Let's say I have a file called credentials.json in my current directory, an environment variable MY_CREDENTIALS=credentials.json and a script main.py that uses this environment variable via os.getenv('MY_CREDENTIALS').
Now suppose I create a subfolder and put something there like this: /subfolder/my_other_script.py.
If I print os.getenv('MY_CREDENTIALS') then I get indeed credentials.json but I can't use this file as it is in my root directory (not in /subfolder). So, how can I use this file although it is in the root directory? The only thing that works for me is to make a copy of credentials.json in /subfolder, but then I would have multiple copies of this file and I don't want that.
Thanks for your response!
Something like this could work:
from pathlib import Path
import os
FILENAME = os.getenv('MY_CREDENTIALS')
filePath = Path(FILENAME)
if filePath.exists() and filePath.is_file():
print("Success: File exists")
print(filePath.read_text())
else:
print("Error: File does not exist. Getting file from level below.")
print((filePath.absolute().parent.parent / filePath.name).read_text())
Basically, you check whether your file exists in the current folder. This will be the case, if your script is in your root folder. If it is not, you assume that you are in a subfolder. So you try to get the file from one level below (your root).
It's not totally production ready, but for the specific case you mentioned it should work. In production you should think about cases where you might have nested subfolder or your file is missing for good.
You can create symbolic link
os.symlink('<root>', '<subfolder>')
https://docs.python.org/3/library/os.html#os.symlink
Related
After getting the path to the current working directory using:
cwd = os.getcwd()
How would one go up one folder: C:/project/analysis/ to C:/project/ and enter a folder called data (C:/project/data/)?
In general it a bad idea to 'enter' a directory (ie change the current directory), unless that is explicity part of the behaviour of the program.
In general to open a file in one directory 'over from where you are you can do .. to navigate up one level.
In your case you can open a file using the path ../data/<filename> - in other words use relative file names.
If you really need to change the current working directory you can use os.chdir() but remember this could well have side effects - for example if you import modules from your local directory then using os.chdir() will probably impact that import.
As per Python documentation, you could try this:
os.chdir("../data")
Data visualisation: I want to import a file using pandas. I assigned a variable and file name given is correct but it shows a error message as file does not exist
Code: sample_data = pd.read_csv('sample_data.csv')
You're probably in the wrong working directory. Run os.getcwd() to check. If so, you can either change working directories with os.chdir() or give an absolute path to the file instead of a relative path.
If you're already in the right working directory, run os.listdir() to make sure the file is actually there.
Ok, in this case, what you will have to do is get the path file by right-clicking on the file and going to properties(Windows, don't know about Mac). Then just copy the file path and paste it instead of the file name. So for now, it should be something like this(as I don't know your file path):
sample_data = pd.read_csv('C:\Users\SVISHWANATH\Downloads\datasets')
Now, after the last folder, give in your file name. So now, it should look like this:
sample_data = pd.read_csv('C:\Users\SVISHWANATH\Downloads\datasets\sample_data.csv')
However, this will still not work as the slashes need to be the other way. Because of this, there will have to be an r before the quotes.
sample_data = pd.read_csv(r'C:\Users\SVISHWANATH\Downloads\datasets\sample_data.csv')
Now this should work as all the requirements are met.
You need to store the .csv file in a folder where the actual program is stored, otherwise you have to give a proper path to the file:
sample_data = pd.read_csv('filepath')
I'm trying to build a file transfer system with python3 sockets. I have the connection and sending down but my issue right now is that the file being sent has to be in the same directory as the program, and when you receive the file, it just puts the file into the same directory as the program. How can I get a user to input the location of the file to be sent and select the location of the file to be sent to?
I assume you're opening files with:
open("filename","r")
If you do not provide an absolute path, the open function will always default to a relative path. So, if I wanted to open a file such as /mnt/storage/dnd/5th_edition.txt, I would have to use:
open("/mnt/storage/dnd/4p5_edition","r")
And if I wanted to copy this file to /mnt/storage/trash/ I would have to use the absolute path as well:
open("/mnt/storage/trash/4p5_edition","w")
If instead, I decided to use this:
open("mnt/storage/trash/4p5_edition","w")
Then I would get an IOError if there wasn't a directory named mnt with the directories storage/trash in my present folder. If those folders did exist in my present folder, then it would end up in /whatever/the/path/is/to/my/current/directory/mnt/storage/trash/4p5_edition, rather than /mnt/storage/trash/4p5_edition.
since you said that the file will be placed in the same path where the program is, the following code might work
import os
filename = "name.txt"
f = open(os.path.join(os.path.dirname(__file__),filename))
Its pretty simple just get the path from user
subpath = raw_input("File path = ")
print subpath
file=open(subpath+str(file_name),'w+')
file.write(content)
file.close()
I think thats all you need let me know if you need something else.
like you say, the file should be in the same folder of the project so you have to replace it, or to define a function that return the right file path into your open() function, It's a way that you can use to reduce the time of searching a solution to your problem brother.
It should be something like :
import os
filename = "the_full_path_of_the_fil/name.txt"
f = open(os.path.join(os.path.dirname(__file__),filename))
then you can use the value of the f variable as a path to the directory of where the file is in.
Does anyone know how I can copy/duplicate a file from one directory into another without specification of src path? I got it to work with "shutil.copy2" but it's not exactly what I am looking for since the src argument asks for the path.
My goal is to be able to copy/duplicate a file from one directory into another by filename. Has anyone done this before, if so can you guide me in the right direction? - Thanks
#----------------------------------------------------------------------------------------------------------------#
# These params will be used for specifying which template you want to copy and where to output
#----------------------------------------------------------------------------------------------------------------#
'''Load file from x directory into current working directory '''
#PullTemplate: Specify which template you want to copy, by directory path
TemplateRepo = ("/home/hadoop/BackupFolders/Case_Project/scripts")
#OutputTemplate: Let's you specify where you want to output the copied template.
#Originally set to your current working directory (u".")
OutputTemplate = (u".")
shutil.copy2(TemplateRepo, OutputTemplate)
Well if you are trying to load a file in the same project you need to have at least the folder name inside that project.
You can use json
Something like this.
import json
#someFiles is just a fold name inside the projects main folder.
with open("someFiles\\file_name", "r") as whatever_u_want:
var_of_choice = json.load(whatever_u_want)
print (var_of_choice)
once the file is open you can save the variable var_of_choice as any file name you wish where you wish using the json dump method.
Click the file you want to copy, create a duplicate of the file by choosing Duplicate key under File ( below the Jupyter logo at the top ).
Choose the file copied(file_copy), and choose Move key under File.
Choose the file path where you want to paste/move the copied file.
Rename the copied file name as you wish.
For more info, you can refer to here: https://subscription.packtpub.com/book/big_data_and_business_intelligence/9781785884870/1/ch01lvl1sec12/basic-notebook-operations
I have created a small python script. With that I am trying to read a txt file but my access is denied resolving to an no.13 error, here is my code:
import time
import os
destPath = 'C:\Users\PC\Desktop\New folder(13)'
for root, dirs, files in os.walk(destPath):
f=open(destPath, 'r')
.....
Based on the name, I'm guessing that destPath is a directory, not a file. You can do a os.walk or a os.listdir on the directory, but you can't open it for reading. You can only call open on a file.
Maybe you meant to call open on one or more of the items from files
1:
I take it you are trying to access a file to get what's inside but don't want to use a direct path and instead want a variable to denote the path. This is why you did the destPath I'm assuming.
From what I've experienced the issue is that you are skipping a simple step. What you have to do is INPUT the location then use os.CHDIR to go to that location. and finally you can use your 'open()'.
From there you can either use open('[direct path]','r') or destPath2 = 'something' then open(destPath2, 'r').
To summarize: You want to get the path then NAVIGATE to the path, then get the 'filename' (can be done sooner or not at all if using a direct path for this), then open the file.
2: You can also try adding an "r" in front of your path. r'[path]' for the raw line in case python is using the "\" for something else.
3: Try deleting the "c:/" and switching the / to \ or vice versa.
That's all I got, hope one of them helps! :-)
I got this issue when trying to create a file in the path -C:/Users/anshu/Documents/Python_files/Test_files . I discovered python couldn't really access the directory that was under the user's name.
So, I tried creating the file under the directory - C:/Users/anshu/Desktop .
I was able to create files in this directory through python without any issue.