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- s=[(2,4),(4,5),(2,8),(3,4),(4,6),(3,8),(4,4),(4,7),(5,8),(6,4),(4,8),(6,8)]
- a=set()
- a_len=[]
- for y in s:
- tmp=y[0]
- for x in s:
- if x[0]==tmp:
- a.add(x)
- a_len.append(len(a))
- a.clear()
- max_len=max(a_len) #能够以元组第一元素组成的列表的最大长度
- a_len=[]
- for y in s:
- tmp=y[0]
- for x in s:
- if x[0]==tmp:
- a.add(x)
- if len(a)==max_len: #再次循环 以便找到这组元组
- break
- a.clear()
- print(a) #输出某元素为元组第一个元素 且 含有它的数量最多的集合
- for x in a:
- if x in s:
- s.remove(x) #剔除上述元素
- print(s) #输出剩余元素
- '''
- PS D:\wpp> & C:/Programs/Python/python.exe d:/wpp/jihe1.py
- 输出以4开头的元组,因为以4开头的元组数量最多
- {(4, 4), (4, 6), (4, 5), (4, 8), (4, 7)}
- 输出扣除上面的元素后剩余的列表内容
- [(2, 4), (2, 8), (3, 4), (3, 8), (5, 8), (6, 4), (6, 8)]
- PS D:\wpp>
- '''
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- s = [(2, 4), (4, 5), (2, 8), (3, 4), (4, 6), (3, 8), (4, 4), (4, 7), (5, 8), (6, 4), (4, 8), (6, 8)]
- counts = dict()
- for i, _ in s:
- counts[i] = counts.get(i, 0) + 1
- max_len_num = max(counts, key=lambda x: counts[x])
- max_len_nums = (x for x in s if x[0] == max_len_num)
- other_numbers = (x for x in s if x[0] != max_len_num)
- print(list(max_len_nums))
- print(list(other_numbers))
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