Question

我想使用datetime模块（而不是时间）从时间戳中提取月份和日期，然后根据固定日期确定它是否属于给定的季节（秋季，夏季，冬季，春季）。至日和昼夜平分点。

例如，如果日期在3月21日到6月20日之间，那就是春天。无论一年如何。我希望它只是查看月和日，并忽略计算中的年份。

由于the month is not being extracted properly from my data，for this reason，我一直遇到麻烦。

Answer 1

使用年份参数可能更容易。它与你的方法没什么不同，但可能比神奇的数字更容易理解。

# get the current day of the year
doy = datetime.today().timetuple().tm_yday

# "day of year" ranges for the northern hemisphere
spring = range(80, 172)
summer = range(172, 264)
fall = range(264, 355)
# winter = everything else

if doy in spring:
  season = 'spring'
elif doy in summer:
  season = 'summer'
elif doy in fall:
  season = 'fall'
else:
  season = 'winter'

Answer 2

如果日期在3月21日到6月20日之间，那就是春天。无论一年如何。我希望它只是看月和日并忽略此计算中的年份。

#!/usr/bin/env python
from datetime import date, datetime

Y = 2000 # dummy leap year to allow input X-02-29 (leap day)
seasons = [('winter', (date(Y,  1,  1),  date(Y,  3, 20))),
           ('spring', (date(Y,  3, 21),  date(Y,  6, 20))),
           ('summer', (date(Y,  6, 21),  date(Y,  9, 22))),
           ('autumn', (date(Y,  9, 23),  date(Y, 12, 20))),
           ('winter', (date(Y, 12, 21),  date(Y, 12, 31)))]

def get_season(now):
    if isinstance(now, datetime):
        now = now.date()
    now = now.replace(year=Y)
    return next(season for season, (start, end) in seasons
                if start <= now <= end)

print(get_season(date.today()))

它是@Manuel G answer的扩展版本，可以支持任何一年。

Answer 3

这就是我最终解决它的方式。我怀疑这是最好的解决方案，但它确实有效。随意提供更好的解决方案。

import datetime

def get_season(date):
    """
    convert date to month and day as integer (md), e.g. 4/21 = 421, 11/17 = 1117, etc.
    """
    m = date.month * 100
    d = date.day
    md = m + d

    if ((md >= 301) and (md <= 531)):
        s = 0  # spring
    elif ((md > 531) and (md < 901)):
        s = 1  # summer
    elif ((md >= 901) and (md <= 1130)):
        s = 2  # fall
    elif ((md > 1130) and (md <= 0229)):
        s = 3  # winter
    else:
        raise IndexError("Invalid date")

    return s

season = get_season(dt.date())

Answer 4

您所在的半球必须加以考虑。您必须自己使用地理位置确定半球。

def season(self, HEMISPHERE):
    date = self.now()
    md = date.month * 100 + date.day

    if ((md > 320) and (md < 621)):
        s = 0 #spring
    elif ((md > 620) and (md < 923)):
        s = 1 #summer
    elif ((md > 922) and (md < 1223)):
        s = 2 #fall
    else:
        s = 3 #winter

    if not HEMISPHERE == 'north':
        s = (s + 2) % 3
    return s

Answer 5

我来到这里的目的是如何将日期映射到季节，并基于this answer，我终于通过以下方式解决了该问题：

def season_of_date(date):
    year = str(date.year)
    seasons = {'spring': pd.date_range(start='21/03/'+year, end='20/06/'+year),
               'summer': pd.date_range(start='21/06/'+year, end='22/09/'+year),
               'autumn': pd.date_range(start='23/09/'+year, end='20/12/'+year)}
    if date in seasons['spring']:
        return 'spring'
    if date in seasons['summer']:
        return 'summer'
    if date in seasons['autumn']:
        return 'autumn'
    else:
        return 'winter'

# Assuming df has a date column of type `datetime`
df['season'] = df.date.map(season_of_date)

因此，原则上，只要有datetime，它就可以使用一年。

Answer 6

我认为您可以使用pandas.Series.dt.quarter？例如，

datetime = pd.Series(pd.to_datetime(['2010-09-30', '2010-04-25', '2010-01-25', '2010-10-29', '2010-12-25']))
seasons = datetime.dt.quarter

seasons:
0    3
1    2
2    1
3    4
4    4

季节就是您想要的吗？

Answer 7

这是我通常使用的：

seasons = {'Summer':(datetime(2014,6,21), datetime(2014,9,22)),
           'Autumn':(datetime(2014,9,23), datetime(2014,12,20)),
           'Spring':(datetime(2014,3,21), datetime(2014,6,20))}

def get_season(date):
    for season,(season_start, season_end) in seasons.items():
        if date>=season_start and date<= season_end:
            return season
    else:
        return 'Winter'

Answer 8

我的评论太新了，我的编辑被拒绝了，所以这里有更正的@adsf响应代码。

def season(date, hemisphere):
    ''' date is a datetime object
        hemisphere is either 'north' or 'south', dependent on long/lat.
    '''
    md = date.month * 100 + date.day

    if ((md > 320) and (md < 621)):
        s = 0 #spring
    elif ((md > 620) and (md < 923)):
        s = 1 #summer
    elif ((md > 922) and (md < 1223)):
        s = 2 #fall
    else:
        s = 3 #winter

    if hemisphere != 'north':
        if s < 2:
            s += 2 
        else:
            s -= 2

    return s

使用datetime确定Python中给定时间戳的季节

8 个答案: