在2元组序列中测试与第一项元组匹配的最Pythonic方法是什么？

Question

假设你有一系列2元组:

seq_of_tups = (('a', 1), ('b', 2), ('c', 3))

并且您想测试if 'a'是序列中任何元组的第一项.

什么是最恐怖的方式？

转换为字典并测试密钥,这似乎很容易理解？即

'a' in dict(seq_of_tups)

使用可爱的拉链技巧,除非你知道这个技巧,否则不是特别清楚？即

'a' in zip(*seq_of_tups)[0]

或者真的明确地图？即

'a' in map(lambda tup: tup[0], seq_of_tups)

还是有比这些选择更好的方法？

Answer 1

>>> seq_of_tups = (('a', 1), ('b', 2), ('c', 3))
>>> any(x == 'a' for x, y in seq_of_tups)
True

对于任何大小的元组,您可以使用它:

any(x[0] == 'a' for x in seq_of_tups)

这里还有一些有趣的时间:

>python -m timeit -s "seq_of_tups = (('a', 1), ('b', 2), ('c', 3))" 
                 "any(x == 'a' for x, y in seq_of_tups)"
1000000 loops, best of 3: 0.564 usec per loop

>python -m timeit -s "seq_of_tups = (('a', 1), ('b', 2), ('c', 3))" 
                 "'a' in (x[0] for x in seq_of_tups)"
1000000 loops, best of 3: 0.526 usec per loop

>python -m timeit -s "seq_of_tups = (('a', 1), ('b', 2), ('c', 3)); 
                      from operator import itemgetter; from itertools import imap" 
                 "'a' in imap(itemgetter(0), seq_of_tups)"
1000000 loops, best of 3: 0.343 usec per loop

Answer 2

>>> tups = (('a', 1), ('b', 2), ('c', 3))

>>> 'a' in (x[0] for x in tups)
True
>>> 'd' in (x[0] for x in tups)
False

上述解决方案将a在找到后立即退出,证明:

>>> tups = (('a', 1),('a',5), ('b', 2), ('c', 3))
>>> gen=(x[0] for x in tups)
>>> 'a' in gen
True
>>> list(gen)
['a', 'b', 'c']  #this result means generator stopped at first 'a'