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更新:我尝试了@Sunitha解决方案,但在itertools中没有积累-可能是因为运行2.7。
我已经使用Python 2.7.15和Python 3.6.5测试了此代码。此代码从列表中的第二个子列表(索引1,如果适用)开始,并向后看前一个子列表,以累积值,如您的示例一样。
Python 2.7.15rc1 (default, Apr 15 2018, 21:51:34)
[GCC 7.3.0] on linux2
Type "help", "copyright", "credits" or "license" for more information.
>>> hmm = [['x', 1, 2, 3], ['y', 2, 5, 4], ['z', 6, 2, 1]]
>>> for i in range(1, len(hmm)):
... prev = hmm[i - 1][1:]
... current = iter(hmm[i])
... hmm[i] = [next(current)] + [a + b for a, b in zip(prev, current)]
...
>>> hmm
[['x', 1, 2, 3], ['y', 3, 7, 7], ['z', 9, 9, 8]]
它在Python 3中的编写也可能略有不同:
Python 3.6.5 (default, Jun 14 2018, 13:19:33)
[GCC 7.3.0] on linux
Type "help", "copyright", "credits" or "license" for more information.
>>> hmm = [['x', 1, 2, 3], ['y', 2, 5, 4], ['z', 6, 2, 1]]
>>> for i in range(1, len(hmm)):
... _, *prev = hmm[i - 1]
... letter, *current = hmm[i]
... hmm[i] = [letter] + [a + b for a, b in zip(prev, current)]
...
>>> hmm
[['x', 1, 2, 3], ['y', 3, 7, 7], ['z', 9, 9, 8]]
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