python-os

问题 I have a python program, and I want to get the path to the program from within the program, but INCLUDING the file name itself. The name of my file is PyWrapper.py. Right now I'm doing this: import sys,os pathname = os.path.dirname(sys.argv[0]) fullpath = os.path.abspath(pathname) print fullpath The output is: /home/mrpickles/Desktop/HANSTUFF/securesdk/src/ This is the path to the directory in which my file is saved, but I would like it to output: /home/mrpickles/Desktop/HANSTUFF/securesk/src

问题 I am practicing with the os module and more specifically os.walk() . I am wondering if there is an easier/more efficient way to find the actual path to a file considering this produces a path that suggests the file is in the original folder when os.walk() is first ran: import os threshold_size = 500 for folder, subfolders, files in os.walk(os.getcwd()): for file in files: filePath = os.path.abspath(file) if os.path.getsize(filePath) >= threshold_size: print filePath, str(os.path.getsize

问题 Is it possible for a jupyter notebook to get the name of its own file, similarly to what we would do from a python script? os.path.basename(__file__) doesn't seem to work, at least for me on jupyterlab sys.argv[0] returns my_home/anaconda3/lib/python3.6/site-packages/ipykernel_launcher.py 回答1: The only way I've found is through JavaScritp as in this answer. The compact form is a cell like this: %%javascript IPython.notebook.kernel.execute(`notebookName = '${window.document.getElementById(

问题 Is it possible for a jupyter notebook to get the name of its own file, similarly to what we would do from a python script? os.path.basename(__file__) doesn't seem to work, at least for me on jupyterlab sys.argv[0] returns my_home/anaconda3/lib/python3.6/site-packages/ipykernel_launcher.py 回答1: The only way I've found is through JavaScritp as in this answer. The compact form is a cell like this: %%javascript IPython.notebook.kernel.execute(`notebookName = '${window.document.getElementById(

问题 I am new to python and encountering some issues while executing os commands. I have set my environment variables like as shown below SPARK_HOME = '/opt/spark' HAIL_HOME = '/opt/hail/hail' When I type os.getenv('SPARK_HOME') , I get the below output '/opt/spark/' But when I type os.getenv('HAIL_HOME') , I get blank output Please note that I type the above two commands from a virtual environment using jupyter notebook. Why it works for spark and returns empty for hail Can guide me with this

问题 I am new to python and encountering some issues while executing os commands. I have set my environment variables like as shown below SPARK_HOME = '/opt/spark' HAIL_HOME = '/opt/hail/hail' When I type os.getenv('SPARK_HOME') , I get the below output '/opt/spark/' But when I type os.getenv('HAIL_HOME') , I get blank output Please note that I type the above two commands from a virtual environment using jupyter notebook. Why it works for spark and returns empty for hail Can guide me with this

问题 I recently created two environment variables in my terminal as shown below export SPARK_HOME='/opt/spark/' export HAIL_HOME='/home/ABCD/.pyenv/versions/3.7.2/envs/bio/lib/python3.7/site-packages/hail/' When I use echo $SPARK_HOME or echo $HAIL_HOME , I am able to see the path as output But, when I use the below os commands in jupyter notebook os.getenv('SPARK_HOME') # able to get the output /opt/spark/ os.getenv('HAIL_HOME') # returns no output I also tried defining the same variables from

问题 I want to close some files like .txt, .csv, .xlsx that I have opened using os.startfile(). I know this question asked earlier but I did not find any useful script for this. I use windows 10 Environment 回答1: I believe the question wording is a bit misleading - in reality you want to close the app you opend with the os.startfile(file_name) Unfortunately, os.startfile does not give you any handle to the returned process. help(os.startfile) startfile returns as soon as the associated application

问题 my main file uses several other files to complete a 3D render in Paraview. I have been using Pythons os library to get the file path that I need. For some reason Paraview breaks when I try to retrieve my path. I am using Paraview's build in shell to run my script. My code goes as follows: import os dir = os.path.dirname(os.path.realpath(__file__)) print(dir) This line of code works inconsistently and I want to understand why this isn't working If my file is in a directory named C:\Test\Test1

问题 I have the following code (simplified for clarity): import os import errno import imp lib_dir = os.path.expanduser('~/.brian/cython_extensions') module_name = '_cython_magic_5' module_path = os.path.join(lib_dir, module_name + '.so') code = 'some code' have_module = os.path.isfile(module_path) if not have_module: pyx_file = os.path.join(lib_dir, module_name + '.pyx') # THIS IS WHERE EACH PROCESS TRIES TO WRITE TO THE FILE. THE CODE HERE # PREVENTS A RACE CONDITION. try: fd = os.open(pyx_file,