我必须在两个日期之间创建一个范围,在几分钟的日期之间的间隔,在Julian日期,我创建一个代码,但是花了很多时间(15分钟,前者)
我的代码是:
from astropy.time import Time
import pandas as pd
timedelta = "600s"
start = "2018-01-01"
end = "2018-06-30"
dateslist = pd.date_range(start,end, freq =timedelta ).tolist()
dates = pd.DataFrame({'col':dateslist})
dates["col2"] =""
for i in range(len(dateslist)):
#print(i," / ", len(dateslist))
dates["col2"][i] = (Time(str(dateslist[i]).replace(" ", "T"), format="fits").jd)
我尝试使用Time for for,但是收到错误
time = str(list(dates['col'])).replace("[Timestamp('","").replace(" Timestamp('","").replace("')","").replace(" ","T").split(",")
time
Time(time, format="fits")
ValueError:输入值与格式类匹配
不匹配
有没有办法快速做到这一点?
现在谢谢,
答案 0 :(得分:4)
使用DatetimeIndex.to_julian_date
:
dates["col2"] = pd.date_range(start,end, freq = timedelta).to_julian_date()
答案 1 :(得分:0)
天蝎的等效方式是:
from astropy.time import Time
import astropy.units as u
timedelta = 600 * u.s
start = "2018-01-01"
end = "2018-06-30"
dates["col2"] = np.arange(Time(start).jd, Time(end).jd, timedelta.to_value('day'))
另一种(可能更常用的玄武方式)是:
start = Time("2018-01-01")
end = Time("2018-06-30")
timedelta = 600 * u.s
dates = start + timedelta * np.arange((end - start) / timedelta)
这会为您提供一个向量Time
对象,您可以通过jd
属性将其转换为JD。