I am working in a set of directories with the following structure:
Master/
Subfolder_1/file_1_1.py
file_1_2.txt
Subfolder_2/file_2_1.py
I import file_1_1 in file_2_1 as follows:
import sys
sys.path.append('../file_1_1')
file_1_1 is reading file_1_2.txt which is in the same directory. However, when I call the function that reads file_1_2.txt from file_2_1.py, it says no such file and directory and it gives me the path of file_1_2.txt as:
Master/Subfolder_2/file_1_2.txt
which is a wrong path. It seems like python in this case is using the working directory as a reference. How can I solve this error given that I don't want to include the absolute path for each file I read.
Don't mess with sys.path, and don't think of imports as working against files and directories. Ultimately everything has to live in a file somewhere, but the module hierarchy is a little more subtle than "replace dot with slash and stick a .py at the end".
You almost certainly want to be in Master and run python -m Subfolder_1.file_1_1. You can use pkg_resources to get the text file:
pkg_resources.resource_string('Subfolder_1', 'file_1_1.txt')
info=[]
with open(os.path.realpath('yourtextfile.txt','r') as myfile:
for line in myfile:
info.append(line)
Related
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'.
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
Sorry for this beginner question, but...I'm a Python beginner. Still, I can't seem to find a proper answer for loadtxt not 'finding my file'...
import os
print(os.getcwd())
returns, I suppose, my current working directory.
In this case:C:\Users\danie\Desktop\python
So, when I place my csv file in it and run:
import numpy as np
dataset=np.loadtxt('Desktop/python/pima-indians-diabetes.csv', delimiter=",")
I still get
OSError: Desktop/python/pima-indians-diabetes.csv not found.
I have tried relative paths, absolute paths, f=open(..), paths with '/' and paths with '\' or with '\'...but nothing seems to make it work..
Any ideas ?
**RESOLVED: I tried Max L's hint: print(os.listdir(os.getcwd()))
and I saw the list of files in my current directory:...'pima-indians-diabetes.csv.csv' ....turns out I had put the csv extension on the file name myself **
If your working directory is C:\Users\danie\Desktop\python, that means that is where Python will start to look for files to import when using a relative path.
What is a relative path? It's the path to the file you want, relative to your current working directory. If a file is in the same directory, no prefix should be needed so it should just be
np.loadtxt('pima-indians-diabetes.csv', ...
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)
Probably a simple query.. But basically, I have data in directory "/foo/bar/foobar.txt"
and I am working in directory "/some/path/read_foobar.py"..
Now I want to read the file "foobar.txt" but rather than giving full path, I thought of adding /foo/bar/ to the path..
So, added the following at the start of read_foobar.py
import sys
sys.path.append("/foo/bar")
But when I try to read open("foobar.txt","r"), it is not able to find the file?
how do I do this?
Thanks
You can do it like this:
import os
os.chdir('/foo/bar')
f = open('foobar.txt', 'r')
sys.path is used to set the path used to look for python modules. Short of you writing some helper function that has a list of directories to search in when opening a file, I don't believe there is a standard module that provides this functionality.
From what I gathered from here and some quick tests, appending a path to sys.path will make python search in that path when you import a file/module, but not when open-ing it. Let's say we have a file called foo.py in /foo/bar/
import sys
sys.path.append("/foo/bar/")
try:
f = open('foo.py', 'r')
except:
print('this did not work') # this will print
import foo # no problems here
Unfortunately you can't. The PATH environment variable is only used by the operating system to search for executable files, and python uses it (along with the environment variable PYTHONPATH) to search for python modules to import.
You may want to consider setting a symbolic link to that file from your current working directory
ln -s /foo/bar/foobar.txt /some/path/foobar.text