我知道set实施很棒。但是，它有一个巨大的限制：它不能处理项目的重复。
让我们使用整数：
tuple1 = (100, 100, 375)
layer = [(100, 100, 100, 375), (20, 100, 100, 375), (20, 20, 100, 375), (20, 20, 20, 375), (20, 20, 30, 375), (20, 20, 40, 375), (20, 20, 50, 375), (20, 20, 60, 375), (20, 20, 70, 375), (20, 20, 80, 375), (20, 30, 100, 375), (20, 30, 30, 375), (20, 30, 40, 375), (20, 30, 50, 375), (20, 30, 60, 375), (20, 30, 70, 375), (20, 30, 80, 375), (20, 40, 100, 375), (20, 40, 40, 375), (20, 40, 50, 375), (20, 40, 60, 375), (20, 40, 70, 375), (20, 40, 80, 375), (20, 50, 100, 375), (20, 50, 50, 375), (20, 50, 60, 375), (20, 50, 70, 375), (20, 50, 80, 375), (20, 60, 100, 375), (20, 60, 60, 375), (20, 60, 70, 375), (20, 60, 80, 375), (20, 70, 100, 375), (20, 70, 70, 375), (20, 70, 80, 375), (20, 80, 100, 375), (20, 80, 80, 375), (30, 100, 100, 375), (30, 30, 100, 375), (30, 30, 30, 375), (30, 30, 40, 375), (30, 30, 50, 375), (30, 30, 60, 375), (30, 30, 70, 375), (30, 30, 80, 375), (30, 40, 100, 375), (30, 40, 40, 375), (30, 40, 50, 375), (30, 40, 60, 375), (30, 40, 70, 375), (30, 40, 80, 375), (30, 50, 100, 375), (30, 50, 50, 375), (30, 50, 60, 375), (30, 50, 70, 375), (30, 50, 80, 375), (30, 60, 100, 375), (30, 60, 60, 375), (30, 60, 70, 375), (30, 60, 80, 375), (30, 70, 100, 375), (30, 70, 70, 375), (30, 70, 80, 375), (30, 80, 100, 100), (30, 80, 100, 375), (30, 80, 80, 375), (40, 100, 100, 375), (40, 40, 100, 375), (40, 40, 40, 375), (40, 40, 50, 375), (40, 40, 60, 375), (40, 40, 70, 375), (40, 40, 80, 375), (40, 50, 100, 375), (40, 50, 50, 375), (40, 50, 60, 375), (40, 50, 70, 375), (40, 50, 80, 375), (40, 60, 100, 375), (40, 60, 60, 375), (40, 60, 70, 375), (40, 60, 80, 375), (40, 70, 100, 375), (40, 70, 70, 375), (40, 70, 80, 375), (40, 80, 100, 375), (40, 80, 80, 375), (50, 100, 100, 375), (50, 50, 100, 375), (50, 50, 50, 375), (50, 50, 60, 375), (50, 50, 70, 375), (50, 50, 80, 375), (50, 50, 80, 80), (50, 60, 100, 375), (50, 60, 60, 375), (50, 60, 70, 375), (50, 60, 80, 375), (50, 70, 100, 375), (50, 70, 70, 375), (50, 70, 70, 80), (50, 70, 80, 375), (50, 80, 100, 375), (50, 80, 80, 375), (50, 80, 80, 80), (60, 100, 100, 375), (60, 60, 100, 375), (60, 60, 60, 375), (60, 60, 70, 375), (60, 60, 80, 375), (60, 70, 100, 375), (60, 70, 70, 375), (60, 70, 80, 375), (60, 80, 100, 375), (60, 80, 80, 375), (60, 80, 80, 80), (70, 100, 100, 375), (70, 70, 100, 375), (70, 70, 70, 375), (70, 70, 80, 375), (70, 80, 100, 100), (70, 80, 100, 375), (70, 80, 80, 375), (80, 100, 100, 375), (80, 80, 100, 375), (80, 80, 80, 100), (80, 80, 80, 375)]
我希望将layer中的项目保留为tuple1的子集。例如，必须保留2个第一个。
现在我正在使用for循环：
new_layer = list()
for elt in layer:
copy = list(elt)
for x in tuple1:
if x in copy:
copy.remove(x)
if len(copy) == 1:
continue
else:
new_layer.append(elt)
这将是一个比这更好的解决方案......而且最后的问题实际上更复杂。
我有5层：
len的元素：2 len：3 len：4 len：5 len：6 目标是摆脱层N中派生（由子集构成）的元素的层N-k。
感谢您的帮助：）
答案 0 :(得分：1)
这是一种方法，子类化Counter并实施__contains__（对于in运算符）：
from collections import Counter
class TupleCounter(Counter):
def __init__(self, t):
super().__init__(t)
def __contains__(self, other):
if not isinstance(other, self.__class__):
other = self.__class__(other)
for (v,c) in other.items():
if self.get(v,0) < c:
return False
return True
示例：
needle = (100, 100, 375)
layer = [(100, 100, 100, 375), (20, 100, 100, 375), (20, 20, 100, 375), (20, 20, 20, 375), (20, 20, 30, 375), (20, 20, 40, 375), (20, 20, 50, 375), (20, 20, 60, 375), (20, 20, 70, 375), (20, 20, 80, 375), (20, 30, 100, 375), (20, 30, 30, 375), (20, 30, 40, 375), (20, 30, 50, 375), (20, 30, 60, 375), (20, 30, 70, 375), (20, 30, 80, 375), (20, 40, 100, 375), (20, 40, 40, 375), (20, 40, 50, 375), (20, 40, 60, 375), (20, 40, 70, 375), (20, 40, 80, 375), (20, 50, 100, 375), (20, 50, 50, 375), (20, 50, 60, 375), (20, 50, 70, 375), (20, 50, 80, 375), (20, 60, 100, 375), (20, 60, 60, 375), (20, 60, 70, 375), (20, 60, 80, 375), (20, 70, 100, 375), (20, 70, 70, 375), (20, 70, 80, 375), (20, 80, 100, 375), (20, 80, 80, 375), (30, 100, 100, 375), (30, 30, 100, 375), (30, 30, 30, 375), (30, 30, 40, 375), (30, 30, 50, 375), (30, 30, 60, 375), (30, 30, 70, 375), (30, 30, 80, 375), (30, 40, 100, 375), (30, 40, 40, 375), (30, 40, 50, 375), (30, 40, 60, 375), (30, 40, 70, 375), (30, 40, 80, 375), (30, 50, 100, 375), (30, 50, 50, 375), (30, 50, 60, 375), (30, 50, 70, 375), (30, 50, 80, 375), (30, 60, 100, 375), (30, 60, 60, 375), (30, 60, 70, 375), (30, 60, 80, 375), (30, 70, 100, 375), (30, 70, 70, 375), (30, 70, 80, 375), (30, 80, 100, 100), (30, 80, 100, 375), (30, 80, 80, 375), (40, 100, 100, 375), (40, 40, 100, 375), (40, 40, 40, 375), (40, 40, 50, 375), (40, 40, 60, 375), (40, 40, 70, 375), (40, 40, 80, 375), (40, 50, 100, 375), (40, 50, 50, 375), (40, 50, 60, 375), (40, 50, 70, 375), (40, 50, 80, 375), (40, 60, 100, 375), (40, 60, 60, 375), (40, 60, 70, 375), (40, 60, 80, 375), (40, 70, 100, 375), (40, 70, 70, 375), (40, 70, 80, 375), (40, 80, 100, 375), (40, 80, 80, 375), (50, 100, 100, 375), (50, 50, 100, 375), (50, 50, 50, 375), (50, 50, 60, 375), (50, 50, 70, 375), (50, 50, 80, 375), (50, 50, 80, 80), (50, 60, 100, 375), (50, 60, 60, 375), (50, 60, 70, 375), (50, 60, 80, 375), (50, 70, 100, 375), (50, 70, 70, 375), (50, 70, 70, 80), (50, 70, 80, 375), (50, 80, 100, 375), (50, 80, 80, 375), (50, 80, 80, 80), (60, 100, 100, 375), (60, 60, 100, 375), (60, 60, 60, 375), (60, 60, 70, 375), (60, 60, 80, 375), (60, 70, 100, 375), (60, 70, 70, 375), (60, 70, 80, 375), (60, 80, 100, 375), (60, 80, 80, 375), (60, 80, 80, 80), (70, 100, 100, 375), (70, 70, 100, 375), (70, 70, 70, 375), (70, 70, 80, 375), (70, 80, 100, 100), (70, 80, 100, 375), (70, 80, 80, 375), (80, 100, 100, 375), (80, 80, 100, 375), (80, 80, 80, 100), (80, 80, 80, 375)]
needle = TupleCounter(needle)
filtered = [t for t in layer if needle in TupleCounter(t)]
print(filtered)
输出：
[(100, 100, 100, 375), (20, 100, 100, 375), (30, 100, 100, 375), (40, 100, 100, 375), (50, 100, 100, 375), (60, 100, 100, 375), (70, 100, 100, 375), (80, 100, 100, 375) ]
编辑：请注意，这只关心元组中值的频率，而不是它们在其中的相对位置。因此，即使元组的元素顺序相反，它也会匹配（375,100,100）。