比较 python 中的“列表”

Question

我有多个清单。我需要找到一种方法来生成每个列表中与所有列表相比的唯一项目列表。有没有简单或直接的方法可以做到这一点。我知道这些列表基本上可以用作集合。

Answer 1

import collections

def uniques(*args):
    """For an arbitrary number of sequences,
           return the items in each sequence which
            do not occur in any of the other sequences
    """

    # ensure that each value only occurs once in each sequence
    args = [set(a) for a in args]

    seen = collections.defaultdict(int)
    for a in args:
        for i in a:
            seen[i] += 1
    # seen[i] = number of sequences in which value i occurs

    # for each sequence, return items
    #  which only occur in one sequence (ie this one)
    return [[i for i in a if seen[i]==1] for a in args]

所以

uniques([1,1,2,3,5], [2,3,4,5], [3,3,3,9])  ->  [[1], [4], [9]]

Answer 2

使用集合类和其中定义的集合操作：

>>> l1 = [1,2,3,4,5,5]
>>> l2 = [3,4,4,6,7]
>>> set(l1) ^ set(l2)    # symmetric difference
set([1, 2, 5, 6, 7])

编辑：啊，误读了你的问题。如果您的意思是“l1 中的唯一元素不在 l2、l3、...、ln 中的任何一个中，那么：

l1set = set(l1)
for L in list_of_lists:   # list_of_lists = [l2, l3, ..., ln]
    l1set = l1set - set(L)

Answer 3

l1 = [4, 6, 3, 7]
l2 = [5, 5, 3, 1]
l3 = [2, 5, 4, 3]
l4 = [9, 8, 7, 6]

# first find the union of the "other" lists
l_union = reduce(set.union, (map(set, (l1, l2, l3))))

# then subtract the union from the remaining list
uniques = set(l4) - l_union

print uniques

结果：

>>> set([8, 9])

Answer 4

for l in lists:
  s = set(l)
  for o in lists:
    if o != l:
      s -= set(o)

  # At this point, s holds the items unique to l

为了提高效率，您可以将所有列表一次转换为集合。

Answer 5

import itertools

# Test set
lists = []
lists.append([1,2,3,4,5,5])
lists.append([3,4,4,6,7])
lists.append([7,])
lists.append([8,9])
lists.append([10,10])

# Join all the lists removing the duplicates in each list
join_lists = []
for list_ in lists:
    join_lists.extend(set(list_))

# First, sort
join_lists.sort()

# Then, list only the groups with one element
print [ key for key, grp in itertools.groupby(join_lists) if len(list(grp)) < 2 ]

#>>> [1, 2, 6, 8, 9]

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