List of the most recently updated files in Python

Question

So using the following piece of code I can easily find the most recently updated file in a folder:

files = os.listdir(UPLOAD_DIR+"/"+configData[obj]["client_name"])
paths = [os.path.join(UPLOAD_DIR+"/"+configData[obj]["client_name"], basename) for basename in files]
file = max(paths, key=os.path.getctime)

But what if there are two or more files that have the exact same updated time? How do I get a list of such files?

Totally unrelated, but hardcoding path seprators (`UPLOAD_DIR+"/"+configData[obj]["client_name"]) kind of defeats the whole point of `os.path.join()`. This should be `os.path.join(UPLOAD_DIR, configData[obj]["client_name"], basename)`. — bruno desthuilliers, Oct 18 '18 at 12:12
Possible duplicate of [How to get all the maximums max function](https://stackoverflow.com/questions/10823227/how-to-get-all-the-maximums-max-function) — Georgy, Oct 18 '18 at 12:42

Thierry Lathuille · Accepted Answer · 2018-10-18T13:00:06.210

The shortest code: find the latest ctime, then get all files having this latest ctime:

def most_recent(paths):
    if not paths:
        return []
    latest_ctime = max(os.path.getctime(p) for p in paths)
    most_recent_files = [p for p in paths if os.path.getctime(p)==latest_ctime]
    return most_recent_files

We loop twice over the list of paths, though, and there is a risk of race condition if the ctime of the most recent file changes between the two loops: in this case, it wouldn't be found again in the second loop.

We can do it in one loop, with a little bit more code, eliminating the race condition:

def most_recent_one_loop(paths):
    out = []
    latest_ctime = 0
    for p in paths:
        ct = os.path.getctime(p)
        if ct > latest_ctime:
            latest_ctime = ct
            out = [p]
        elif ct == latest_ctime:
            out.append(p)
    return out

As we can expect, this is about twice as fast (about 100 paths in the folder for the test):

%timeit most_recent(paths)
# 1000 loops, best of 3: 477 µs per loop

%timeit most_recent_one_loop(paths)
# 1000 loops, best of 3: 239 µs per loop

@AzatIbrakov Yes, I added a way to do it in one loop to avoid that. — Thierry Lathuille, Oct 18 '18 at 12:53
it will be great to add that race conditions are unavoidable here without some sort of mechanism for "blocking" files, but this is a different and IMO harder question — Azat Ibrakov, Oct 18 '18 at 13:03

score 0 · Answer 2 · answered Oct 18 '18 at 12:23

0

Probably not the tidiest way of doing it but:

maxval = os.path.getctime(max(paths, key=os.path.getctime))

indices = [index for index, val in enumerate(paths) if os.path.getctime(val) == maxval]
for index in indices:
    print(paths[index])

answered Oct 18 '18 at 12:23

Andrew McDowell

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score 0 · Answer 3 · answered Oct 18 '18 at 12:27

For Python 3, look like the max method can't fix your issue , as the Python 3 docs explicitly state:

If multiple items are maximal, the function returns the first one encountered. This is consistent with other sort-stability preserving tools such as sorted(iterable, key=keyfunc, reverse=True)[0] and heapq.nlargest(1, iterable, key=keyfunc).

You may need to use sorted command to find the multi max value

list = sorted(paths, key=os.path.getctime, reverse=True)
files=[]
for i in list:
  if os.path.getctime(list[0]) == os.path.getctime(i):
    files.append(i)
  else:
    break

List of the most recently updated files in Python

3 Answers3