GROUPBY（），給出一個空列表

Question

我執行下面的腳本：

from itertools import groupby 
from pprint import pprint as prnt 
dt = [('23271800', 0.00066790780636275307), 
('23271812', 0.0010018617095441298), 
('26112103', 0.00066790780636275307), 
('27111616', 0.0056772163540834012), 
# ... many lines deleted ... 
('40161500', 0.00040074468381765189) 
] 

agg = groupby(dt, lambda x: x[0]) 
lst = list(agg) 
lst1 = map(lambda x: (x[0], list(x[1])), lst) 
prnt(lst1) 

對於項目'23271800'它應該作爲其相應的GROUPBY項目報告[('23271800', 0.00066790780636275307)]。但是，我收到了錯誤的輸出。

[('23271800', []), 
('23271812', []), 
('26112103', []), 
('27111616', []), 
# ... many lines deleted ... 
('40161500', [('40161500', 0.00040074468381765189)])] 

需要幫助瞭解我是否在這裏做了任何不正確的事情。

PS：粘貼代碼：http://codepad.org/cCd8DfoT

Answer 1

d = [(key, list(group)) for key, group in groupby(dt, lambda x: x[0])] 
prnt(d) 

groupby將返回鍵和該組的發電機，每個發現組。

輸出

[('23271800', [('23271800', 0.0006679078063627531)]), 
('23271812', [('23271812', 0.0010018617095441298)]), 
('26112103', [('26112103', 0.0006679078063627531)]), 
('27111616', [('27111616', 0.005677216354083401)]), 
('30101600', 
    [('30101600', 1.3909064158636346e-05), ('30101600', 0.002002905238843634)]), 
('30102200', [('30102200', 0.00013358156127255062)]), 
('31100000', [('31100000', 2.1849453575689805e-05)]), 
('31161500', [('31161500', 0.0005180729752775727)]), 
('31161501', [('31161501', 0.00012902764441098641)]), 
('31161505', [('31161505', 0.013866049271881438)]), 
('31161513', [('31161513', 0.021559049445886335)]), 
('31161518', [('31161518', 0.0011596016382808651)]), 
('31161520', [('31161520', 0.022263593545425106)]), 
('31161600', [('31161600', 0.003930380552826971)]), 
('31161618', [('31161618', 0.0016029787352706075)]), 
('31161620', [('31161620', 0.0008462931211056002)]), 
('31161700', [('31161700', 0.0008833842874611101)]), 
('31161716', [('31161716', 7.067074299688881e-05)]), 
('31161717', [('31161717', 0.0014193040885208503)]), 
('31161727', [('31161727', 0.01364664212812536)]), 
('31161801', [('31161801', 0.000179280516444739)]), 
('31161900', 
    [('31161900', 1.6624352427769844e-05), ('31161900', 0.0001496191718499286)]), 
('31161904', [('31161904', 6.666007460763289e-05)]), 
('31162409', [('31162409', 0.007129527514430318)]), 
('31162800', 
    [('31162800', 0.0002625302360269781), 
    ('31162800', 0.359403893120933), 
    ('31162800', 0.2207879284986886), 
    ('31162800', 0.0002625302360269781)]), 
('31163200', 
    [('31163200', 0.00037295581888139136), 
    ('31163200', 4.1439535431265705e-05)]), 
('31163201', [('31163201', 0.011292216638533014)]), 
('31163202', 
    [('31163202', 4.5417730832667214e-05), 
    ('31163202', 4.5417730832667214e-05)]), 
('31163203', [('31163203', 0.003471418917146539)]), 
('31163204', [('31163204', 0.0002962025923869601)]), 
('31163214', [('31163214', 0.0014119501813264418)]), 
('31163215', [('31163215', 0.017772155543217604)]), 
('31171504', [('31171504', 0.05423235622453355)]), 
('31181600', [('31181600', 5.262772981769086e-05)]), 
('31181602', [('31181602', 0.00019920057382748777)]), 
('31191518', [('31191518', 0.0014972878296483697)]), 
('39121719', [('39121719', 0.0022708865416333607)]), 
('40141600', [('40141600', 5.0614113112184855e-05)]), 
('40141607', [('40141607', 0.0958030259751574)]), 
('40141616', 
    [('40141616', 0.005499007768977646), ('40141616', 0.00015275021580493458)]), 
('40141636', [('40141636', 0.0007247510239255406)]), 
('40141680', [('40141680', 0.12267031972561518)]), 
('40142000', [('40142000', 0.0002962025923869601)]), 
('40142100', [('40142100', 8.188292818389522e-05)]), 
('40142315', [('40142315', 0.00034758467473980007)]), 
('40142323', [('40142323', 0.0006308018171203779)]), 
('40161500', [('40161500', 0.0004007446838176519)])] 

Answer 2

通過itertools.groupby()返回的迭代器是需要一點技巧的使用。 As it says in the documentation：

返回的基團本身是與groupby()共享迭代底層的迭代器。由於源是共享的，因此當groupby()對象進階時，先前的組不再可見。

這意味着您必須處理每個組，因爲它是由groupby()對象生成的。如果稍後查看該組，則會發現它的內容已被跳過。例如：

>>> from itertools import groupby 
>>> groups = list(groupby('AAABBBCCC')) 
>>> groups 
[('A', <itertools._grouper object at 0x107155490>), 
('B', <itertools._grouper object at 0x1071553d0>), 
('C', <itertools._grouper object at 0x107155d50>)] 
>>> list(groups[0][1]) 
[] 

文檔說：

所以，如果以後需要這些數據，它應該被存儲爲一個列表。

例如：

>>> groups = [(key, list(group)) for key, group in groupby('AAABBBCCC')] 
>>> groups[0][1] 
['A', 'A', 'A'] 

但它通常是更好地嘗試重新組織你的代碼，以便您可以處理依次在每個組，而無需將其存儲在列表中。例如，像這樣：

for key, group in groupby('AAABBBCCC'): 
    for item in group: 
     # do something with item 

Answer 3

這聽起來像你想要一個非流式groupby操作。 itertools中的實現可能對您的應用程序是過度的。你可以試試implementation in toolz。