Python列表按每个唯一值获取前n个元素
我有一个类似的清单Python列表按每个唯一值获取前n个元素,python,list,unique,Python,List,Unique,我有一个类似的清单 [('10', '100'), ('11', '100'), ('18', '108'), ('22', '100'), ('12', '102'), ('15', '104'), ('21', '100'), ('25', '108'), ('20', '102'), ('24', '104'), ('105', '108'), ('35', '100'),
[('10', '100'),
('11', '100'),
('18', '108'),
('22', '100'),
('12', '102'),
('15', '104'),
('21', '100'),
('25', '108'),
('20', '102'),
('24', '104'),
('105', '108'),
('35', '100'),
('14', '104'),
('96', '100'),
('100', '104'),
('26', '100'),
('19', '100'),
('110', '108'),
('36', '102'),
('30', '104')]
所有项目的第二个值都是唯一的“100”、“102”、“104”和“108”
我想取'100','102','104','108'组的前3个值
输出应如下所示:
[('10', '100'),
('11', '100'),
('22', '100'),
('18', '108'),
('25', '108'),
('105', '108'),
('12', '102'),
('20', '102'),
('36', '102'),
('15', '104'),
('24', '104'),
('14', '104')]
我不想将列表更改为dataframe并使dataframe正常工作。您可以使用
itertools.groupby
:
from itertools import groupby
new_d = [(a, list(b)) for a, b in groupby(sorted(d, key=lambda x:int(x[-1])), key=lambda x:int(x[-1]))]
result = [b for _, c in new_d for b in c[:3]]
输出:
[('10', '100'),
('11', '100'),
('22', '100'),
('12', '102'),
('20', '102'),
('36', '102'),
('15', '104'),
('24', '104'),
('14', '104'),
('18', '108'),
('25', '108'),
('105', '108')]
您可以迭代元组列表,并使用字典跟踪元组中的第二个元素出现了多少次。然后,如果第二个值出现的次数少于
3次,只需向结果列表中添加一个元组:
d = {}
n = 3
out = []
for i,j in l:
if d.setdefault(j,0) < n:
d[j]+= 1
out.append((i,j))
另一种方法是在字典中保留键出现的简单计数,然后对其进行排序,例如(假设输入数据如问题中所示):
有关更多详细信息,请参阅和。您在尝试此操作时忘记了包含您编写的代码。到目前为止,您在解决家庭作业方面做了哪些尝试?不要忘记您可以投票并接受答案,请参阅。谢谢
print(out)
[('10', '100'),
('11', '100'),
('18', '108'),
('22', '100'),
('12', '102'),
('15', '104'),
('25', '108'),
('20', '102'),
('24', '104'),
('105', '108'),
('14', '104'),
('36', '102')]
from collections import defaultdict
from itertools import count
from operator import itemgetter
counts = defaultdict(lambda: count(0))
result = [(value, key) for value, key in data if next(counts[key]) < 3]
print(sorted(result, key=itemgetter(1)))
[('10', '100'), ('11', '100'), ('22', '100'), ('12', '102'), ('20', '102'), ('36', '102'), ('15', '104'), ('24', '104'), ('14', '104'), ('18', '108'), ('25', '108'), ('105', '108')]