Develop Reference

Path.replace(target) does not work with my approach

Pathlib is suggested instead of os.module because is newer. However I find less documentation about Pathlib if I look for. I show an issue that give me problems.
Initial condition
In a directory I have the following files:
I want to change the prefix of the files "comune_min_corr_" in "zp".
OS approach
With the following code where df4 is dataframe that contain "comune":
for comune in df4:
os.replace('./Spedire/comune_min_corr_'+str(comune)+'.gpkg', './Spedire/zp_'+str(comune)+'.gpkg')
To be clear, "comune" in the following image are three items: Airolo, Alto Malcantone, Aranno. They are part of the filename.
The result is right, the files "zp_comune.gpkg" are replaced by the files "comune_min_corr_comune.gpkg" and "comune_min_corr_comune.gpkg" disappear:
Pathlib approach
I create again the "initial condition".
Then, I try to do the same thing than before, adapting for the "pathlib.replace(target)".
They say here https://realpython.com/python-pathlib/ that we have to use "source.replace(destination)" structure. I did with the following code.
for comune in df4:
Path=('Spedire')
('/comune_min_corr_'+str(comune)+'.gpkg').replace('/comune_min_corr_'+str(comune)+'.gpkg','/zp_'+str(comune)+'.gpkg')
It give me no error but it does not work, the content of the directory does not change
Question
Could you provide the solution to obtain the same result for pathlib that I obtain with os? Please explain your step and why it does not work for pathlib.
Thank you.

#Barmar I changed now and it works. Thank you
To resume:
with os.replace() it work without defining a variable because replace() is a function of the module "os".
with "path.replace" we have to do so:
for comune in df4:
variable = Path('Spedire')/f'comune_min_corr_{comune}.gpkg'
variable.replace(Path('Spedire')/f'zp_{comune}.gpkg')
In fact, we have to build an object (here called "variable") because "path.replace" is a class of the module "pathlib". Class always needs to have an object to obtain something because otherwise python does not know what we are talking about.
To be noted: I probably have done something similar, but I repeateadly worked confusing "gpkg" with "gpgk". This trivial mistake leads me in the wrong way.
If I told something wrong, please correct me.

How to print COCOeval results to file, or get the printed str?

Following the coco api toturial, using cocoEval.evaluate(), cocoEval.accumulate() and cocoEval.summarize() can have the ecaluation result printed in the console.
I wonder if there is a way one can derive the str object of evaluation result, rather than have it printed, so that I can record it to my history file. Better if no need to modify the original code in cocoapi. I'm not sure if cocoapi has some relative interfaces.
I've checked the source code in COCOeval and found it's possible if I make cocoEval.evaluate(), cocoEval.accumulate() and cocoEval.summarize() return the strs. But these methods do return something else. I'm afraid there will be some potential instability after changing or extending their original return value, since some other functions may need them.

Instead of doing all this, you can actually do redirection. In your terminal you can run your script and redirect the output to a file. The output of the console will be logged to a file.
E.g. python eval_model.py > results.txt

The results have also been saved to stats, more specifically, you can directly use coco_eval.stats to extract the mAP results.

One-line "for" loop using list comprehension

Someone has challenged me to create a program that sorts their pictures into folders based on the month they were taken, and I want to do it in one line (I know, it's inefficient and unreadable, but I still want to do it because one-liners are cool)
I needed a for loop to accomplish this, but the only way I know of to use a for loop in one line is list comprehension, so that's what I did, but it creates an empty list, and doesn't print anything from the list or anything.
What I'm doing is renaming the file to be the month created + original filename (ex: bacon.jpg --> May\bacon.jpg)
Here is my code (Python 3.7.3):
import time
import os.path
[os.rename(str(os.fspath(f)), str(time.ctime(os.path.getctime(str(os.fspath(f))))).split()[1] + '\\' + str(os.fspath(f))) for f in os.listdir() if f.endswith('.jpg')]
and the more readable, non-list-comprehension version:
import time
import os.path
for f in os.listdir():
fn = str(os.fspath(f))
dateCreated = str(time.ctime(os.path.getctime(fn)))
monthCreated = dateCreated.split()[1]
os.rename(fn, monthCreated + '\\' + fn)
Is list comprehension a bad way to do it? Also, is there a reason why, if I print the list it's [] instead of [None, None, None, None, None, (continuing "None"s for every image moved)]?
Please note: I understand that it's inefficient and bad practice. If I were doing this for purposes other than just for fun to see if I could do it, I would obviously not try to do it in one line.

This is bad in two immediate respects:
You're using a list comprehension when you're not actually interested in constructing a list -- you ignore the object you just constructed.
Your construction has an ugly side effect in the OS.
Your purpose appears to be renaming a sequence of files, not constructing a list. The Python facility you want is, I believe, the map function. Write a function to change one file name, and then use map on a list of file names -- or tuples of old, new file names -- to run through the sequence of desired changes.

Is list comprehension a bad way to do it?
YES. But if you want to do it in one line, it is either that or using ";". For instance:
for x in range(5): print(x);print(x+2)
And, by the way, just renaming a file including a slash will not create a folder. You have to use os.mkdir('foldername').
In the end, if you really want to do that, I would just recommend doing it normally in many lines and then separating it with semicolons in a single line.

Checking a file exists (and ignoring case) in Python

I have a Python script and I want to check if a file exists, but I want to ignore case
eg.
path = '/Path/To/File.log'
if os.path.isfile(path):
return true
The directory may look like this "/path/TO/fILe.log". But the above should still return true.

Generate one-time a set S of all absolute paths in the filesystem using os.walk, lowering them all as you collect them using str.lower.
Iterate through your large list of paths to check for existing, checking with if my_path.lower() in S.
(Optional) Go and interrogate whoever provided you the list with inconsistent cases. It sounds like an XY problem, there may be some strange reason for this and an easier way out.

How do I parse a listing of files to get just the filenames in Python?

So lets say I'm using Python's ftplib to retrieve a list of log files from an FTP server. How would I parse that list of files to get just the file names (the last column) inside a list? See the link above for example output.

Using retrlines() probably isn't the best idea there, since it just prints to the console and so you'd have to do tricky things to even get at that output. A likely better bet would be to use the nlst() method, which returns exactly what you want: a list of the file names.

This best answer
You may want to use ftp.nlst() instead of ftp.retrlines(). It will give you exactly what you want.
If you can't, read the following :
Generators for sysadmin processes
In his now famous review, Generator Tricks For Systems Programmers An Introduction, David M. Beazley gives a lot of receipes to answer to this kind of data problem with wuick and reusable code.
E.G :
# empty list that will receive all the log entry
log = []
# we pass a callback function bypass the print_line that would be called by retrlines
# we do that only because we cannot use something better than retrlines
ftp.retrlines('LIST', callback=log.append)
# we use rsplit because it more efficient in our case if we have a big file
files = (line.rsplit(None, 1)[1] for line in log)
# get you file list
files_list = list(files)
Why don't we generate immediately the list ?
Well, it's because doing it this way offer you much flexibility : you can apply any intermediate generator to filter files before turning it into files_list : it's just like pipe, add a line, you add a process without overheat (since it's generators). And if you get rid off retrlines, it still work be it's even better because you don't store the list even one time.
EDIT : well, I read the comment to the other answer and it says that this won't work if there is any space in the name.
Cool, this will illustrate why this method is handy. If you want to change something in the process, you just change a line. Swap :
files = (line.rsplit(None, 1)[1] for line in log)
and
# join split the line, get all the item from the field 8 then join them
files = (' '.join(line.split()[8:]) for line in log)
Ok, this may no be obvious here, but for huge batch process scripts, it's nice :-)

And a slightly less-optimal method, by the way, if you're stuck using retrlines() for some reason, is to pass a function as the second argument to retrlines(); it'll be called for each item in the list. So something like this (assuming you have an FTP object named 'ftp') would work as well:
filenames = []
ftp.retrlines('LIST', lambda line: filenames.append(line.split()[-1]))
The list 'filenames' will then be a list of the file names.

Is there any reason why ftplib.FTP.nlst() won't work for you? I just checked and it returns only names of the files in a given directory.

Since every filename in the output starts at the same column, all you have to do is get the position of the dot on the first line:
drwxrwsr-x 5 ftp-usr pdmaint 1536 Mar 20 09:48 .
Then slice the filename out of the other lines using the position of that dot as the starting index.
Since the dot is the last character on the line, you can use the length of the line minus 1 as the index. So the final code is something like this:
lines = ftp.retrlines('LIST')
lines = lines.split("\n") # This should split the string into an array of lines
filename_index = len(lines[0]) - 1
files = []
for line in lines:
files.append(line[filename_index:])

If the FTP server supports the MLSD command, then please see section “single directory case” from that answer.
Use an instance (say ftpd) of the FTPDirectory class, call its .getdata method with connected ftplib.FTP instance in the correct folder, then you can:
directory_filenames= [ftpfile.name for ftpfile in ftpd.files]

I believe it should work for you.
file_name_list = [' '.join(each_file.split()).split()[-1] for each_file_detail in file_list_from_log]
NOTES -
Here I am making a assumption that you want the data in the program (as list), not on console.
each_file_detail is each line that is being produced by the program.
' '.join(each_file.split())
To replace multiple spaces by 1 space.