您可以使用
zip
函数:
>>> set(zip(*mytuple)[1])
set(['Sarah@gmail.com', 'Jim@gmail.com', 'Andrew@gmail.com'])
或者使用性能较低的方法,您可以使用map
和operator.itemgetter
,并使用set
来获取唯一的元组:
>>> from operator import itemgetter
>>> tuple(set(map(lambda x:itemgetter(1)(x),mytuple)))
('Sarah@gmail.com', 'Jim@gmail.com', 'Andrew@gmail.com')
一些答案的基准测试:
我的答案:
s = """\
mytuple = [('Andrew','Andrew@gmail.com','20'),('Jim',"Jim@gmail.com",'12'),("Sarah","Sarah@gmail.com",'43'),("Jim","Jim@gmail.com",'15'),("Andrew","Andrew@gmail.com",'56')]
set(zip(*mytuple)[1])
"""
print timeit.timeit(stmt=s, number=100000)
0.0740020275116
icodez answer :
s = """\
mytuple = [('Andrew','Andrew@gmail.com','20'),('Jim',"Jim@gmail.com",'12'),("Sarah","Sarah@gmail.com",'43'),("Jim","Jim@gmail.com",'15'),("Andrew","Andrew@gmail.com",'56')]
seen = set()
[x[1] for x in mytuple if x[1] not in seen and not seen.add(x[1])]
"""
print timeit.timeit(stmt=s, number=100000)
0.0938332080841
Hasan的回答:
s = """\
mytuple = [('Andrew','Andrew@gmail.com','20'),('Jim',"Jim@gmail.com",'12'),("Sarah","Sarah@gmail.com",'43'),("Jim","Jim@gmail.com",'15'),("Andrew","Andrew@gmail.com",'56')]
set([k[1] for k in mytuple])
"""
print timeit.timeit(stmt=s, number=100000)
0.0699651241302
Adem的回答:
s = """
from itertools import izip
mytuple = [('Andrew','Andrew@gmail.com','20'),('Jim',"Jim@gmail.com",'12'),("Sarah","Sarah@gmail.com",'43'),("Jim","Jim@gmail.com",'15'),("Andrew","Andrew@gmail.com",'56')]
set(map(lambda x: x[1], mytuple))
"""
print timeit.timeit(stmt=s, number=100000)
0.237300872803 !!!