添加额外的信息，以拉链

Question

from itertools import groupby 
#input 
l = [['Cautus B.V.', 'plein 92', '1129008', '10', 'AVB', 'Geachte mevrouw Daa', 'Mevrouw C.P. Daa'] , 
['Cautus B.V.', 'Wei 9-11', '1019123', '10', 'AVB', 'Geachte mevrouw Daa', 'Mevrouw C.P. Daa'] , 
['Cautus B.V.', 'plein 92', '1129008', '10', 'BEDR', 'Geachte mevrouw Daa', 'Mevrouw C.P. Daa'] , 
['Cautus B.V.', 'Wei 9-11', '1019123', '10', 'BEDR', 'Geachte mevrouw Daa', 'Mevrouw C.P. Daa'] , 
['De company', 'tiellaan 42', 'KD0022232', '13', 'AVB', 'Geachte heer Tigch', 'De heer I. Tigch'] , 
['De company', 'tiellaan 42', 'KD0022232', '13', 'DAS', 'Geachte heer Tigch', 'De heer I. Tigch'] , 
['Slever ', 'klopt 42', 'KD2220115', '17', 'AVB', 'Geachte heer Slever', 'De heer T. Slever']] 
#script 
l_clean = sorted(zip(zip(*l)[1], zip(*l)[4],)) 

l_final = [(k, zip(*v)[1]) for k,v in groupby(l_clean, key = lambda x:x[0])] 

for k,v in l_final: 

    print k,list(v) 

#My output is: 

Wei 9-11 ['AVB', 'BEDR'] 

klopt 42 ['AVB'] 

plein 92 ['AVB', 'BEDR'] 

tiellaan 42 ['AVB', 'DAS'] 

问题

我的问题是，我似乎无法到其他数据添加到输出的数据我也想在输出为：

['Cautus B.V.','1019123', '10', 'Geachte mevrouw Daa', 'Mevrouw C.P. Daa'](not only for this entry but for all entry's) 

我尝试了一切更改/添加密钥等，但它似乎并没有工作。 我知道我的英语含糊不清，所以如果我需要澄清一些词或任何东西只是这样说。 已经提前。

Answer 1

我有困难的时候理解你想要做什么，但也许这将帮助：

from itertools import groupby 
import operator 
#input 
l = [['Cautus B.V.', 'plein 92', '1129008', '10', 'AVB', 'Geachte mevrouw Daa', 'Mevrouw C.P. Daa'] , 
['Cautus B.V.', 'Wei 9-11', '1019123', '10', 'AVB', 'Geachte mevrouw Daa', 'Mevrouw C.P. Daa'] , 
['Cautus B.V.', 'plein 92', '1129008', '10', 'BEDR', 'Geachte mevrouw Daa', 'Mevrouw C.P. Daa'] , 
['Cautus B.V.', 'Wei 9-11', '1019123', '10', 'BEDR', 'Geachte mevrouw Daa', 'Mevrouw C.P. Daa'] , 
['De company', 'tiellaan 42', 'KD0022232', '13', 'AVB', 'Geachte heer Tigch', 'De heer I. Tigch'] , 
['De company', 'tiellaan 42', 'KD0022232', '13', 'DAS', 'Geachte heer Tigch', 'De heer I. Tigch'] , 
['Slever ', 'klopt 42', 'KD2220115', '17', 'AVB', 'Geachte heer Slever', 'De heer T. Slever']] 
#script 
sortkey = operator.itemgetter(1,4) 
l_clean = sorted(l,key=sortkey) 

l_final = [(k, list(v)) for k,v in groupby(l_clean, key = operator.itemgetter(1))] 

for k,v in l_final: 
    info_rest = v[0][:4]+v[0][5:] 
    info_combine = map(operator.itemgetter(4),v) 
    print k,info_combine,info_rest 

基本上，我排序的所有基于特定的关键数据。这样你就不会在排序阶段丢失任何数据。然后我将groupby中的密钥更改为与新数据布局一致并打印出结果。