lunar_days = get_syear_days(lyear) + get_days_of_syear(syear, smonth, sday) \ - get_days_of_syear(lyear, spring_month, spring_day) else: lunar_days = get_days_of_syear(syear, smonth, sday) \ - get_days_of_syear(syear, spring_month, spring_day) lunar_days += 1 # ...
140deflunar_year_days(year): 141days = 0 142foriinrange(1, 13): 143(high, low) = lunar_month_days(year, i) 144days += high 145days += low 146returndays 147 148defget_ludar_date(tm): 149span_days = date_diff(tm) 150 151#阳历1901年2月19日为阴历1901年正月初一 152#阳历1901...
vmin=-30,vmax=20)plt.title(f'GridMet tmin on {target_date.strftime("%Y-%m-%d")}')plt.subplot(1,2,2)ds_CPC_interp.sel(time=target_date).tmin.plot(cmap='nipy_spectral',vmin=-30,vmax=20)plt.title(f'CPC-Globe tmin on
def lunar_year_days(year): days = 0 for i in range(1, 13): (high, low) = lunar_month_days(year, i) days += high days += low return days def get_ludar_date(tm): span_days = date_diff(tm) #阳历1901年2月19日为阴历1901年正月初一 #阳历1901年1月1日到2月19日共有49天 if ...
['Test Statistic','p-value','#Lags Used','NumberofObservations Used']) for key,value in dftest[4].items(): dfoutput['CriticalValue(%s)'%key] = value return dfoutput # 自相关和偏相关图,默认阶数为31阶 def draw_acf_pacf(ts, lags=31): f = plt.figure(facecolor='white')ax1=f....
- datetime.datetime.strptime(date2, datetimeFormat) print('Difference:', diff) print('Days:', diff.days) print('Microseconds:', diff.microseconds) print('Seconds:', diff.seconds) Output: Difference: 37 days, 0:05:00.518000 Days: 37 ...
notdesired.To learn more about the frequency strings, please see `this link<https://pandas.pydata.org/pandas-docs/stable/user_guide/timeseries.html#offset-aliases>`__.Examples---Note how the two weekend days are skipped in the result.>>> pd.bdate_range(start='1/1/2018', end='1/0...
The output should be similar to 31 days, 3:00:00 This indicates there is a 31 day and 3-hour difference between the two timestamps. What if we just want to get a number back indicating a given unit of time? The variable type returned when we subtract one time from another is timede...
future = now + timedelta(days=3) # 3天后 print(future) # 输出: 2023-10-28 14:30:00.123456 datetime.date / datetime.time 分别表示纯日期或纯时间。 python today = date.today() # 当前日期 print(today) # 输出: 2023-10-25 datetime.timedelta ...
本文实例讲述了java计算时间差的方法.分享给大家供大家参考.具体如下: /** * 计算时间差 * @param begin * @param end * @return 返回格式,"hh:mm:ss" */ public String getTimeDifference(Date begin,Date end) { long between=(end.getTime()-begin.getTime())/1000; //除以1000是为了转换成秒 lo...