二维列表排序python 3.6.1 anaconda

Question

lijst = [[], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [],
         [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [], [],
         [], [], [], [], [], [], [], [], [], [], [], [],
         ['/vacatures/oracle-plsql-ontwikkelaar-1/'], [], [], [], [],
         ['/vacatures/oracle-plsql-ontwikkelaar-1/'],
         ['/vacatures/business-intelligence-developer-1/'], [], [], [], [], [],
         ['/vacatures/business-intelligence-developer-1/'],
         ['/vacatures/oracle-dba/'], [], [], ['/vacatures/oracle-dba/'],
         ['/vacatures/database-beheerder/'], [], [], [],
         ['/vacatures/database-beheerder/'],
         ['/vacatures/sql-server-dba-powershell/'], [], [], [],
         ['/vacatures/sql-server-dba-powershell/'],
         ['/vacatures/junior-msbi-consultant/'], [], [], [], [], [],
         ['/vacatures/junior-msbi-consultant/'],
         ['/vacatures/senior-msbi-consultant/'], [], [], [], [], [],
         ['/vacatures/senior-msbi-consultant/'],
         ['/vacatures/medior-msbi-consultant/'], [], [], [], [],
         ['/vacatures/medior-msbi-consultant/'],
         ['/vacatures/zos-mainframe-specialist/'], [], [],
         ['/vacatures/zos-mainframe-specialist/'],
         ['/vacatures/junior-business-analyst/'], [], [], [], [],
         ['/vacatures/junior-business-analyst/'], [], [], [], [], [], [], [],
         [], ['/vacatures/oracle-plsql-ontwikkelaar-1/'], [], [],
         ['/vacatures/oracle-dba/'], [], [],
         ['/vacatures/business-intelligence-developer-1/'], [], [],
         ['/vacatures/database-beheerder/'], [], [],
         ['/vacatures/sql-server-dba-powershell/'], [], [], [], [], [], [], [],
         [], [], []]

我有个问题。如何过滤空列表并删除二维列表中的重复项？

Answer 1

这很简单，就像做

new_list0 = list(filter(len, lijst))

然后，要删除重复项，可以将new_list转换为set，然后将其转换回列表。如下所示

new_list1 = list(set(tuple(x) for x in new_list0))

如果您想将new_list1的元素(现在是元组)转换回列表，您可以这样做

new_list2 = list(map(list, new_list1))

​

但是，考虑到上面执行的来回次数(从生成器到列表，到设置，...，到列表，等等)，在性能方面看起来更好的东西可能是

​

new_list = []
for el in lijst:
    if el and el not in new_list:
        new_list.append(el)            
#print(new_list)

最后，请注意，new_list仍将是二维的，就像原始的一样。如果想要使其成为一维的，可以使用making it flat，如下所示

import itertools
new_list = list(itertools.chain.from_iterable(new_list))

或者直接将其创建为一维列表，并将时间复杂度降低到O(n) (通过避免in运算符而不是O(n**2) )

new_set = set()
for el in lijst:
    if el:
        new_set.update(el)        
new_list = list(new_set)

​

答案经过测试且功能正常

​

Answer 2

你的列表并不是真正的二维。每个列表要么有0个元素，要么有1个元素。

在这种情况下，您可以只提取字符串并将它们放入一个集合中：

print({l[0] for l in lijst if l})

它输出：

set(['/vacatures/junior-msbi-consultant/', '/vacatures/junior-business-analyst/', '/vacatures/business-intelligence-developer-1/', '/vacatures/zos-mainframe-specialist/', '/vacatures/sql-server-dba-powershell/', '/vacatures/database-beheerder/', '/vacatures/medior-msbi-consultant/', '/vacatures/oracle-dba/', '/vacatures/oracle-plsql-ontwikkelaar-1/', '/vacatures/senior-msbi-consultant/'])

它既简洁又快速(O(n))。