Group Liste Python par date

Question

Dire que j'ai une liste ressemble à ceci:

[(datetime.datetime(2013, 8, 8, 1, 20, 15), 2060), (datetime.datetime(2013, 8, 9, 1, 6, 14), 2055), (datetime.datetime(2013, 8, 9, 1, 21, 1), 2050), (datetime.datetime(2013, 8, 10, 1, 5, 49), 2050), (datetime.datetime(2013, 8, 10, 1, 19, 51), 2050), (datetime.datetime(2013, 8, 11, 2, 4, 53), 2050), (datetime.datetime(2013, 8, 12, 0, 29, 45), 2050), (datetime.datetime(2013, 8, 12, 0, 44, 13), 2050), (datetime.datetime(2013, 8, 13, 0, 34, 13), 2050), (datetime.datetime(2013, 8, 13, 0, 47, 29), 2050), (datetime.datetime(2013, 8, 14, 1, 30, 39), 2050), (datetime.datetime(2013, 8, 14, 1, 33, 51), 2050), (datetime.datetime(2013, 8, 15, 0, 41, 1), 2050), (datetime.datetime(2013, 8, 15, 0, 54, 45), 2050), (datetime.datetime(2013, 8, 16, 0, 29, 57), 1950), (datetime.datetime(2013, 8, 16, 0, 43, 11), 1950), (datetime.datetime(2013, 8, 17, 0, 27, 4), 1950), (datetime.datetime(2013, 8, 17, 0, 42, 30), 1950), (datetime.datetime(2013, 8, 18, 0, 26, 26), 1950), (datetime.datetime(2013, 8, 18, 0, 43, 11), 1950), (datetime.datetime(2013, 8, 19, 0, 41, 49), 1950), (datetime.datetime(2013, 8, 20, 1, 10, 23), 1950), (datetime.datetime(2013, 8, 20, 1, 23, 44), 1950), (datetime.datetime(2013, 8, 21, 0, 47, 25), 1950), (datetime.datetime(2013, 8, 21, 1, 0, 12), 1950), (datetime.datetime(2013, 8, 22, 0, 45, 21), 1950), (datetime.datetime(2013, 8, 22, 1, 4, 33), 1950), (datetime.datetime(2013, 8, 23, 0, 51, 27), 1950), (datetime.datetime(2013, 8, 23, 1, 6, 36), 1950), (datetime.datetime(2013, 8, 24, 0, 41, 3), 1950), (datetime.datetime(2013, 8, 24, 0, 53, 14), 1950), (datetime.datetime(2013, 8, 25, 0, 29, 24), 1950), (datetime.datetime(2013, 8, 25, 0, 42, 40), 1950), (datetime.datetime(2013, 8, 26, 0, 28, 13), 1950), (datetime.datetime(2013, 8, 26, 0, 43, 30), 1950), (datetime.datetime(2013, 8, 27, 0, 30, 1), 1950), (datetime.datetime(2013, 8, 27, 0, 43, 43), 1950), (datetime.datetime(2013, 8, 28, 0, 33, 19), 1950), (datetime.datetime(2013, 8, 28, 0, 49, 11), 1950), (datetime.datetime(2013, 8, 29, 0, 26, 49), 1950), (datetime.datetime(2013, 8, 29, 0, 41, 21), 1950), (datetime.datetime(2013, 8, 30, 0, 26, 13), 1950), (datetime.datetime(2013, 8, 30, 0, 42, 9), 1950), (datetime.datetime(2013, 8, 31, 0, 23, 40), 1950), (datetime.datetime(2013, 8, 31, 0, 39, 49), 1950), (datetime.datetime(2013, 9, 1, 0, 22, 2), 1950), (datetime.datetime(2013, 9, 1, 0, 38, 16), 1950), (datetime.datetime(2013, 9, 2, 0, 21, 2), 1950), (datetime.datetime(2013, 9, 2, 0, 36, 19), 1950), (datetime.datetime(2013, 9, 3, 0, 22, 16), 1950), (datetime.datetime(2013, 9, 3, 0, 39, 2), 1900)] 

vous pourriez voir clairement que ceci est une liste de tuple et le premier élément de chaque tuple est un horodatage. Déjà en bon format, généré par:

datetime.strptime(record[0], timeFormat) 

Et le deuxième élément est la valeur de surveillance. Cependant, il peut y avoir plusieurs enregistrements par jour. Par exemple, il existe deux enregistrements sur datetime.datetime (2013, 8, 9 ..), qui ont deux valeurs différentes 2055 et 2050. Ce que je veux, c'est réellement le maximum de chaque jour. Donc, dans ce cas. 2055 serait les seuls enregistrements pour (2013, 8, 9).

Je me demande s'il y aurait un moyen pratique en Python de le faire. Quelque chose semblable comme mysql:

select 
    date(timestamp), 
    max(value) 
from table 
group by date(timestamp) 

La déclaration de MySQL est juste pour montrer l'idée et je veux vraiment une solution de python.

Answer 1

Utilisation itertools.groupby:

>>> records = [(datetime.datetime(2013, 8, 8, 1, 20, 15), 2060), ....] 
>>> import itertools 
>>> [(dt, max(v for d, v in grp)) for dt, grp in itertools.groupby(records, key=lambda x: x[0].date())] 
[(datetime.date(2013, 8, 8), 2060), 
(datetime.date(2013, 8, 9), 2055), 
(datetime.date(2013, 8, 10), 2050), 
... 
] 

NOTE: suppose que les enregistrements sont triés. Sinon, vous devriez les trier d'abord par dates.

Answer 2

Vous pouvez utiliser collections.defaultdict (Cela fonctionne pour les deux triés et des données non triées en O(N) temps):

>>> from collections import defaultdict 
>>> lis = [(datetime.datetime(2013, 8, 8, 1, 20, 15), 2060), (datetime.datetime(2013, 8, 9, 1, 6, 14), 2055), (datetime.datetime(2013, 8, 9, 1, 21, 1), 2050), (datetime.datetime(2013, 8, 10, 1, 5, 49), 2050), (datetime.datetime(2013, 8, 10, 1, 19, 51), 2050), (datetime.datetime(2013, 8, 11, 2, 4, 53), 2050), (datetime.datetime(2013, 8, 12, 0, 29, 45), 2050), (datetime.datetime(2013, 8, 12, 0, 44, 13), 2050), (datetime.datetime(2013, 8, 13, 0, 34, 13), 2050), (datetime.datetime(2013, 8, 13, 0, 47, 29), 2050), (datetime.datetime(2013, 8, 14, 1, 30, 39), 2050), (datetime.datetime(2013, 8, 14, 1, 33, 51), 2050), (datetime.datetime(2013, 8, 15, 0, 41, 1), 2050), (datetime.datetime(2013, 8, 15, 0, 54, 45), 2050), (datetime.datetime(2013, 8, 16, 0, 29, 57), 1950), (datetime.datetime(2013, 8, 16, 0, 43, 11), 1950), (datetime.datetime(2013, 8, 17, 0, 27, 4), 1950), (datetime.datetime(2013, 8, 17, 0, 42, 30), 1950), (datetime.datetime(2013, 8, 18, 0, 26, 26), 1950), (datetime.datetime(2013, 8, 18, 0, 43, 11), 1950), (datetime.datetime(2013, 8, 19, 0, 41, 49), 1950), (datetime.datetime(2013, 8, 20, 1, 10, 23), 1950), (datetime.datetime(2013, 8, 20, 1, 23, 44), 1950), (datetime.datetime(2013, 8, 21, 0, 47, 25), 1950), (datetime.datetime(2013, 8, 21, 1, 0, 12), 1950), (datetime.datetime(2013, 8, 22, 0, 45, 21), 1950), (datetime.datetime(2013, 8, 22, 1, 4, 33), 1950), (datetime.datetime(2013, 8, 23, 0, 51, 27), 1950), (datetime.datetime(2013, 8, 23, 1, 6, 36), 1950), (datetime.datetime(2013, 8, 24, 0, 41, 3), 1950), (datetime.datetime(2013, 8, 24, 0, 53, 14), 1950), (datetime.datetime(2013, 8, 25, 0, 29, 24), 1950), (datetime.datetime(2013, 8, 25, 0, 42, 40), 1950), (datetime.datetime(2013, 8, 26, 0, 28, 13), 1950), (datetime.datetime(2013, 8, 26, 0, 43, 30), 1950), (datetime.datetime(2013, 8, 27, 0, 30, 1), 1950), (datetime.datetime(2013, 8, 27, 0, 43, 43), 1950), (datetime.datetime(2013, 8, 28, 0, 33, 19), 1950), (datetime.datetime(2013, 8, 28, 0, 49, 11), 1950), (datetime.datetime(2013, 8, 29, 0, 26, 49), 1950), (datetime.datetime(2013, 8, 29, 0, 41, 21), 1950), (datetime.datetime(2013, 8, 30, 0, 26, 13), 1950), (datetime.datetime(2013, 8, 30, 0, 42, 9), 1950), (datetime.datetime(2013, 8, 31, 0, 23, 40), 1950), (datetime.datetime(2013, 8, 31, 0, 39, 49), 1950), (datetime.datetime(2013, 9, 1, 0, 22, 2), 1950), (datetime.datetime(2013, 9, 1, 0, 38, 16), 1950), (datetime.datetime(2013, 9, 2, 0, 21, 2), 1950), (datetime.datetime(2013, 9, 2, 0, 36, 19), 1950), (datetime.datetime(2013, 9, 3, 0, 22, 16), 1950), (datetime.datetime(2013, 9, 3, 0, 39, 2), 1900)] 
>>> dic = defaultdict(list) 
for dt, val in lis: 
    dic[dt.date()].append(val) 
...  
>>> for k, v in dic.iteritems(): 
    print k, max(v) 
...  
2013-08-20 1950 
2013-08-15 2050 
2013-08-22 1950 
2013-08-09 2055 
2013-08-16 1950 
2013-08-11 2050 
2013-08-18 1950 
2013-09-03 1950 
2013-09-01 1950 
... 

Comme mentionné par @hughdbrown comme meilleur moyen serait:

>>> dic = {} 
>>> for dt, val in lis: 
...  dt = dt.date() 
...  dic[dt] = max(dic.get(dt,0), val) 
...  
>>> for k, v in dic.iteritems(): 
...  print k,v 
...  
2013-08-20 1950 
2013-08-15 2050 
2013-08-22 1950 
2013-08-09 2055 
2013-08-16 1950 
2013-08-11 2050 
2013-08-18 1950 
2013-09-03 1950 
2013-09-01 1950 
...