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Python多层嵌套list的递归处理方法(推荐)

WBOY
WBOYOriginal
2016-06-16 08:47:581428browse

问题:用Python处理一个多层嵌套list

['and', 'B', ['not', 'A'],[1,2,1,[2,1],[1,1,[2,2,1]]], ['not', 'A', 'A'],['or', 'A', 'B' ,'A'] , 'B']

需求1)如何展开成一层?

需求2)如何删除重复的元素? 包括重复的list, 要考虑子list的重复元素删除后造成的子list重复

#!/usr/bin/env python
# -*- coding: utf-8 -*-


def unilist(ll):
  
  """
  功能:用递归方法删除多层列表中重复元素
  """
  
  result = []
  for i in ll:
    if isinstance(i, list):
      if unilist(i) not in result:
        result.append(unilist(i))
    else:
      if i not in result:
        result.append(i)
  return result


def flatten(ll):
  """
  功能:用递归方法展开多层列表,以生成器方式输出
  """
  if isinstance(ll, list):
    for i in ll:
      for element in flatten(i):
        yield element
  else:
    yield ll


testcase= ['and', 'B', ['not', 'A'],[1,2,1,[2,1],[1,1,[2,2,1]]], ['not', 'A', 'A'],['or', 'A', 'B' ,'A'] , 'B']

print unilist(testcase)

print list(flatten(testcase))

运行结果

['and', 'B', ['not', 'A'], [1, 2, [2, 1], [1, [2, 1]]], ['or', 'A', 'B']]

['and', 'B', 'not', 'A', 1, 2, 1, 2, 1, 1, 1, 2, 2, 1, 'not', 'A', 'A', 'or', 'A', 'B', 'A', 'B']

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