Question

我尝试使用python的set()方法来查找List中的唯一元素。它可以正常删除所有重复项。但这是我的要求，我想通过使用set()方法获取被删除的元素。有人可以帮助我吗？

a=[1,2,3,1,4]
b=set(a)
Output:[1,2,3,4]

我的预期输出为[1]。从set()方法中删除的元素

Answer 1

collections.Counter在这里很有用。

from collections import Counter
counts = Counter(a)
b = set(counts.keys())
for x, count in counts.items():
    if count > 1:
        print('%d appearances of %s were removed in the set' % (count-1, x))

Answer 2

你甚至不需要设置。您希望每个元素的计数具有多个匹配项。来自收藏品的反击以及字典理解应该可以帮助你。

from collections import Counter

a = [1, 1, 1, 2, 2, 3, 4]    
removed = {k: v-1 for k, v in Counter(a).iteritems() if v > 1}

>>> removed
Out[8]: {1: 2, 2: 1}

Answer 3

使用Counter

尝试此操作

from collections import Counter
a = [1, 2, 3, 1, 4]
>>>[i for i in Counter(a) if Counter(a)[i] > 1]
[1]

Answer 4

您可以扩展Set类（让您自己的Set类说MySet）并覆盖此函数

def _update(self, iterable):
    # The main loop for update() and the subclass __init__() methods.
    data = self._data

    # Use the fast update() method when a dictionary is available.
    if isinstance(iterable, BaseSet):
        data.update(iterable._data)
        return

    value = True

    if type(iterable) in (list, tuple, xrange):
        # Optimized: we know that __iter__() and next() can't
        # raise TypeError, so we can move 'try:' out of the loop.
        it = iter(iterable)
        while True:
            try:
                for element in it:
                    data[element] = value
                return
            except TypeError:
                transform = getattr(element, "__as_immutable__", None)
                if transform is None:
                    raise # re-raise the TypeError exception we caught
                data[transform()] = value
    else:
        # Safe: only catch TypeError where intended
        for element in iterable:
            try:
                data[element] = value
            except TypeError:
                transform = getattr(element, "__as_immutable__", None)
                if transform is None:
                    raise # re-raise the TypeError exception we caught
                data[transform()] = value

Answer 5

这将返回一个仅包含从原始集中删除的项目的集合：

>>> a = [1, 2, 3, 4, 1, 1, 5]

>>> set(i for i in a if a.count(i) > 1)

>>> {1}

Answer 6

我认为你是以略微混乱的方式解决问题。我不是试图让set()做一些不打算做的事情（返回重复列表），而是使用collections.Counter()来收集重复项，然后从中获取集合。

以下是一些代码：

#!python
from collections import Counter
c = Counter([1,2,3,1,4])
dupes = [k for k,v in c.items() if v>1]
b = set(c.keys())

Answer 7

简单的python示例，

for result in soup.select('.sold-results__normal-hit'):
    nodes = result.select('.sold-property-listing__location h2 + div span')
    if len(nodes)==2:
        place = nodes[1].text.strip()
    else:
        place = 'not specified'    
    print(place)

产生，

def finder(s):
    seen,yields=set(),set()
    for i in s:
      if i in seen and i not in yields:
          yield i
          yields.add(i)
      else:
          seen.add(i)
    print(type(seen), seen )
    
a = [1,2,3,1,4] 
print(list(finder(a)))

如何识别从python中的set（）中删除的元素？

7 个答案: