days

可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效，请关闭广告屏蔽插件后再试): 问题: I am having an issue trying to get my date ticks rotated in matplotlib. A small sample program is below. If I try to rotate the ticks at the end, the ticks do not get rotated. If I try to rotate the ticks as shown under the comment 'crashes', then matplot lib crashes. This only happens if the x-values are dates. If I replaces the variable dates with the variable t in the call to avail_plot , the xticks(rotation=70) call works just fine inside avail_plot . Any ideas? import numpy as np import matplotlib.pyplot as plt import datetime as dt def

可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效，请关闭广告屏蔽插件后再试): 问题: Quite simply, how would I find out if a specific file in a folder is older than X hours? The file name and location will always be the same. Thanks 回答1: On comparing file times with batch code following must be taken into account: Date and time format settings File system of storage media Time zone, daylight saving time Automatic adjustment for daylight saving time Version of Windows 1. Date and time format settings There are the environment variables DATE and TIME which on access (and not on start of batch execution) return current date and

可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效，请关闭广告屏蔽插件后再试): 问题: I have a date "10/10/11(m-d-y)" and I want to add 5 days to it using a Python script. Please consider a general solution that works on the month ends also. I am using following code: import re from datetime import datetime StartDate = "10/10/11" Date = datetime.strptime(StartDate, "%m/%d/%y") print Date -> is printing '2011-10-10 00:00:00' Now I want to add 5 days to this date. I used the following code: EndDate = Date.today()+timedelta(days=10) Which returned this error: name 'timedelta' is not defined 回答1: The previous answers are correct

可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效，请关闭广告屏蔽插件后再试): 问题: I'm a Python newbie (2 weeks) and I'm having trouble formatting a datetime.timedelta object. Here's what I'm trying to do: I have a list of objects and one of the members of the class of the object is a timedelta object that shows the duration of an event. I would like to display that duration in the format of hours:minutes. I have tried a variety of methods for doing this and I'm having difficulty. My current approach is to add methods to the class for my objects that return hours and minutes. I can get the hours by dividing the timedelta

可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效，请关闭广告屏蔽插件后再试): 问题: If i had an array of days names and i wanted to check for example if sunday - first letter capital or small - in this array what would be the best thing to do ? 回答1: You may also use Array.indexOf : var days = ["monday", "tuesday", "wednesday", "thursday", "friday", "saturday", "sunday"]; function isInArray(days, day) { return days.indexOf(day.toLowerCase()) > -1; } isInArray(days, "Sunday"); // true isInArray(days, "sunday"); // true isInArray(days, "sUnDaY"); // true isInArray(days, "Anyday"); // false Check the browser compatibility in

可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效，请关闭广告屏蔽插件后再试): 问题: This is my first time using netCDF and I'm trying to wrap my head around working with it. I have multiple version 3 netcdf files (NOAA NARR air.2m daily averages for an entire year). Each file spans a year between 1979 - 2012. They are 349 x 277 grids with approximately a 32km resolution. Data was downloaded from here . The dimension is time (hours since 1/1/1800) and my variable of interest is air. I need to calculate accumulated days with a temperature Day 1 = +4 degrees, accumulated days = 0 Day 2 = -1 degrees, accumulated days = 1 Day 3

可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效，请关闭广告屏蔽插件后再试): 问题: I am trying to convert a timestamp like this: 2015-06-27T09:34:22+00:00 to a time since format so it would say like 9 months ago 1 day 2 hours 30 minutes 2 seconds. something like that. I used time.Parse and time.Since to get to this: 6915h7m47.6901559s But how do I convert from there on? Something like this is what I thought: for hours > 24 { days++ hours -= 24 } But the issue with this is that this won't be accurate for months because months can have 28, 30 and 31 days. Is there a better way of achieving what I want? 回答1: The days in a

可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效，请关闭广告屏蔽插件后再试): 问题: I am getting the current date (in format 12/31/1999 i.e. mm/dd/yyyy) as using the below code: Textview txtViewData; txtViewDate.setText("Today is " + android.text.format.DateFormat.getDateFormat(this).format(new Date())); and I am having another date in format as: 2010-08-25 (i.e. yyyy/mm/dd) , so I want to find the difference between date in number of days, how do I find difference in days? (In other words, I want to find the difference between CURRENT DATE - yyyy/mm/dd formatted date ) 回答1: Not really a reliable method, better of using

可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效，请关闭广告屏蔽插件后再试): 问题: Hi i am using jquery-ui datepicker to select date and date.js to find difference between 2 dates. Right now the problem is I want to exclude weekend days from calculation (saturday and sunday). How should i do that? For example the user select start date (13/8/2010) and end date (16/8/2010). Since 14/8/2010 and 15/8/2010 is in week days, instead of 4 days total, i want it to be only 2 days. This is the code im using right now: 回答1: Maybe someone else can help you converting this function into JQuery's framework... In raw javascript i will

可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效，请关闭广告屏蔽插件后再试): 问题: How do I count the number of days between these two dates? start_date = Date.parse "2012-03-02 14:46:21 +0100" end_date = Date.parse "2012-04-02 14:46:21 +0200" 回答1: With the Date (and DateTime) classes you can do (end_date - start_date).to_i to get the number of days difference. 回答2: Assuming that end_date and start_date are both of class ActiveSupport::TimeWithZone in Rails, then you can use: (end_date.to_date - start_date.to_date).to_i 回答3: Rails has some built in helpers that might solve this for you. One thing to keep in mind is that