通过理解检查列表项是否在其他2D列表中丢失

Question

The first list contains filenames with extensions:

afiles = [['file1', '.exe'], ['file2', '.txt']]

The second list contains filenames without extension. eg:

bfiles = ['file1', 'file2', 'file3', 'file4']

I know want to know which files of bfiles are missing in afiles. The expected result should be:

['file3', 'file4']

I'd like to realize this with comprehension.

Answer 1

You could just try this simple list comprehension :)

>>> afiles = [['file1', '.exe'], ['file2', '.txt']]
>>> bfiles = ['file1', 'file2', 'file3', 'file4']
>>> [x for x in bfiles if x not in (y[0] for y in afiles)]
['file3', 'file4']

or better, you could just assign the afiles files without extensions like,

>>> afiles_names = [x[0] for x in afiles]
>>> [x for x in bfiles if x not in afiles_names] # so you won't have to compute that each time
['file3', 'file4']

Answer 2

Just check if items of bfiles fail to be found in any zero indexes in afiles.

afiles = [['file1', '.exe'], ['file2', '.txt']]
bfiles = ['file1', 'file2', 'file3', 'file4']

result = [file for file in bfiles if not any(file == afile[0] for afile in afiles)]
['file3', 'file4']

Answer 3

List-comprehension with itertools.chain :

from itertools import chain

afiles = [['file1', '.exe'], ['file2', '.txt']]
bfiles = ['file1', 'file2', 'file3', 'file4']

print([x for x in bfiles if x not in chain.from_iterable(afiles)])
# ['file3', 'file4']

Answer 4

You can use the set method difference() :

set(bfiles).difference(i[0] for i in afiles)
# {'file4', 'file3'}