如何在python中合并.dat文件中的列表

Question

我有很多包含数千行数据的列表。我想将这些列表并排合并到一个像1::2::F::4::990 这样的.dat 文件中。将这些单个数据视为不同的列表，我想完全按照此合并数据。你能帮忙吗

到目前为止，我已经这样做了。

with open("mergefile.dat", "w") as outfile:
    outfile.write("\n".join(users_id)+'::'+'\n'.join(movie_id)+'::'+'\n'.join(ratings)+'::'+'\n'.join(timestamps)+'::'+'\n'.join(genders)+'::'+'\n'.join(age)+'::'+'\n'.join(occupation)+'::'+'\n'.join(zipp)+'::'+'\n'.join(moviename)+'::'+'\n'.join(yearOfMovie))

这是一个示例输出

我想合并这样的数据，我有 user_id、movie_id、评级等列表，每一列都是一个列表

这就是我的列表的样子

这些是我的示例列表

user_id=['1','2', '3','4', '5','6','7']
movie_id=['1246','1247', '2081', '1240', '714']
ratings=['5','6','2','7']
timestamp=['9999','0000','5555','2222','1111']
gender=['F','G','F','F','G']
age=['1','1','2']
occupation=['10','10','10','10','10','10']
zip=['67890','45600']
title=['titl1','title2']
genres=['genre1','genre2','genre3']
year=['1999','1990']

我试过了，但没有得到我想要的

for user,movie,rating,timestamp,gender,age,occupations,zipps,title,genre,year in zip(users_id,movies_id,ratings,timestamps,genders,age,occupation,zipp,moviename,genres,yearOfMovie):
    with open("mergefile1.dat", "w") as outfile:
        outfile.write(user+'::'+movie+'::'+rating+'::'+timestamp+'::'+gender+'::'+age+'::'+occupations+'::'+zipps+'::'+title+'::'+genre+'::'+year)

这是我的输出只有一行

1::1246::4::978302091::F::1::1::48067::Toy Story::Animation|Children's|Comedy::1995

Answer 1

在您发布的示例中存在各种问题，列表和变量的名称，缺少换行符（\n），您也不需要每次都打开文件。请注意，当列表长度不同时，zip 迭代器只迭代最少数量的元素。 使用您的示例但解决上述问题，这可行：

user_ids=['1','2', '3','4', '5','6','7']
movie_ids=['1246','1247', '2081', '1240', '714']
ratings=['5','6','2','7']
timestamps=['9999','0000','5555','2222','1111']
genders=['F','G','F','F','G']
ages=['1','1','2']
occupations=['10','10','10','10','10','10']
zipps=['67890','45600']
titles=['titl1','title2']
genres=['genre1','genre2','genre3']
yearOfMovie=['1999','1990']

with open("mergefile1.dat", "w") as outfile:
    for user,movie,rating,timestamp,gender,age,occupation,zipp,title,genre,yearof in zip(user_ids,movie_ids,ratings,timestamps,genders,ages,occupations,zipps,titles,genres,yearOfMovie):
        outfile.write(user+'::'+movie+'::'+rating+'::'+timestamp+'::'+gender+'::'+age+'::'+occupation+'::'+zipp+'::'+title+'::'+genre+'::'+yearof+'\n')

输出：

1::1246::5::9999::F::1::10::67890::titl1::genre1::1999
2::1247::6::0000::G::1::10::45600::title2::genre2::1990