如何使用熊猫季节性地分组4年的数据

时间:2017-06-07 15:05:00

标签: python-2.7 pandas group-by

我有一个包含4年数据的csv文件,我需要在4年内对每个季节的数据进行分组:这里有我的数据:

timestamp,heure,lat,lon,impact,type
2006-01-01 00:00:00,13:58:43,33.837,-9.205,10.3,1
2006-01-02 00:00:00,00:07:28,34.5293,-10.2384,17.7,1
2007-02-01 00:00:00,23:01:03,35.0617,-1.435,-17.1,2
2007-02-02 00:00:00,01:14:29,36.5685,0.9043,36.8,1
2008-01-01 00:00:00,05:03:51,34.1919,-12.5061,-48.9,1
2008-01-02 00:00:00,05:03:51,34.1919,-12.5061,-48.9,1
....
2011-12-31 00:00:00,05:03:51,34.1919,-12.5061,-48.9,1

这是我想要的输出:

winter     (the mean value of impacts)
summer     (the mean value of impacts)
autumn      ....
spring      .....

所以我期待在4个赛季中总共4行总结。 我从下面开始:

data['impact'] = data['impact'].abs()
yearly = data.groupby(data.index.month)['impact'].mean()

任何想法??

2 个答案:

答案 0 :(得分:2)

粗略的月份......假设时间戳在索引中。

mlist = [[12, 1, 2], [3, 4, 5], [6, 7, 8], [9, 10, 11]]
slist = ['winter', 'spring', 'summer', 'autum']
sdict = {k: v for v, ks in zip(slist, mlist) for k in ks}

df.groupby(df.index.month.map(sdict.get)).impact.mean()

设置

import pandas as pd
from io import StringIO

txt = """timestamp,heure,lat,lon,impact,type
2006-01-01 00:00:00,13:58:43,33.837,-9.205,10.3,1
2006-01-02 00:00:00,00:07:28,34.5293,-10.2384,17.7,1
2007-02-01 00:00:00,23:01:03,35.0617,-1.435,-17.1,2
2007-02-02 00:00:00,01:14:29,36.5685,0.9043,36.8,1
2008-01-01 00:00:00,05:03:51,34.1919,-12.5061,-48.9,1
2008-01-02 00:00:00,05:03:51,34.1919,-12.5061,-48.9,1
2011-12-31 00:00:00,05:03:51,34.1919,-12.5061,-48.9,1
"""

df = pd.read_csv(StringIO(txt), parse_dates=[0], index_col=0)

答案 1 :(得分:1)

确切日期

import pandas as pd
spring = range(80, 172)
summer = range(172, 264)
fall = range(264, 355)

def season(x):
    if x in spring:
        return 'Spring'
    if x in summer:
        return 'Summer'
    if x in fall:
        return 'Fall'
    else :
        return 'Winter'

df = pd.DataFrame({'_date' :pd.date_range(start=pd.datetime(2016,1,1), end=pd.datetime(2016,12,31), freq='D'),'impact' : range(0,366)})    

df['SEASON'] = df['_date'].dt.dayofyear.apply(lambda x : season(x))
df.groupby('SEASON')['impact'].mean()