如何使用Python将嵌套列表中的每两个值组合起来？

大家好,， 我有一个上面显示的嵌套列表，我想将该嵌套列表中的每两个元素组合起来。预期的产出将是：

nested_lst = [['AAA, CEO','BBB, Global Head','CCC, Regional Manager','DDD, Analyst','People also report to CCC','XXX, Analyst','VVV, Analyst'],
['AAA, CEO','BBB, Global Head','EEE, Regional Manager','FFF, Analyst','People also report to EEE','SSS, Analyst','LLL, Analyst'],
['AAA, CEO','BBB, Global Head','PPP, Regional Manager','MMM, Manager','People report to MMM','GGG, Associate','People also report to EEE','ZZZ, Junior Analyst','UUU, Contractor']]

我尝试了

['&'.join（x）for x in zip（nested_lst[2][0:：2]，nested_lst[2][1:：2]）

，但它不适用于嵌套列表，也不合并元素

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有人能帮我回答这个问题吗？谢谢你的帮助

您的列表比较接近。它需要一个嵌套循环和一些更改

列出公司：

[['AAA, CEO & BBB, Global Head','BBB, Global Head & CCC, Regional Manager','CCC, Regional Manager & DDD, Analyst','DDD, Analyst & People also report to CCC',...],
['AAA, CEO & BBB, Global Head','BBB, Global Head & EEE, Regional Manager','EEE, Regional Manager & FFF, Analyst',...],
['AAA, CEO & BBB, Global Head','BBB, Global Head & PPP, Regional Manager',...]]

您可以使用配方：

或不使用itertools：

result = [list(map(' & '.join, pairwise(l))) for l in nested_lst]

结果:

result = [list(map(' & '.join, zip(l, l[1:]))) for l in nested_lst]

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谢谢你，布莱恩。你的回答非常有用！谢谢你，杰普。itertools.pairwise看起来棒极了！

result = [list(map(' & '.join, zip(l, l[1:]))) for l in nested_lst]

[['AAA, CEO & BBB, Global Head', 'CCC, Regional Manager & DDD, Analyst', 'People also report to CCC & XXX, Analyst', 'VVV,Analyst'], ['AAA, CEO & BBB, Global Head', 'EEE, Regional Manager & FFF, Analyst', 'People also report to EEE & SSS, Analyst', 'LLL, Analyst'], ['AAA, CEO & BBB, Global Head', 'PPP, Regional Manager & MMM, Manager', 'People report to MMM & GGG, Associate', 'People also report to EEE & ZZZ, Junior Analyst', 'UUU, Contractor']]