timedelta

SDL_GetTicks() accuracy below the millisecond level

依然范特西╮ 提交于 2019-12-01 20:15:15
I program currently something with SDL2. All works fine, but I have a problem with the SDL_GetTicks() method. Normally it should return the total application time in milliseconds, but it always returns most of the time the value 0 and sometimes the value 1. I initialized SDL with SDL_INIT_EVERYTHING flag. The problem with the following code is the loop is too fast, so the delta time is smaller than 1 ms. Is there a method to achieve a higher precision? #include "Application.hpp" void Application::Initialize() { int sdl_initialize_result = SDL_Init(SDL_INIT_EVERYTHING); if(sdl_initialize_result

Python's timedelta: can't I just get in whatever time unit I want the value of the entire difference?

被刻印的时光 ゝ 提交于 2019-12-01 15:09:10
I am trying to have some clever dates since a post has been made on my site ("seconds since, hours since, weeks since, etc..") and I'm using datetime.timedelta difference between utcnow and utc dated stored in the database for a post. Looks like, according to the docs, I have to use the days attribute AND the seconds attribute, to get the fancy date strings I want. Can't I just get in whatever time unit I want the value of the entire difference? Am I missing something? It would be perfect if I could just get the entire difference in seconds. It seems that Python 2.7 has introduced a total

Years between two date column = 'Timedelta' object has no attribute 'item'

不羁的心 提交于 2019-12-01 09:15:15
Afternoon All, Looking for the number of years between two dates to 4 decimal places. My Data: df_Years = df[ df['state'].str.contains('Done') ][[ 'maturity_date' ]].copy() df_Years['maturity_date'] = pd.to_datetime(df_Date['maturity_date']) df_Years['Today'] = pd.to_datetime('today') display(df_Years.head(6)) maturity_date Today 13 2022-12-15 2018-03-21 81 2028-02-15 2018-03-21 82 2045-12-01 2018-03-21 100 2025-08-18 2018-03-21 115 2019-01-16 2018-03-21 116 2018-12-21 2018-03-21 display(df_Years.dtypes) maturity_date datetime64[ns] Today datetime64[ns] dtype: object #Dataframe types Attempt 1

Calculate seconds from now to specified time today or tomorrow [closed]

江枫思渺然 提交于 2019-12-01 08:17:52
问题 Closed . This question needs to be more focused. It is not currently accepting answers. Want to improve this question? Update the question so it focuses on one problem only by editing this post. Closed 3 years ago . I would have thought this question would already have an answer, but it doesn't seem to. In Python, I want to calculate the number of seconds from "now" to a specified time "today" or "tomorrow," whenever it next occurs (e.g., the next occurrence of 6:30 AM). The plan is to count

Years between two date column = 'Timedelta' object has no attribute 'item'

心已入冬 提交于 2019-12-01 06:45:48
问题 Afternoon All, Looking for the number of years between two dates to 4 decimal places. My Data: df_Years = df[ df['state'].str.contains('Done') ][[ 'maturity_date' ]].copy() df_Years['maturity_date'] = pd.to_datetime(df_Date['maturity_date']) df_Years['Today'] = pd.to_datetime('today') display(df_Years.head(6)) maturity_date Today 13 2022-12-15 2018-03-21 81 2028-02-15 2018-03-21 82 2045-12-01 2018-03-21 100 2025-08-18 2018-03-21 115 2019-01-16 2018-03-21 116 2018-12-21 2018-03-21 display(df

python: which file is newer & by how much time

隐身守侯 提交于 2019-12-01 05:24:54
I am trying to create a filedate comparison routine. I suspect that the following is a rather clunky approach. I had some difficulty finding info about timedelta's attributes or methods, or whatever they are called; hence, I measured the datetime difference below only in terms of days, minutes and seconds, and there is no list item representing years. Any suggestions for an alternative, would be much appreciated. import os import datetime from datetime import datetime import sys def datetime_filedif(filepath1e, filepath2e): filelpath1 = str(filepath1e) filepath1 = str(filepath1e) filepath2 =

Delta value in render method libgdx

我只是一个虾纸丫 提交于 2019-12-01 04:02:12
I have check delta value in render method in Screen class.. I saw it is not constant. Can any body tell where it come from and what it is? And does it differ in different screen sizes? If it is so how can we overcome this? I am asking this cause my player jump depends upon delta time and sometimes it jumps too high ..... The deltaTime has nothing to do with screen sizes. It is the amount of time the last frame took to be rendered. Rendered in case of LibGDX also includes all the logic you execute in your render() method. Usually you want a game to run at the same speed on different devices. If

python: which file is newer & by how much time

白昼怎懂夜的黑 提交于 2019-12-01 03:41:43
问题 I am trying to create a filedate comparison routine. I suspect that the following is a rather clunky approach. I had some difficulty finding info about timedelta's attributes or methods, or whatever they are called; hence, I measured the datetime difference below only in terms of days, minutes and seconds, and there is no list item representing years. Any suggestions for an alternative, would be much appreciated. import os import datetime from datetime import datetime import sys def datetime

Delta value in render method libgdx

好久不见. 提交于 2019-12-01 02:29:03
问题 I have check delta value in render method in Screen class.. I saw it is not constant. Can any body tell where it come from and what it is? And does it differ in different screen sizes? If it is so how can we overcome this? I am asking this cause my player jump depends upon delta time and sometimes it jumps too high ..... 回答1: The deltaTime has nothing to do with screen sizes. It is the amount of time the last frame took to be rendered. Rendered in case of LibGDX also includes all the logic

From TimeDelta to float days in Pandas

北城余情 提交于 2019-11-30 18:49:52
I have a TimeDelta column with values that look like this: 2 days 21:54:00.000000000 I would like to have a float representing the number of days, let's say here 2+21/24 = 2.875, neglecting the minutes. Is there a simple way to do this ? I saw an answer suggesting res['Ecart_lacher_collecte'].apply(lambda x: float(x.item().days+x.item().hours/24.)) But I get "AttributeError: 'str' object has no attribute 'item' " Numpy version is '1.10.4' Pandas version is u'0.17.1' The columns has originally been obtained with: lac['DateHeureLacher'] = pd.to_datetime(lac['Date lacher']+' '+lac['Heure lacher']