4

假设您有一个 2 元组序列：

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

并且您想测试是否'a'是序列中任何元组的第一项。

最Pythonic的方式是什么？

转换为字典并测试键，这似乎很容易理解？IE

'a' in dict(seq_of_tups)

使用一个不是特别清楚的可爱拉链技巧，除非你知道这个技巧？IE

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

还是对地图非常明确？IE

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

还是有比这些选择更好的方法？

4

2 回答 2

11 
>>> 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，您都可以使用它：

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
于 2012-08-12T14:24:28.417 回答
5 
>>> 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

上述解决方案一找到就会退出a，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'
于 2012-08-12T14:29:15.807 回答