How can I replace a substring in a Python pathlib.Path?
Is there an easy way to replace a substring within a pathlib.Path
object in Python? The pathlib module is nicer in many ways than storing a path as a str
and using os.path
, glob.glob
etc, which are built in to pathlib
. But I often use files that follow a pattern, and often replace substrings in a path to access other files:
data/demo_img.png
data/demo_img_processed.png
data/demo_spreadsheet.csv
Previously I could do:
img_file_path = "data/demo_img.png"
proc_img_file_path = img_file_path.replace("_img.png", "_img_proc.png")
data_file_path = img_file_path.replace("_img.png", "_spreadsheet.csv")
pathlib
can replace the file extension with the with_suffix()
method, but only accepts extensions as valid suffixes. The workarounds are:
import pathlib
import os
img_file_path = pathlib.Path("data/demo_img.png")
proc_img_file_path = pathlib.Path(str(img_file_path).replace("_img.png", "_img_proc.png"))
# os.fspath() is available in Python 3.6+ and is apparently safer than str()
data_file_path = pathlib.Path(os.fspath(img_file_path).replace("_img.png", "_img_proc.png"))
Converting to a string to do the replacement and reconverting to a Path
object seems laborious. Assume that I never have a copy of the string form of img_file_path
, and have to convert the type as needed.
python python-3.x pathlib
add a comment |
Is there an easy way to replace a substring within a pathlib.Path
object in Python? The pathlib module is nicer in many ways than storing a path as a str
and using os.path
, glob.glob
etc, which are built in to pathlib
. But I often use files that follow a pattern, and often replace substrings in a path to access other files:
data/demo_img.png
data/demo_img_processed.png
data/demo_spreadsheet.csv
Previously I could do:
img_file_path = "data/demo_img.png"
proc_img_file_path = img_file_path.replace("_img.png", "_img_proc.png")
data_file_path = img_file_path.replace("_img.png", "_spreadsheet.csv")
pathlib
can replace the file extension with the with_suffix()
method, but only accepts extensions as valid suffixes. The workarounds are:
import pathlib
import os
img_file_path = pathlib.Path("data/demo_img.png")
proc_img_file_path = pathlib.Path(str(img_file_path).replace("_img.png", "_img_proc.png"))
# os.fspath() is available in Python 3.6+ and is apparently safer than str()
data_file_path = pathlib.Path(os.fspath(img_file_path).replace("_img.png", "_img_proc.png"))
Converting to a string to do the replacement and reconverting to a Path
object seems laborious. Assume that I never have a copy of the string form of img_file_path
, and have to convert the type as needed.
python python-3.x pathlib
Whatever you do, beware of usingPath.replace
as as attempted substitute - not the same thing, and can clobber existing data on filesystem!
– wim
Nov 20 at 19:05
That's right. I luckily read the documentation before trying.replace()
will rename the current file to the target, and replace it if the file already exists.
– Hector
Nov 20 at 20:31
add a comment |
Is there an easy way to replace a substring within a pathlib.Path
object in Python? The pathlib module is nicer in many ways than storing a path as a str
and using os.path
, glob.glob
etc, which are built in to pathlib
. But I often use files that follow a pattern, and often replace substrings in a path to access other files:
data/demo_img.png
data/demo_img_processed.png
data/demo_spreadsheet.csv
Previously I could do:
img_file_path = "data/demo_img.png"
proc_img_file_path = img_file_path.replace("_img.png", "_img_proc.png")
data_file_path = img_file_path.replace("_img.png", "_spreadsheet.csv")
pathlib
can replace the file extension with the with_suffix()
method, but only accepts extensions as valid suffixes. The workarounds are:
import pathlib
import os
img_file_path = pathlib.Path("data/demo_img.png")
proc_img_file_path = pathlib.Path(str(img_file_path).replace("_img.png", "_img_proc.png"))
# os.fspath() is available in Python 3.6+ and is apparently safer than str()
data_file_path = pathlib.Path(os.fspath(img_file_path).replace("_img.png", "_img_proc.png"))
Converting to a string to do the replacement and reconverting to a Path
object seems laborious. Assume that I never have a copy of the string form of img_file_path
, and have to convert the type as needed.
python python-3.x pathlib
Is there an easy way to replace a substring within a pathlib.Path
object in Python? The pathlib module is nicer in many ways than storing a path as a str
and using os.path
, glob.glob
etc, which are built in to pathlib
. But I often use files that follow a pattern, and often replace substrings in a path to access other files:
data/demo_img.png
data/demo_img_processed.png
data/demo_spreadsheet.csv
Previously I could do:
img_file_path = "data/demo_img.png"
proc_img_file_path = img_file_path.replace("_img.png", "_img_proc.png")
data_file_path = img_file_path.replace("_img.png", "_spreadsheet.csv")
pathlib
can replace the file extension with the with_suffix()
method, but only accepts extensions as valid suffixes. The workarounds are:
import pathlib
import os
img_file_path = pathlib.Path("data/demo_img.png")
proc_img_file_path = pathlib.Path(str(img_file_path).replace("_img.png", "_img_proc.png"))
# os.fspath() is available in Python 3.6+ and is apparently safer than str()
data_file_path = pathlib.Path(os.fspath(img_file_path).replace("_img.png", "_img_proc.png"))
Converting to a string to do the replacement and reconverting to a Path
object seems laborious. Assume that I never have a copy of the string form of img_file_path
, and have to convert the type as needed.
python python-3.x pathlib
python python-3.x pathlib
asked Nov 20 at 18:58
Hector
9828
9828
Whatever you do, beware of usingPath.replace
as as attempted substitute - not the same thing, and can clobber existing data on filesystem!
– wim
Nov 20 at 19:05
That's right. I luckily read the documentation before trying.replace()
will rename the current file to the target, and replace it if the file already exists.
– Hector
Nov 20 at 20:31
add a comment |
Whatever you do, beware of usingPath.replace
as as attempted substitute - not the same thing, and can clobber existing data on filesystem!
– wim
Nov 20 at 19:05
That's right. I luckily read the documentation before trying.replace()
will rename the current file to the target, and replace it if the file already exists.
– Hector
Nov 20 at 20:31
Whatever you do, beware of using
Path.replace
as as attempted substitute - not the same thing, and can clobber existing data on filesystem!– wim
Nov 20 at 19:05
Whatever you do, beware of using
Path.replace
as as attempted substitute - not the same thing, and can clobber existing data on filesystem!– wim
Nov 20 at 19:05
That's right. I luckily read the documentation before trying.
replace()
will rename the current file to the target, and replace it if the file already exists.– Hector
Nov 20 at 20:31
That's right. I luckily read the documentation before trying.
replace()
will rename the current file to the target, and replace it if the file already exists.– Hector
Nov 20 at 20:31
add a comment |
1 Answer
1
active
oldest
votes
You are correct. To replace old with new in Path p, you need:
p = Path(str(p).replace(old, new))
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
You are correct. To replace old with new in Path p, you need:
p = Path(str(p).replace(old, new))
add a comment |
You are correct. To replace old with new in Path p, you need:
p = Path(str(p).replace(old, new))
add a comment |
You are correct. To replace old with new in Path p, you need:
p = Path(str(p).replace(old, new))
You are correct. To replace old with new in Path p, you need:
p = Path(str(p).replace(old, new))
answered Nov 20 at 21:42
J_H
3,1781616
3,1781616
add a comment |
add a comment |
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Whatever you do, beware of using
Path.replace
as as attempted substitute - not the same thing, and can clobber existing data on filesystem!– wim
Nov 20 at 19:05
That's right. I luckily read the documentation before trying.
replace()
will rename the current file to the target, and replace it if the file already exists.– Hector
Nov 20 at 20:31