解决方案
如果您的子列表仅包含数字并因此是可散列的,您可以使用collections.Counter:
>>> from collections import Counter
>>> lis = [[12, 34, 56], [45, 78, 334], [56, 90, 78], [12, 34, 56]]
>>> sum(y for y in Counter(tuple(x) for x in lis).values() if y > 1)
2
>>> lis = [[12, 34, 56], [45, 78, 334], [56, 90, 78], [12, 34, 56], [56, 90, 78], [12, 34, 56]]
>>> sum(y for y in Counter(tuple(x) for x in lis).values() if y > 1)
5
按步骤
将您的子列表转换为元组:
tuple(x) for x in lis
数一数:
>>> Counter(tuple(x) for x in lis)
Counter({(12, 34, 56): 3, (45, 78, 334): 1, (56, 90, 78): 2})
只取值:
>>> Counter(tuple(x) for x in lis).values()
dict_values([3, 1, 2])
最后,只求计数大于 1 的那些:
> sum(y for y in Counter(tuple(x) for x in lis).values() if y > 1)
5
使其可重复使用
将其放入一个函数中,添加一个文档字符串和一个文档测试:
"""Count duplicates of sub-lists.
"""
from collections import Counter
def count_duplicates(lis):
"""Count duplicates of sub-lists.
Assumption: Sub-list contain only hashable elements.
Result: If a sub-list appreas twice the result is 2.
If a sub-list aprears three time and a other twice the result is 5.
>>> count_duplicates([[12, 34, 56], [45, 78, 334], [56, 90, 78],
... [12, 34, 56]])
2
>>> count_duplicates([[12, 34, 56], [45, 78, 334], [56, 90, 78],
... [12, 34, 56], [56, 90, 78], [12, 34, 56]])
...
5
"""
# Make it a bit more verbose than necessary for readability and
# educational purposes.
tuples = (tuple(elem) for elem in lis)
counts = Counter(tuples).values()
return sum(elem for elem in counts if elem > 1)
if __name__ == '__main__':
import doctest
doctest.testmod(verbose=True)
运行测试:
python count_dupes.py
Trying:
count_duplicates([[12, 34, 56], [45, 78, 334], [56, 90, 78],
[12, 34, 56]])
Expecting:
2
ok
Trying:
count_duplicates([[12, 34, 56], [45, 78, 334], [56, 90, 78],
[12, 34, 56], [56, 90, 78], [12, 34, 56]])
Expecting:
5
ok
1 items had no tests:
__main__
1 items passed all tests:
2 tests in __main__.count_duplicates
2 tests in 2 items.
2 passed and 0 failed.
Test passed.