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Open file in a relative location in Python
Suppose python code is executed in not known by prior windows directory say 'main' , and wherever code is installed when it runs it needs to access to directory 'main/2091/data.txt' .
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How do I get the path and name of the file that is currently executing?
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p1.py:
execfile("p2.py")
p2.py:
import inspect, os
print (inspect.getfile(inspect.currentframe()) # script filename (usually with path)
print (os.path.dirname(os.path.abspath(inspect.getfile(inspect.currentframe())))) # script directory
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os.walk without digging into directories below
How do I limit os.walk to only return files in the directory I provide it?
20 Answers
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`from … import` vs `import .` [duplicate]
...urself when you import for simplicity or to avoid masking built ins:
from os import open as open_
# lets you use os.open without destroying the
# built in open() which returns file handles.
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How to check if a file exists in Go?
... function solely intended to check if a file exists or not (like Python's os.path.exists ). What is the idiomatic way to do it?
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How can I safely create a nested directory?
What is the most elegant way to check if the directory a file is going to be written to exists, and if not, create the directory using Python? Here is what I tried:
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How to find the operating system version using JavaScript?
How can I find the OS name and OS version using JavaScript?
13 Answers
13
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How to access environment variable values?
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Environment variables are accessed through os.environ
import os
print(os.environ['HOME'])
Or you can see a list of all the environment variables using:
os.environ
As sometimes you might need to see a complete list!
# using get will return `None` if a key is not...
Python list directory, subdirectory, and files
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Use os.path.join to concatenate the directory and file name:
for path, subdirs, files in os.walk(root):
for name in files:
print os.path.join(path, name)
Note the usage of path and not root in the concatenation, sinc...
How do I check whether a file exists without exceptions?
...open it.
If you're not planning to open the file immediately, you can use os.path.isfile
Return True if path is an existing regular file. This follows symbolic links, so both islink() and isfile() can be true for the same path.
import os.path
os.path.isfile(fname)
if you need to be sure it...