nothing is displayed when I do this. The timedelta object is shown as follows on the errors page if I use Assert False: time datetime.timedelta(0, 38, 132827) This displays the time difference as: 0:00:38.132827 I would like to only show the hours, minutes, and seconds for each timedelta object. Does anyone have any suggestions on how I can do this? I followed Peter's advice and wrote a custom template filter. Here's the steps I took. First I followed this

How to add weekly timedeltas with regards to daylight saving timezones

こ雲淡風輕ζ 提交于 2019-12-04 19:07:18
问题 I want to add or subtract weeks (or days or month or years) to localized datetime objects. The problem is, that the naive approach will result in 1 hour shifts due to daylight saving timezones. 2014-03-27 12:00 is right before the switch from winter to summer time. If I add a timedelta of one week to this date localized in timezone Europe/Berlin for example, the result will be 2014-04-03 13:00. I would like to have the same hour of day, 2014-04-03 12:00. I found a solution: from datetime

Python Pandas: TypeError: unsupported operand type(s) for +: 'datetime.time' and 'Timedelta'

独自空忆成欢 提交于 2019-12-04 12:14:24
问题 I am attempting to add two series in a dataframe in pandas with the first series being a 24-hr time value (e.g. 17:30) exported from an excel file and the second series being a series of the same length in Timedelta format converted from floats with the 'pd.Timedelta' command. The desired resulting third column would be a 24-hr time regardless of day change (e.g. 22:00 + 4 hours = 02:00). I created the Delta series like this: delta = pd.Series(0 for x in range(0, len(df.Time_In_Hours))) for j

TimeDelta java?

梦想与她 提交于 2019-12-04 07:04:53
问题 I am trying to convert code from Python to Java. I need to rewrite the timeDelta function in Java. Here is the code in Python: def timeDate(date): return (timedelta(seconds=date * 3600 % 86400)) Does anyone have any idea on how to make a function that acts the same? 回答1: double hours = 21.37865107050986; long nanos = Math.round(hours * TimeUnit.HOURS.toNanos(1)); Duration d = Duration.ofNanos(nanos); // Delete any whole days d = d.minusDays(d.toDays()); System.out.println(d); This prints:

Python 2.6.5: Divide timedelta with timedelta

Deadly 提交于 2019-12-03 23:19:48
I'm trying to divide one timedelta object with another to calculate a server uptime: >>> import datetime >>> installation_date=datetime.datetime(2010,8,01) >>> down_time=datetime.timedelta(seconds=1400) >>> server_life_period=datetime.datetime.now()-installation_date >>> down_time_percentage=down_time/server_life_period Traceback (most recent call last): File "<stdin>", line 1, in <module> TypeError: unsupported operand type(s) for /: 'datetime.timedelta' and 'datetime.timedelta' I know this has been solved in Python 3.2 , but is there a convenient way to handle it in prior versions of Python,

How do I remove the microseconds from a timedelta object?

牧云@^-^@ 提交于 2019-12-03 22:08:23
I do a calculation of average time, and I would like to display the resulted average without microseconds. avg = sum(datetimes, datetime.timedelta(0)) / len(datetimes) Hobbestigrou Take the timedetla and remove its own microseconds, as microseconds and read-only attribute: avg = sum(datetimes, datetime.timedelta(0)) / len(datetimes) avg = avg - datetime.timedelta(microseconds=avg.microseconds) You can make your own little function if it is a recurring need: import datetime def chop_microseconds(delta): return delta - datetime.timedelta(microseconds=delta.microseconds) I have not found a better

Loop forever and provide delta time

若如初见. 提交于 2019-12-03 20:44:56
I'm writing an HTML5 Game Development Javascript framework and I want to provide the user the difference in time between the last tick and the current one. setInterval(tick, 16.6666666); function tick() { update(); draw(); } That's what I have, but I want to have: while (true) { /* Calculate delta time */ tick(dt); } function tick(dt) { update(dt); draw(); } I tried that, using date.getTime(); to calculate delta time, but Firefox said the script crashed. Obviously, an infinite loop will crash. Got any suggestions for how I can go about this? I want an infinite loop, that can be stopped with

Querying timedelta column in pandas, and filtering rows

泄露秘密 提交于 2019-12-01 21:56:24
I have a column of timedelta in pandas. It is in the format x days 00:00:00. I want to filter out and flag the rows which have a value >=30 minutes. I have no clue how to do that using pandas. I tried booleans and if statements but it didn't work. Any help would be appreciated. You can convert timedelta s to seconds by total_seconds and compare with scalar: df = df[df['col'].dt.total_seconds() < 30] Or compare with Timedelta : df = df[df['col'] < pd.Timedelta(30, unit='s')] Sample : df = pd.DataFrame({'col':pd.to_timedelta(['25:10:01','00:01:20','00:00:20'])}) print (df) col 0 1 days 01:10:01

Querying timedelta column in pandas, and filtering rows

拜拜、爱过 提交于 2019-12-01 21:47:43
问题 I have a column of timedelta in pandas. It is in the format x days 00:00:00. I want to filter out and flag the rows which have a value >=30 minutes. I have no clue how to do that using pandas. I tried booleans and if statements but it didn't work. Any help would be appreciated. 回答1: You can convert timedelta s to seconds by total_seconds and compare with scalar: df = df[df['col'].dt.total_seconds() < 30] Or compare with Timedelta : df = df[df['col'] < pd.Timedelta(30, unit='s')] Sample : df =