Python 检查是否有任何列表不为空，如果不为空，则对链接到该列表的相应文件夹执行某些操作

Question

So here I am trying to optimize some code I have.

Code checks if any of lists is empty, and if it is not, corresponding folder that is linked to that list will be saved in external program, so to make it more clear this is code:

if len(match_FIRST)>0 and len(match_SECOND)>0 and len(match_THIRD)>0 and len(match_FOURTH)>0:
  keep = ['FIRST', 'SECOND', 'THIRD', 'FOURTH']
elif len(match_FIRST)>0 and len(match_SECOND)>0 and len(match_THIRD)>0 and len(match_FOURTH)==0:
  keep = ['FIRST', 'SECOND', 'THIRD']
elif len(match_FIRST)>0 and len(match_SECOND)>0 and len(match_THIRD)==0 and len(match_FOURTH)>0:
  keep = ['FIRST', 'SECOND', 'FOURTH']
elif len(match_FIRST)>0 and len(match_SECOND)==0 and len(match_THIRD)>0 and len(match_FOURTH)>0:
  keep = ['FIRST', 'THIRD', 'FOURTH']
elif len(match_FIRST)>0 and len(match_SECOND)>0 and len(match_THIRD)==0 and len(match_FOURTH)==0:
  keep = ['FIRST', 'SECOND']
elif len(match_FIRST)>0 and len(match_SECOND)==0 and len(match_THIRD)>0 and len(match_FOURTH)==0:
  keep = ['FIRST', 'THIRD']
... etc

So basically we will in this way keep in keep= [] list only names that are linked to lists that are not empty...

What do you suggest any algorithm that can optimize this since now when I have only four lists it is easy to manually check it this way... but better solution is needed...

Please advice!

Thanks,

Salute

Answer 1

You're probably looking to filter only non-empty lists from your list of lists.

Another way to do the same thing is to use a comprehension with if as filter:

non_empty_lists = [l for l in lists if len(l) > 0]

For example:

>>> lists = [
...     ['a', 'b', 'c'],
...     [],
...     [],
...     ['d', 'e']
... ]
>>> non_empty_lists = [l for l in lists if len(l) > 0]
>>> non_empty_lists
[['a', 'b', 'c'], ['d', 'e']]