Loop over a directory, but starting from the end? - python-3.x

Question

I have a directory with many sub-directories, each of which follow the same naming convention; the day's date. Today a folder was made: 2021-04-22
I occasionally need to go through these directories and read a file from one, but once I've read it I don't need to again.
li = []
for root, dirs, files in os.walk(path):
for f in files:
li.append(f)
The list shows me the order the files are read, which is an alphabetic(numeric?) order. I know the newest files are going to be towards the bottom because of the naming convention.
How can I start my for loop from the 'end' rather than the 'beginning'?
If this is possible, I'd then exit the loop when my criteria are met, else, what would be the point of starting at the end?
EDIT: My original naming convention was mistyped. It is YYYY-MM-DD thank you #null

Answer 1

To reverse any iterable or iterator in python warp it in reversed().
In your code:
li = []
for root, dirs, files in os.walk(path):
for f in reversed(files):
li.append(f)

Answer 2

Suppose you have this tree of directories:
.
├── 1
│   ├── a
│   │   ├── 03-01-2021
│   │   └── 04-22-2021
│   ├── b
│   │   └── 04-21-2021
│   └── c
├── 2
│   ├── a
│   │   └── 05-01-2020
│   ├── b
│   └── c
│   └── 01-01-1966
└── 3
├── a
│   ├── 12-15-2001
│   └── 12-15-2001_blah
├── b
└── c
You can use pathlib with a recursive glob to get your directories. Then use a regex to reverse the date pattern to ISO 8601 format of YYYY-MM-DD and sort in reverse fashion:
import re
from pathlib import Path
p=Path('/tmp/test/')
my_glob='**/[0-9][0-9]-[0-9][0-9]-[0-9][0-9][0-9][0-9]*'
my_regex=r'.*/(\d{2})-(\d{2})-(\d{4}).*'
for pa in sorted(
[pa for pa in p.glob(my_glob) if pa.is_dir()],
key=lambda pa: re.sub(my_regex,r'\3-\2-\1', str(pa)), reverse=True):
print(pa)
Prints:
/tmp/test/1/a/04-22-2021
/tmp/test/1/b/04-21-2021
/tmp/test/1/a/03-01-2021
/tmp/test/2/a/05-01-2020
/tmp/test/3/a/12-15-2001_blah
/tmp/test/3/a/12-15-2001
/tmp/test/2/c/01-01-1966
The glob of '**/*' makes the search recursive and adding:
**/[0-9][0-9]-[0-9][0-9]-[0-9][0-9][0-9][0-9]*
will only return files and directories that match that naming pattern. By adding the test if pa.is_dir() we are only looking at directories -- not files.
The regex:
my_regex=r'.*/(\d{2})-(\d{2})-(\d{4})/'
re.sub(my_regex,r'\3-\2-\1', str(pa))
Removes everything other than the date and reverses it to ISO 8601 for the key passed to sorted.
You asked the default order files are returned. Usually files are breadth first oldest to newest. That said, it is OS and implementation dependent.
You updated the question that your files DO have the YYYY-MM-DD naming convention. If so, just change or remove the regex. Same basic method handles both.

Answer 3

Since files is a list, you can use extended list slicing to reverse the list:
li = []
for root, dirs, files in os.walk(path):
for f in files[::-1]:
li.append(f)