Suppose I have a filename ./file and I want to remove the leading ./ (if it starts with ./). What would be the best python way to do this? My take at this is:
if filename.startswith("./"):
# remove leading ./
filename = filename[2:]
Isn't there a more elegant way?
PS: it should remove the leading part only if it matches exactly, think of the following filenames:
./filename
.configfile
../file
./.configfile
filenames = ['.configfile', './file1', './dir/file2']
filenames = [fn[2:] if fn.startswith("./") else fn for fn in filenames]
borrowing from your idea but doing it in one line
My personal preference would be the solution by @user1443098 But for completnes - using re
import re
pattern = re.compile(r'^\./')
files = ['./filename', '.configfile', '../file']
new_files = [re.sub(pattern, '', fname) for fname in files]
print(new_files)
The way you do it is the ways it's usually done. Of course, you can always move it into a function:
def remove_prefix(s, prefix):
if s.startswith(prefix):
s = s[len(prefix):]
return s
I would consider writing a helper function:
def strip_prefix(s, prefix):
if s.startswith(prefix):
return s[len(prefix):]
else:
return s
print(strip_prefix('./filename', './'))
print(strip_prefix('.configfile', './'))
print(strip_prefix('../file', './'))
Assuming you are using python3, the best way to deal with paths is trough the pathlib module:
if you only want the filename
from pathlib import Path
paths = ["./filename", ".configfile", "../file"]
only_filenames = [Path(f).name for f in paths]
./ should be removed, something like ../ should stay the same. Nowhere is stated that only the filename is wantedif filename.startswith("./"):
filename.lstrip("./")
./.configfile
new_list = [element.replace("./", "") for element in old_list]if element.startswith(".\")if it's needed. It's just quick comment, not an nswer, feel the deffierenceregex:re.sub(r'^./', '', file)