n级嵌套元组数据，以逗号分隔带引号的字符串

Question

I have below string:

companies = [(('Paramount Pictures Corporation (Canada)',),), (('20th Century Fox Home Entertainment',),), (('Shanghai Epic Music Entertainment',),), (('Paramount Pictures',),), (('Fox Filmes do Brasil',),), (('Gativideo',),), (('Abril V\xc3\xaddeo',),), (('20th Century Fox Netherlands',),), (('20th Century Fox Home Entertainment',),), (('Nelonen',),), (('20th Century Fox',),), (('Finnkino',),), (('Modhumita Movies',),), (('Twentieth Century Fox Home Entertainment',),), (('Egmont Entertainment',),), (('National Broadcasting Company (NBC)',),), (('MTV3',),), (('Bandur Film',),), (('Westec Media Limited',),), (('20th Century Fox Home Entertainment',),), (('Turner Classic Movies (TCM)',),), (('Lightstorm Entertainment',),), (('20th Century Fox',),), (('Pioneer Entertainment',),), (('Odeon',),), (('FS Film',),), (('Filmes Castello Lopes',),), (('Paramount Home Entertainment',),), (('Big Picture 2 Films',),), (('Hispano Foxfilms S.A.E.',),), (('TV3 Norge',),), (('Paramount Home Video',),), (('Warner Bros.',),), (('Sub',),), (('20th Century Fox de Argentina',),), (('Twentieth Century Fox Film Corporation',),), (('UGC-Fox Distribution (UFD)',),), (('PFC Vid\xc3\xa9o',),), (('Columbia TriStar Egmont Film Distributors',),), (('20th Century Fox Home Entertainment',),), (('Syrena Entertainment Group',),), (('AVA',),), (('Forum Cinemas',),)]

I would like to convert it into a string which looks like:

 companies = 'Paramount Pictures Corporation (Canada)', '20th Century Fox Home Entertainment', 'Shanghai Epic Music Entertainment', 'Twentieth Century Fox Film Corporation',  ....

I asked this question just 2 days ago which is similar but the solutions provided there do not work for the above string.

Answer 1

For each item in companies what you want is item[0][0] . using a generator expression you can do:

res = (item[0][0] for item in companies)

To get the final result, apply tuple() like this:

tuple(res)

Output:

>>> companies = [(('Paramount Pictures Corporation (Canada)',),), (('20th Century Fox Home Entertainment',),), (('Shanghai Epic Music Entertainment',),)]
>>> 
>>> res = (item[0][0] for item in companies)
>>> res
<generator object <genexpr> at 0x023AAE68>
>>> tuple(res)
('Paramount Pictures Corporation (Canada)', '20th Century Fox Home Entertainment', 'Shanghai Epic Music Entertainment')

Answer 2

Once you defined the nested get [0] as such:

first = lambda x: first(x[0]) if type(x)==tuple else x

It will work no matter what is the depth of the tuple nesting.

All we have left is to apply it to the companies list

companies = [(('Paramount Pictures Corporation (Canada)',),), (('20th Century Fox Home Entertainment',),), (('Shanghai Epic Music Entertainment',),), (('Paramount Pictures',),), (('Fox Filmes do Brasil',),), (('Gativideo',),), (('Abril V\xc3\xaddeo',),), (('20th Century Fox Netherlands',),), (('20th Century Fox Home Entertainment',),), (('Nelonen',),), (('20th Century Fox',),), (('Finnkino',),), (('Modhumita Movies',),), (('Twentieth Century Fox Home Entertainment',),), (('Egmont Entertainment',),), (('National Broadcasting Company (NBC)',),), (('MTV3',),), (('Bandur Film',),), (('Westec Media Limited',),), (('20th Century Fox Home Entertainment',),), (('Turner Classic Movies (TCM)',),), (('Lightstorm Entertainment',),), (('20th Century Fox',),), (('Pioneer Entertainment',),), (('Odeon',),), (('FS Film',),), (('Filmes Castello Lopes',),), (('Paramount Home Entertainment',),), (('Big Picture 2 Films',),), (('Hispano Foxfilms S.A.E.',),), (('TV3 Norge',),), (('Paramount Home Video',),), (('Warner Bros.',),), (('Sub',),), (('20th Century Fox de Argentina',),), (('Twentieth Century Fox Film Corporation',),), (('UGC-Fox Distribution (UFD)',),), (('PFC Vid\xc3\xa9o',),), (('Columbia TriStar Egmont Film Distributors',),), (('20th Century Fox Home Entertainment',),), (('Syrena Entertainment Group',),), (('AVA',),), (('Forum Cinemas',),)]
companies = map(first,companies)

And if you want to concatenate it all to one string:

s = ",".join(companies)

Answer 3

如果只在寻找一个字符串作为输出，这应该可以工作：

str = ", ".join(["'"+company[0][0]+"'" for company in companies])