first post here so sorry if it's hard to understand. Is it possible to shorten the directory in python to the location of the .py file?. For example, if I wanted to grab something from the directory "C:\Users\Person\Desktop\Code\Data\test.txt", and if the .py was located in the Code folder, could I shorten it to "\data\test.txt". I'm new to python so sorry if this is something really basic and I just didn't understand it correctly.
I forgot to add i plan to use this with multiple files, for example: "\data\test.txt" and \data\test2.txt
import os
CUR_FILE = os.path.abspath(__file__)
TARGET_FILE = "./data/test.txt"
print(os.path.join(CUR_FILE, TARGET_FILE))
With this, you can move around your Code directory anywhere and not have to worry about getting the full path to the file.
Also, you can run the script from anywhere and it will work (you don't have to move to Code's location to run the script.
You can import os and get current working directory ,this will give you the location of python file and then you can add the location of folder data and the file stored in that ,code is given below
import os
path=os.getcwd()
full_path1=path+"\data\test.txt"
full_path2=path+"\data\test2.txt"
print(full_path1)
print(full_path2)
I think this will work for your case and if it doesn't work then add a comment
Related
I have a script in python, I want to import a csv from another folder. how can I do this? (for example, my .py is in a folder and I want to reach the data from the desktop)
First of all, you need to understand how relative and absolute paths work.
I write an example using relative paths. I have two folders in desktop called scripts which includes python files and csvs which includes csv files. So, the code would be:
df = pd.read_csv('../csvs/file.csv)
The path means:
.. (previous folder, in this case, desktop folder).
/csvs (csvs folder).
/file.csv (the csv file).
If you are on Windows:
Right-click on the file on your desktop, and go to its properties.
You should see a Location: tag that has a structure similar to this: C:\Users\<user_name>\Desktop
Then you can define the file path as a variable in Python as:
file_path = r'C:\Users\<your_user_name>\Desktop\<your_file_name>.csv'
To read it:
df = pd.read_csv(file_path)
Obviously, always try to use relative paths instead of absolute paths like this in your code. Investing some time into learning the Pathlib module would greatly help you.
I would like to take a screenshot for my selenium driver and save it to a specific directory. Right now, I can run:
driver.save_screenshot('1.png')
and it saves the screenshot within the same directory as my python script. However, I would like to save it within a subdirectory of my script.
I tried the following for each attempt, I have no idea where the screenshot was saved on my machine:
path = os.path.join(os.getcwd(), 'Screenshots', '1.png')
driver.save_screenshot(path)
driver.save_screenshot('./Screenshots/1.png')
driver.save_screenshot('Screenshots/1.png')
Here's a kinda hacky way, but it ought to work for your end result...
driver.save_screenshot('1.png')
os.system("mv 1.png /directory/you/want/")
You might need to use the absolute path for your file and/or directory in the command above, not 100% sure on that.
You can parse the file path you want to the save_screenshot function.
As your doing this already a good thing to check is that os.getcwd is the same as the location of the script (may be different if your calling it from somewhere else) and that the directory exists, this can be created via os.makedirs.
import os
from os import path
file_dir = path.join(os.getcwd(), "screenshots")
os.makedirs(file_dir, exist_ok=True)
file_path = path.join(file_dir, "screenshot_one.png")
driver.save_screenshot(file_path)
If os.getcwd is not the right location, the following will get the directory of the current script..
from os import path
file_dir = path.dirname(path.realpath(__file__))
i just downloaded a file called "N_PR_8705_004A_.doc" in my "Downloads" folder and i want to put it into my "Stage NLP" folder using os. I know how to do it without os but i'd like that shit to work it's faster and it simply doesnt. First i tried to get the path of my file doing this:
import os
os.path.dirname(os.path.abspath("N_PR_8705_004A_.doc"))
# or os.path.realpath it's the same
and the result i get is:
'C:\\Users\\f002722\\Stage NLP'
whereas when i do list all the files in this folder doing:
os.listdir("C:\\Users\\f002722\\Stage NLP")
you clearly see it is simply not there:
['.ipynb_checkpoints',
'ADR service study - D2 (1st part).pdf',
'basetal.py',
'Codes test',
'Cours NLP.ipynb',
'e Deorbit',
'edot CDF study.pdf',
'edot_v5.pdf',
'Entrainement.ipynb',
'ESA edot workshop May 6th 2014 - Summary.msg',
'ESA_edotWorkshop-_Envisat_attitude-Copy1',
'ESA_edotWorkshop-_Envisat_attitude.pdf',
'ESA_edotWorkshop_GNC_.pdf',
'ESA_INNOCENTI_Challenges.pdf',
'ESA_Robin_Biesbroek_edot.pdf',
'GMV_edot_Symposium.pdf',
'JOP_edotWorkshop.pdf',
'KT_HAARMANN_Edot.pdf',
'MDA_edot_Symposium_-_Robotic_Capture.pdf',
'MDA_eDot_Symposium_-_Robotic_Capture.pdf.kx2zd5w.partial',
'Note_Ariane_NLP.ipynb',
'Note_Ariane_NLP_2.ipynb',
'Note_Ariane_NLP_3.ipynb',
'OHB_eDotWorkshop_ADRM.pdf',
'OHB_Sweden_eDotWorkshop_PRISMA_and_IRIDES.pdf',
'SKA_Polska_eDotWorkshop_Net_Simulator.pdf',
'TAS_Carole_Billot_edot.pdf',
'Test.ipynb',
'Text_clustering_v3_2.py',
'Webinar_OOSandADR_7May2020.pdf',
'__pycache__']
So what the hell is going on i'm out of ideas here.
Thx in advance
I think I have a possible answer to your question. Neither realpath nor abspath require their arguments to name existing files. In particular, the documentation for abspath() says: "On most platforms, this is equivalent to calling the function normpath() as follows: normpath(join(os.getcwd(), path))."
This means that if you have a Python script that has a line like,
foo = os.path.dirname(os.path.abspath("doesnotexist"))
then the value of foo will be the current working directory of the script. Since "doesnotexist" isn't the name of a file in this directory, it won't show up if you do os.listdir(foo).
I notice that you wrote that "N_PR_8705_004A_.doc" was in your "Downloads" directory, which is obviously not the same as 'C:\\Users\\f002722\\Stage NLP'. If 'C:\\Users\\f002722\\Stage NLP' is the working directory for your Python script, then running os.path.dirname(os.path.abspath("N_PR_8705_004A_.doc")) is just like writing os.path.dirname(os.path.abspath("doesnotexist")), for the reasons that I just gave.
Python can't automatically figure out the path of a file just by giving it a relative file name. For example, there could be many files named README.txt on a system, each in different directories, so there's no way for os.path.abspath('README.txt') to know which of those directories you want.
To move the file "N_PR_8705_004A_.doc" from the "Downloads" directory to 'C:\\Users\\f002722\\Stage NLP', you'd probably need to do something like this:
import shutil
shutil.move('C:\\Users\\f002722\\Downloads\\N_PR_8705_004A_.doc',
'C:\\Users\\f002722\\Stage NLP')
presuming, of course, that the "Downloads" directory was inside 'C:\\Users\\f002722'.
I've got a task to do that is crushing my head. I have five .py documents and I want to make a menu in another .py so I can run any of them by introducing a string inside an input() but don't really see the way to do that and I don't know if there is somehow I can.
I have tried import every file to the 6th file but I don't even know how to start.
I would like it just to be seen as simple as it can sound, but yet I find it really hard.
If you just want to run them, then try this:-
import os
file_path = input("Enter the path of your file = ")
os.system(file_path)
If the file that you are trying to execute is not in the current
directory, i.e. doesn't exist in the same folder as the currently
executing python file, then you have to provide it's full path.
Path Format:-. C:\Users\lmYoona\OneDrive\Desktop\example.py
If the python file you are trying to execute is in the same directory as
the currently executing python file, then abstract name will also
work
Path Format:- example.py
P.S.:- I would only recommend this method if all you want is just to execute the other python file, rather then importing stuff from it.
Well I searched a lot and found different ways to open program in python,
For example:-
import os
os.startfile(path) # I have to give a whole path that is not possible to give a full path for every program/software in my case.
The second one that I'm currently using
import os
os.system(fileName+'.exe')
In second example problem is:-
If I want to open calculator so its .exe file name is calc.exe and this happen for any other programs too (And i dont know about all the .exe file names of every program).
And assume If I wrote every program name hard coded so, what if user installed any new program. (my program wont able to open that program?)
If there is no other way to open programs in python so Is that possible to get the list of all install program in user's computer.
and there .exe file names (like:- calculator is calc.exe you got the point).
If you want to take a look at code
Note: I want generic solution.
There's always:
from subprocess import call
call(["calc.exe"])
This should allow you to use a dict or list or set to hold your program names and call them at will. This is covered also in this answer by David Cournapeau and chobok.
You can try with os.walk :
import os
exe_list=[]
for root, dirs, files in os.walk("."):
#print (dirs)
for j in dirs:
for i in files:
if i.endswith('.exe'):
#p=os.getcwd()+'/'+j+'/'+i
p=root+'/'+j+'/'+i
#print(p)
exe_list.append(p)
for i in exe_list :
print('index : {} file :{}'.format(exe_list.index(i),i.split('/')[-1]))
ip=int(input('Enter index of file :'))
print('executing {}...'.format(exe_list[ip]))
os.system(exe_list[ip])
os.getcwd()+'/'+i prepends the path of file to the exe file starting from root.
exe_list.index(i),i.split('/')[-1] fetches just the filename.exe
exe_list stores the whole path of an exe file at each index
Can be done with winapps
First install winapps by typing:
pip install winapps
After that use the library:
# This will give you list of installed applications along with some information
import winapps
for app in winapps.list_installed():
print(app)
If you want to search for an app you can simple do:
application = 'chrome'
for app in winapps.search_installed(application):
print(app)