[英]Unique lists from a list
给定一个列表,我需要返回一个唯一项列表。 我想看看是否有比我提出的更多Pythonic方式:
def unique_lists(l):
m = {}
for x in l:
m[x] = (m[x] if m.get(x) != None else []) + [x]
return [x for x in m.values()]
print(unique_lists([1,2,2,3,4,5,5,5,6,7,8,8,9]))
输出:
[[1], [2, 2], [3], [4], [5, 5, 5], [6], [7], [8, 8], [9]]
>>> L=[1,2,2,3,4,5,5,5,6,7,8,8,9]
>>> from collections import Counter
>>> [[k]*v for k,v in Counter(L).items()]
[[1], [2, 2], [3], [4], [5, 5, 5], [6], [7], [8, 8], [9]]
使用默认字典。
>>> from collections import defaultdict
>>> b = defaultdict(list)
>>> a = [1,2,2,3,4,5,5,5,6,7,8,8,9]
>>> for x in a:
... b[x].append(x)
...
>>> b.values()
[[1], [2, 2], [3], [4], [5, 5, 5], [6], [7], [8, 8], [9]]
我发现函数set()
的build很常用:
lst=[1,2,2,3,4,5,5,5,6,7,8,8,9]
def all_eq_elms(lst, elm):
while True:
try:
yield lst.pop(lst.index(elm))
except:
break
[[e for e in all_eq_elms(lst,elm)] for elm in set(lst)]
Out[43]: [[1], [2, 2], [3], [4], [5, 5, 5], [6], [7], [8, 8], [9]]
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