How to create lists with dynamically changing names?

Question

Python

To open a file with a variable name, use the command:

files = [open('textfile_number_{}.txt'.format(i), 'w') for i in list_with_numbers]

You need to create a bunch of lists with different names like this, something like this:

[list_for_new_freq_{}.format(i) = [] for i in len(files)]

However, this code is, of course, not correct. How to fix it to avoid using dictionaries?

Answer 1

["list_for_new_freq_{}".format(i.name) for i in files]

Answer 2

Maybe it's better to write in a dictionary?
If you still need to, use exec() for example.

Answer 3

>>> numbers = (1, 2, 3)
>>> fnames = ('file{}.txt'.format(i) for i in numbers)
>>> files = [open(i, 'w', encoding='utf-8') for i in fnames]
>>> files
[<_io.TextIOWrapper name='file1.txt' mode='w' encoding='utf-8'>, <_io.TextIOWrapper name='file2.txt' mode='w' encoding='utf-8'>, <_io.TextIOWrapper name='file3.txt' mode='w' encoding='utf-8'>]
>>> [i.close() for i in files]
[None, None, None]
>>>

[[email protected] t]$ ls
file1.txt  file2.txt  file3.txt
[[email protected] t]$

In general, it is customary to open files through with. If there is no special need for manual opening and closing, then use with.