Python按顺序返回列表中重复项的列表

Question

您如何快速按照列表出现的顺序返回重复列表？例如，duplicates([2,3,5,5,5,6,6,3]) results in [5,6,3]表示重复的元素仅在其第二个元素出现时才添加到结果重复列表中。到目前为止，我有下面的代码，但是其运行速度不足以通过大型测试用例。有没有进口的更快的选择吗？

def duplicates(L):
    first = set()
    second = []
    for i in L:
        if i in first and i not in second:
            second.append(i)
            continue
        if i not in first and i not in second:
            first.add(i)
            continue
    return second

Answer 1

依赖于字典中的键，对于Python 3.6+，该键由插入顺序排序（或可以使用OrderedDict）

Reference

代码

from collections import defaultdict

def duplicates(lst):
  d = defaultdict(int)
  result = {}    # Relies on dictionary keys tracking the order 
                 # items are added or can use Ordereddict
  for k in lst:
    d[k] += 1
    if d[k] > 1:
      if k not in result:
        result[k] = None

  return list(result.keys())


lst = [2,3,5,5,5,6,6,3]

print(duplicates(lst))

Out

[5, 6, 3]

Answer 2

使用在添加订单项时报告字典键的事实

代码

from collections import defaultdict

def duplicates(lst):
  d = defaultdict(int)
  result = {}    # Relies ondictionary keys tracking the order 
                 # items are added or can use Ordereddict
  for k in lst:
    d[k] += 1
    if d[k] > 1:
      if k not in result:
        result[k] = None

  return list(result.keys())


lst = [2,3,5,5,5,6,6,3]

print(duplicates(lst))

Out

[5, 6, 3]