Question

 [['Demo-Site', '10.227.209.139'], ['Demo-Site', '10.227.215.68'], 
['Demo-Site', '172.18.74.146'], ['Site', '10.152.114.65'], 
['Site', '10.227.211.244'], ['Demo-Site', '10.227.147.98'], 
 ['test', '172.18.74.146']]

如果第一个索引是相同的，我如何连接所有IP以形成一个大字符串？我是否制作了默认词典？

应该是：

["Site", "10.227.211.244, 10.152.114.65"]

Answer 1

您可以使用普通字典setdefault方法：

dic={}
for ele in lis:
    dic.setdefault(ele[0],[]).append(ele[1])
[[a,','.join(b)] for a,b in dic.items()]

Answer 2

您可以使用基于defaultdict的解决方案：

In [1]: list_of_ips = [['Demo-Site', '10.227.209.139'], ['Demo-Site', '10.227.215.68'], 
   ...: ['Demo-Site', '172.18.74.146'], ['test', '10.152.114.65'], 
   ...: ['Site', '10.227.211.244'], ['Demo-Site', '10.227.147.98'], 
   ...:  ['test', '172.18.74.146']]

In [2]: from collections import defaultdict

In [3]: resp_dict = defaultdict(list)

In [4]: for item in list_of_ips:
   ...:     resp_dict[item[0]].append(item[1])
   ...:     

In [5]: result = [[key, ", ".join(resp_dict[key])] for key in resp_dict]

In [6]: result
Out[6]: 
[['test', '10.152.114.65, 172.18.74.146'],
 ['Site', '10.227.211.244'],
 ['Demo-Site', '10.227.209.139, 10.227.215.68, 172.18.74.146, 10.227.147.98']]

Answer 3

您似乎希望根据每个子列表中的第一个元素分组您的记录。这就是groupby的作用。但是有一个重要的初步步骤，它根据每个子列表中的第一个第一个元素对列表进行排序。您可以使用sorted功能执行此操作，并使用itemgetter作为键功能。

from operator import itemgetter
from itertools import groupby
result = []
my_list = [['Demo-Site', '10.227.209.139'], 
['Demo-Site', '10.227.215.68'], 
['Demo-Site', '172.18.74.146'], 
['Site', '10.152.114.65'], 
['Site', '10.227.211.244'], 
['Demo-Site', '10.227.147.98'], 
['test', '172.18.74.146']]

演示groupby

for g, data in groupby(sorted(my_list, key=itemgetter(0)), itemgetter(0)):
     print(g)
     for elt in data:
         print(' ', elt)

的产率：

Demo-Site
  ['Demo-Site', '10.227.209.139']
  ['Demo-Site', '10.227.215.68']
  ['Demo-Site', '172.18.74.146']
  ['Demo-Site', '10.227.147.98']
Site
  ['Site', '10.152.114.65']
  ['Site', '10.227.211.244']
test
  ['test', '172.18.74.146']

如您所见，您的数据按子列表中的第一个元素分组。所以你现在需要的是连接（.join）＆＃34;成员＆＃34;的最后一个元素。相同的组，然后将列表[<given group>, <members string>]附加到result列表。

>>> for g, data in groupby(sorted(my_list, key=itemgetter(0)), itemgetter(0)):
...     result.append([g, ', '.join(elt[1] for elt in data)])
... 
>>> result
[['Demo-Site', '10.227.209.139, 10.227.215.68, 172.18.74.146, 10.227.147.98'], ['Site', '10.152.114.65, 10.227.211.244'], ['test', '172.18.74.146']]

Answer 4

您可以使用字典（和dict.setdefault方法）根据第一个索引保留您的项目，然后加入值：

>>> li=[['Demo-Site', '10.227.209.139'], ['Demo-Site', '10.227.215.68'], 
... ['Demo-Site', '172.18.74.146'], ['Site', '10.152.114.65'], 
... ['Site', '10.227.211.244'], ['Demo-Site', '10.227.147.98'], 
...  ['test', '172.18.74.146']]
>>> 
>>> d={}
>>> 
>>> for i,j in li:
...   d.setdefault(i,[]).append(j)
... 
>>> [[i,','.join(j)] for i,j in d.items()]
[['test', '172.18.74.146'], ['Site', '10.152.114.65,10.227.211.244'], ['Demo-Site', '10.227.209.139,10.227.215.68,172.18.74.146,10.227.147.98']]
>>>

Answer 5

或者，正如您所提到的，使用defaultdict：

from collections import defaultdict

lst = [['Demo-Site', '10.227.209.139'], ['Demo-Site', '10.227.215.68'], 
['Demo-Site', '172.18.74.146'], ['Site', '10.152.114.65'], 
['Site', '10.227.211.244'], ['Demo-Site', '10.227.147.98'], 
 ['test', '172.18.74.146']]

dct = defaultdict(list)

for name, ip in lst:
    dct[name].append(ip)

res = [ [name, ', '.join(ips)] for name, ips in dct.items() ]
print(res)

从列表中的字符串构建python

5 个答案: