I have my data in the file 'Climate\Data\Raw_Flooding\CSV\input\addresses.csv'
My code which needs to access the file, however, is in the folder 'Climate\Code\Flooding\python_code\code.py'
How do I access the data subdirectory when I am in the code subdirectory?
import os
cwd = os.getcwd()
print(cwd)
path = cwd.split('\\')
new_path = ''
for i in range(path.index('Climate')):
new_path += path[i]+'\\'
new_path+='Data\Raw_Flooding\CSV\input\\address.csv'
print(new_path)
This will give you the path to the csv file.
Related
I am currently accessing a script that opens a file in the directory it's located in. I am accessing this file from both the main.py file located in the same directory, as well as a testfile which is located in a "Test" subdirectory. Trying to use a file from the Test subdirectory to call the function that opens the file causes the script to try and open it from the Test directory instead of the super directory, since I am opening the file simply by calling it as following:
with open(filename,"w") as f:
Is there a way to define the location of the file in a way that makes it possible for the script opening it to be called from anywhere?
Use __file__ to get the path to the current script file, then find the file relative to that:
# In main.py: find the file in the same directory as this script
import os.path
open(os.path.join(os.path.dirname(__file__), 'file.txt'))
# In Test/test.py: find the file one directory up from this script
import os.path
open(os.path.join(os.path.dirname(__file__), '..', 'file.txt'))
just give the absolute file path instead of giving a relative one
for eg
abs_path = '/home/user/project/file'
with open(abs_path, 'r') as f:
f.write(data)
Try specifying the path:
import os
path = 'Your path'
path = os.path.abspath(path)
with open(path, 'w') as f:
f.write(data)
From what I understood your file is in a directory parent_directory/file_name.txt
and in another folder parent_directory/sub_directory/file_name.txt. All you have to do is paste the below code in both parent and sub directories.
import os
file_name = 'your_file_name'
# if the file is in current directory set the path to file_name
if file_name in os.listdir(os.getcwd()):
path = file_name
# if the path is not in current directory go back to parent directory set the path to parent directory
else:
path = os.path.abspath(os.path.join(os.getcwd(), os.pardir))
print('from',os.getcwd())
with open(path, 'r') as filename:
print(filename.read())
I would like to use the shutil.move() function to move some files which match a certain pattern to a newly created(inside python script)folder, but it seems that this function only works with existing folders.
For example, I have 'a.txt', 'b.txt', 'c.txt' in folder '/test', and I would like to create a folder '/test/b' in my python script using os.join() and move all .txt files to folder '/test/b'
import os
import shutil
import glob
files = glob.glob('./*.txt') #assume that we in '/test'
for f in files:
shutil.move(f, './b') #assume that './b' already exists
#the above code works as expected, but the following not:
import os
import shutil
import glob
new_dir = 'b'
parent_dir = './'
path = os.path.join(parent_dir, new_dir)
files = glob.glob('./*.txt')
for f in files:
shutil.move(f, path)
#After that, I got only 'b' in '/test', and 'cd b' gives:
#[Errno 20] Not a directory: 'b'
Any suggestion is appreciated!
the problem is that when you create the destination path variable name:
path = os.path.join(parent_dir, new_dir)
the path doesn't exist. So shutil.move works, but not like you're expecting, rather like a standard mv command: it moves each file to the parent directory with the name "b", overwriting each older file, leaving only the last one (very dangerous, because risk of data loss)
Create the directory first if it doesn't exist:
path = os.path.join(parent_dir, new_dir)
if not os.path.exists(path):
os.mkdir(path)
now shutil.move will create files when moving to b because b is a directory.
I have a python script that creates a PDF and saves it in a subfoler of the folder where the script is saved. I have the following that saves the file to the subfolder:
outfilename = "Test" + ".pdf" #in real code there is a var that holds the name of the file
outfiledir = 'C:/Users/JohnDoe/Desktop/dev/PARENTFOLDER/SUBFOLDER/' #parent folder is where the script is - subfolder is where the PDFs get saved to
outfilepath = os.path.join(outfiledir, outfilename)
Is there a way I can save the PDFs to the subfolder without having to specify the full path? Lets say I wanted yto make this script an exe that multiple computers could use, how would I display the path so that the PDFs are just saved in the subfoler?
Thanks!
Try it:
import os
dir_name = os.path.dirname(os.path.abspath(__file__)) + '/subdir'
path = os.path.join(dir_name, 'filename')
This is my current (from a Jupyter notebook) code for renaming some text files.
The issue is when I run the code, the renamed files are placed in my current working Jupyter folder. I would like the files to stay in the original folder
import glob
import os
path = 'C:\data_research\text_test\*.txt'
files = glob.glob(r'C:\data_research\text_test\*.txt')
for file in files:
os.rename(file, file[-27:])
You should only change the name and keep the path the same. Your filename will not always be longer than 27 so putting this into you code is not ideal. What you want is something that just separates the name from the path, no matter the name, no matter the path. Something like:
import os
import glob
path = 'C:\data_research\text_test\*.txt'
files = glob.glob(r'C:\data_research\text_test\*.txt')
for file in files:
old_name = os.path.basename(file) # now this is just the name of your file
# now you can do something with the name... here i'll just add new_ to it.
new_name = 'new_' + old_name # or do something else with it
new_file = os.path.join(os.path.dirname(file), new_name) # now we put the path and the name together again
os.rename(file, new_file) # and now we rename.
If you are using windows you might want to use the ntpath package instead.
file[-27:] takes the last 27 characters of the filename so unless all of your filenames are 27 characters long, it will fail. If it does succeed, you've stripped off the target directory name so the file is moved to your current directory. os.path has utilities to manage file names and you should use them:
import glob
import os
path = 'C:\data_research\text_test*.txt'
files = glob.glob(r'C:\data_research\text_test*.txt')
for file in files:
dirname, basename = os.path.split(file)
# I don't know how you want to rename so I made something up
newname = basename + '.bak'
os.rename(file, os.path.join(dirname, newname))
I'm trying to scrape filenames inside a folder and then make directories for each filename inside another folder. This is what I've got so far but when I run it, it doesn't create the new folders in the destination folder. When I run it in the terminal it doesn't return any errors.
import os
import shutil
folder = "/home/ro/Downloads/uglybettyfanfiction.net/"
destination = "/home/ro/A Python Scripts/dest_test/"
# get each files path name
def ensure_dir(f):
d = os.path.dirname(f)
if not os.path.exists(d):
os.makedirs(d)
for files in os.listdir(folder):
new_path = folder + files
ensure_dir(new_path)
You've got a few mistakes. No need to use dirname and you should write to your destination, not the same folder:
def ensure_dir(f):
if not os.path.exists(f):
os.mkdir(f)
for files in os.listdir(folder):
new_path = destination + files
ensure_dir(new_path)