Relative paths error depending on Terminal's current folder
I've been trying to implement in my scripts that work with files the possibility to run my program regardless of the path the package is located in.
I found that with this code, it would generally work well:
import os.path
file_path = os.path.dirname(__file__)
data_path = file_path + '/myfile's/path/'
If from Terminal I try to execute
python3 path/to/my/script/script.py
I get no problem. However, if I try to run my script within the current folder it is located in (cd path/to/my/script/) an error shows up. Claiming that /myfile's/path is not a valid directory.
I figured that I could implement the following:
file_path = os.path.dirname(__file__)
if file_path:
data_path = file_path + '/myfile's/path'
else:
data_path = 'myfile's/path'
This seems to work out fine, but I wonder if there is a normalized or better way to do this that regards other types of errors I might have not encountered yet.
Thanks ;)
python python-3.x macos
asked Nov 15 '18 at 20:11
MadheolksMadheolks
52
add a comment |
I've been trying to implement in my scripts that work with files the possibility to run my program regardless of the path the package is located in.
I found that with this code, it would generally work well:
import os.path
file_path = os.path.dirname(__file__)
data_path = file_path + '/myfile's/path/'
If from Terminal I try to execute
python3 path/to/my/script/script.py
I get no problem. However, if I try to run my script within the current folder it is located in (cd path/to/my/script/) an error shows up. Claiming that /myfile's/path is not a valid directory.
I figured that I could implement the following:
file_path = os.path.dirname(__file__)
if file_path:
data_path = file_path + '/myfile's/path'
else:
data_path = 'myfile's/path'
This seems to work out fine, but I wonder if there is a normalized or better way to do this that regards other types of errors I might have not encountered yet.
Thanks ;)
python python-3.x macos
asked Nov 15 '18 at 20:11
MadheolksMadheolks
52
Excuse the single quotations marks from the /myfile's/path, I just realized I should've declared the string with these " ". hahaha
– Madheolks
Nov 15 '18 at 20:15
Have you tried usingos.path.join()to createdata_path? It joins components "intelligently" so it may account for something you're missing by using simple concatenation.
– Tim Klein
Nov 15 '18 at 20:30
add a comment |
I've been trying to implement in my scripts that work with files the possibility to run my program regardless of the path the package is located in.
I found that with this code, it would generally work well:
import os.path
file_path = os.path.dirname(__file__)
data_path = file_path + '/myfile's/path/'
If from Terminal I try to execute
python3 path/to/my/script/script.py
I get no problem. However, if I try to run my script within the current folder it is located in (cd path/to/my/script/) an error shows up. Claiming that /myfile's/path is not a valid directory.
I figured that I could implement the following:
file_path = os.path.dirname(__file__)
if file_path:
data_path = file_path + '/myfile's/path'
else:
data_path = 'myfile's/path'
This seems to work out fine, but I wonder if there is a normalized or better way to do this that regards other types of errors I might have not encountered yet.
Thanks ;)
python python-3.x macos
asked Nov 15 '18 at 20:11
MadheolksMadheolks
52
I've been trying to implement in my scripts that work with files the possibility to run my program regardless of the path the package is located in.
I found that with this code, it would generally work well:
import os.path
file_path = os.path.dirname(__file__)
data_path = file_path + '/myfile's/path/'
If from Terminal I try to execute
python3 path/to/my/script/script.py
I get no problem. However, if I try to run my script within the current folder it is located in (cd path/to/my/script/) an error shows up. Claiming that /myfile's/path is not a valid directory.
I figured that I could implement the following:
file_path = os.path.dirname(__file__)
if file_path:
data_path = file_path + '/myfile's/path'
else:
data_path = 'myfile's/path'
This seems to work out fine, but I wonder if there is a normalized or better way to do this that regards other types of errors I might have not encountered yet.
Thanks ;)
python python-3.x macos
python python-3.x macos
asked Nov 15 '18 at 20:11
MadheolksMadheolks
52
asked Nov 15 '18 at 20:11
MadheolksMadheolks
52
asked Nov 15 '18 at 20:11
MadheolksMadheolks
52
asked Nov 15 '18 at 20:11
MadheolksMadheolks
52
asked Nov 15 '18 at 20:11
MadheolksMadheolks
52
52
Excuse the single quotations marks from the /myfile's/path, I just realized I should've declared the string with these " ". hahaha
– Madheolks
Nov 15 '18 at 20:15
Have you tried usingos.path.join()to createdata_path? It joins components "intelligently" so it may account for something you're missing by using simple concatenation.
– Tim Klein
Nov 15 '18 at 20:30
add a comment |
Excuse the single quotations marks from the /myfile's/path, I just realized I should've declared the string with these " ". hahaha
– Madheolks
Nov 15 '18 at 20:15
Have you tried usingos.path.join()to createdata_path? It joins components "intelligently" so it may account for something you're missing by using simple concatenation.
– Tim Klein
Nov 15 '18 at 20:30
Excuse the single quotations marks from the /myfile's/path, I just realized I should've declared the string with these " ". hahaha
– Madheolks
Nov 15 '18 at 20:15
Excuse the single quotations marks from the /myfile's/path, I just realized I should've declared the string with these " ". hahaha
– Madheolks
Nov 15 '18 at 20:15
Have you tried using
os.path.join() to create data_path? It joins components "intelligently" so it may account for something you're missing by using simple concatenation.– Tim Klein
Nov 15 '18 at 20:30
Have you tried using
os.path.join() to create data_path? It joins components "intelligently" so it may account for something you're missing by using simple concatenation.– Tim Klein
Nov 15 '18 at 20:30
add a comment |
1 Answer
1
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not really (this question isnt actually appropriate for SO but here goes)
you need to do this... but when I do i tend to abstract it out
def rel_path(*parts):
BASE = os.path.dirname(__file__)
return os.path.abspath(os.path.join(BASE,*parts))
then I use this function everywhere
my_path = rel_path("myfiles","path","example.jpg")
# /path/to/this/file/myfiles/path/example.jpg
answered Nov 15 '18 at 20:16
Joran BeasleyJoran Beasley
74k682121
add a comment |
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
not really (this question isnt actually appropriate for SO but here goes)
you need to do this... but when I do i tend to abstract it out
def rel_path(*parts):
BASE = os.path.dirname(__file__)
return os.path.abspath(os.path.join(BASE,*parts))
then I use this function everywhere
my_path = rel_path("myfiles","path","example.jpg")
# /path/to/this/file/myfiles/path/example.jpg
answered Nov 15 '18 at 20:16
Joran BeasleyJoran Beasley
74k682121
add a comment |
not really (this question isnt actually appropriate for SO but here goes)
you need to do this... but when I do i tend to abstract it out
def rel_path(*parts):
BASE = os.path.dirname(__file__)
return os.path.abspath(os.path.join(BASE,*parts))
then I use this function everywhere
my_path = rel_path("myfiles","path","example.jpg")
# /path/to/this/file/myfiles/path/example.jpg
answered Nov 15 '18 at 20:16
Joran BeasleyJoran Beasley
74k682121
add a comment |
not really (this question isnt actually appropriate for SO but here goes)
you need to do this... but when I do i tend to abstract it out
def rel_path(*parts):
BASE = os.path.dirname(__file__)
return os.path.abspath(os.path.join(BASE,*parts))
then I use this function everywhere
my_path = rel_path("myfiles","path","example.jpg")
# /path/to/this/file/myfiles/path/example.jpg
answered Nov 15 '18 at 20:16
Joran BeasleyJoran Beasley
74k682121
not really (this question isnt actually appropriate for SO but here goes)
you need to do this... but when I do i tend to abstract it out
def rel_path(*parts):
BASE = os.path.dirname(__file__)
return os.path.abspath(os.path.join(BASE,*parts))
then I use this function everywhere
my_path = rel_path("myfiles","path","example.jpg")
# /path/to/this/file/myfiles/path/example.jpg
answered Nov 15 '18 at 20:16
Joran BeasleyJoran Beasley
74k682121
answered Nov 15 '18 at 20:16
Joran BeasleyJoran Beasley
74k682121
answered Nov 15 '18 at 20:16
Joran BeasleyJoran Beasley
74k682121
answered Nov 15 '18 at 20:16
Joran BeasleyJoran Beasley
74k682121
74k682121
add a comment |
add a comment |
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Excuse the single quotations marks from the /myfile's/path, I just realized I should've declared the string with these " ". hahaha
– Madheolks
Nov 15 '18 at 20:15
Have you tried using
os.path.join()to createdata_path? It joins components "intelligently" so it may account for something you're missing by using simple concatenation.– Tim Klein
Nov 15 '18 at 20:30