什麼是測試與2元組序列中的第一項元組匹配的最Pythonic方法？

Question

假設你有2元組序列：

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

，你想測試，如果'a'是序列中任何元組的第一個項目。

什麼是最Pythonic的方式？

轉換爲字典和測試鍵，這似乎很容易理解？即

'a' in dict(seq_of_tups) 

使用一個可愛的拉鍊招除非你知道的伎倆是不是特別清楚？即

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

還是真的明確地圖？即

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

或者是否有比這些選擇更好的方法？

Answer 1

>>> 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找到a，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' 

Answer 2

>>> 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，你可以使用它代替：

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

這裏也有一些有趣的時刻：

>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