I am doing a project using python in which I got stuck at a point where I want to access some text files which are saved outside the project directory.
The path where my text files are saved:
C:\Users\saqibshakeel035\Desktop\Scientific Project Lithim battery project\text_file_r_w
The path of my python project:
C:\Users\saqibshakeel035\PycharmProjects\Tutorial_1
I want to open/read my text files (external > not included in python project folder)
I already know the Reading/writing etc etc within the same folder where the python project .py file is present but struggling with the different paths.
I tried:
import os
from os import path
print("Your cunrrent directory is : %s" %path.curdir)
strpath = r"C:\Users\saqibshakeel035\Desktop\Scientific Project Lithim battery project\text_file_r_w"
print("Your current directory is %s: " %path.dirname(strpath))
print("Your current directory is : %s" %path.abspath(strpath))
This works fine and it shows my abspath where my text files are stored but when I try to read it with the following command
f = open("file1.txt","r")
It gives error that no such directory or file found
I suggest you try f = open("C:/text/to/path/file1.txt","r") or the code #Jaba has mention. Either works fine
Can you try using the full path to "file1.txt" in the open function.
f = open("Full_path_to_file1.txt", "r")
Another option is to change the current directory,
os.chdir(path)
Related
I am new to VSCODE. I opened a text file via vscode and entered some details. Where can I find the file?
f=open('story.txt','w')
f.write('my name is')
f.write('my age is')
f.close()
f=open('story.txt', 'r')
print(f.readline())
f.close()
this is the output
However I cannot find 'story.txt' in file explorer. I used another text editor and then error came as file not found. but when i reopened the file in vs code I was getting a proper output.
From the screenshot you supplied it looks like you are running the script from C:\Users\<YourName>, then this is where your story.txt file will be.
To specify another location you need to supply the open() method with a full path
Also, it's best practice to close the file before opening it again for reading. also, you might want to use a context manager to help you with this
If you want your file saved in the directory of your script, you can use os and __file__ to locate your script's directory and use that.
import os
my_dir = os.path.dirname(__file__)
new_path = os.path.join(my_dir, 'story.txt')
print(new_path)
os documentation
__file__ explained
When you run python script it will execute from the current working directory by defaut.
If you want to be sure where your file will be, you may pass the complete file path instead of file name only (eg: C:\\filepath\\filename.txt)
or you can move to the desired directory before read/write with os.chdir(filepath)
If you don't know where the script is running you can use os.getcwd() to get this directory from your code
import os
print(os.getcwd()) #will show the current working directory
os.chdir("c:\\") #will move to C:\ directory
f=open('story.txt','w')
f.write('my name is')
f.write('my age is')
f=open('story.txt', 'r')
print(f.readline())
f.close()
When you create a file with Python, the file will be create in the same folder as your script. So, if your folder all_python_scripts contains your script, your file story.txt will be created in this folder. Try to search your file in the script's folder.
I mistakenly, typed the following code:
f = open('\TestFiles\'sample.txt', 'w')
f.write('I just wrote this line')
f.close()
I ran this code and even though I have mistakenly typed the above code, it is however a valid code because the second backslash ignores the single-quote and what I should get, according to my knowledge is a .txt file named "\TestFiles'sample" in my project folder. However when I navigated to the project folder, I could not find a file there.
However, if I do the same thing with a different filename for example. Like,
f = open('sample1.txt', 'w')
f.write('test')
f.close()
I find the 'sample.txt' file created in my folder. Is there a reason for the file to not being created even though the first code was valid according to my knowledge?
Also is there a way to mention a file relative to my project folder rather than mentioning the absolute path to a file? (For example I want to create a file called 'sample.txt' in a folder called 'TestFiles' inside my project folder. So without mentioning the absolute path to TestFiles folder, is there a way to mention the path to TestFiles folder relative to the project folder in Python when opening files?)
I am a beginner in Python and I hope someone could help me.
Thank you.
What you're looking for are relative paths, long story short, if you want to create a file called 'sample.txt' in a folder 'TestFiles' inside your project folder, you can do:
import os
f = open(os.path.join('TestFiles', 'sample1.txt'), 'w')
f.write('test')
f.close()
Or using the more recent pathlib module:
from pathlib import Path
f = open(Path('TestFiles', 'sample1.txt'), 'w')
f.write('test')
f.close()
But you need to keep in mind that it depends on where you started your Python interpreter (which is probably why you're not able to find "\TestFiles'sample" in your project folder, it's created elsewhere), to make sure everything works fine, you can do something like this instead:
from pathlib import Path
sample_path = Path(Path(__file__).parent, 'TestFiles', 'sample1.txt')
with open(sample_path, "w") as f:
f.write('test')
By using a [context manager]{https://book.pythontips.com/en/latest/context_managers.html} you can avoid using f.close()
When you create a file you can specify either an absolute filename or a relative filename.
If you start the file path with '\' (on Win) or '/' it will be an absolute path. So in your first case you specified an absolute path, which is in fact:
from pathlib import Path
Path('\Testfile\'sample.txt').absolute()
WindowsPath("C:/Testfile'sample.txt")
Whenever you run some code in python, the relative paths that will be generate will be composed by your current folder, which is the folder from which you started the python interpreter, which you can check with:
import os
os.getcwd()
and the relative path that you added afterwards, so if you specify:
Path('Testfiles\sample.txt').absolute()
WindowsPath('C:/Users/user/Testfiles/sample.txt')
In general I suggest you use pathlib to handle paths. That makes it safer and cross platform. For example let's say that your scrip is under:
project
src
script.py
testfiles
and you want to store/read a file in project/testfiles. What you can do is get the path for script.py with __file__ and build the path to project/testfiles
from pathlib import Path
src_path = Path(__file__)
testfiles_path = src_path.parent / 'testfiles'
sample_fname = testfiles_path / 'sample.txt'
with sample_fname.open('w') as f:
f.write('yo')
As I am running the first code example in vscode, I'm getting a warning
Anomalous backslash in string: '\T'. String constant might be missing an r prefix.
And when I am running the file, it is also creating a file with the name \TestFiles'sample.txt. And it is being created in the same directory where the .py file is.
now, if your working tree is like this:
project_folder
-testfiles
-sample.txt
-something.py
then you can just say: open("testfiles//hello.txt")
I hope you find it helpful.
I am so new with python and pycharm and i got confuse!!
When I run my project in pycharm it gives me an error about not finding the path of my file. The physical file path is:
'../Project/BC/RequiredFiles/resources/a_reqs.csv'
My project working directory is "Project/BC" and the project running file (startApp.sh) is there too. but the .py file that wants to work with a_req.csv is inside the "RequiredFiles" folder. There is the following code in the .py file:
reqsfile = os.getcwd() + "/resources/a_reqs.csv"
it returns: '../Project/BC/resources/a_reqs.csv'
instead of: '../Project/BC/resources/RequiredFiles/a_reqs.csv'
while the .py file is in "RequiredFiles" the os.getcwd() must include it too. but it does not.
The problem is that i can not change the addressing code. because this code works in another IDE and other people who work with the code in other platform or OS do not have any problem. I am working in mac OS and if i am not mistaken the code works with windows!!
So, how can i tell Pycharm (in mac) to see and load "RequiredFiles" folder as the subfolder of my working directory!!!
os.getcwd returns the current working directory of the process (which may be the directory where startApp.sh is located or another one, depending on the PyCharm's run configuration setting, or, if you start the program from the command line, the directory in which you execute the command).
To make a path independent on the current working directory, you can take the directory where your Python file is located and build the path from it:
os.path.dirname(__file__) + "/resources/a_reqs.csv"
From your question what I see is:
reqsfile = os.getcwd() + "/resources/a_reqs.csv"
Which produces: "../Project/BC/resources/a_reqs.csv", whereas your desired output is
"../Project/BC/resources/RequiredFiles/a_reqs.csv". Since we know os.getcwd is returning "/Project/BC/", then to get your desired result you should be doing:
reqsfile = os.getcwd() + "/resources/RequiredFiles/a_reqs.csv"
But since you want the solution to work with or without the RequiredFiles subdirectory you could apply a conditional solution, ie something like:
import os.path
if os.path.exists(os.getcwd() + "/resources/RequiredFiles/a_reqs.csv"):
reqsfile = os.getcwd() + "/resources/RequiredFiles/a_reqs.csv"
else:
reqsfile = os.getcwd() + "/resources/a_reqs.csv"
This solution will set the reqsfile to the csv in the RequiredFiles directory if the directory exists, and thus will work for you. On the other-hand, if the RequiredFiles directory doesn't exist, it will default to the csv in /resources/.
Typically when groups collaborate on projects, the maintain the same file hierarchy so that these types of issues are avoided, so you might want to consider moving the csv from /RequiredFiles/ to /resources/.
I have a program that reads in several subfolders of files.
I had previously written code that had :
path = 'C:/Users/Me/etc/etc/'
And then I would open several files by saying
file1 = path + str('file1.txt')
How do I change the code so that it can be used by others? I am using Jupyter notebook and so __file __ and sys(argv[0]) don't seem to be working.
I have looked at these:
How to get an absolute file path in Python
How do I get the parent directory in Python?
Use get current working directory,
def join_current_dir(file):
"""Join filepath with current file directory"""
cwd = os.getcwd()
return os.path.join(cwd, file)
How would I go about getting the Windows/System folder paths in python? I need to be able to read from an INI that gets written to the Windows directory.
Use the environment variable %WINDIR%.
import os
winpath = os.environ['WINDIR'] + "\\System\\"
inifile = open(winpath + filename)