在2元組序列中測試與第一項元組匹配的最Pythonic方法是什么？

Question

假設你有一系列2元組：

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

並且您想測試'a'是否是序列中任何元組的第一項。

什么是最恐怖的方式？

轉換為字典並測試密鑰，這似乎很容易理解？ 即

'a' in dict(seq_of_tups)

使用可愛的拉鏈技巧，除非你知道這個技巧，否則不是特別清楚？ 即

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

或者真的明確地圖？ 即

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

還是有比這些選擇更好的方法？

Answer 1

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

對於任何大小的元組，您可以使用它：

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

這里還有一些有趣的時間：

>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

上述解決方案將盡快退出a被發現，證明：

>>> 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'