I am reading in a different machine a file. Therefore, I need to access to the full path of the file. So I tried to use pythons Pathlib module:
a_path = '/dir1/subdir1/sample.txt'
home = str(Path.home())
a_path = str(home) + str(a_path)
Apparently, the above code return me the full path. However, when I read it I get:
FileNotFoundError: [Errno 2] No such file or directory: "/home/user'/dir1/subdir1/sample.txt'"
How can I fix the above error? maybe in the concatenation I am getting problems.
Try this. This uses os.path.join which joins two paths together
import os
import pathlib
a_path = 'dir1/subdir1/sample.txt'
home = str(pathlib.Path.home())
print(os.path.join(home, a_path))
#/home/user/dir1/subdir1/sample.txt
You can use join to paste the string together.
''.join([str(Path.home()), 'path\\t.txt'])
First of all, '/dir1/subdir1/sample.txt' is an absolute path. If you want it to be a relative path (which seems to be the case) you should use 'dir1/subdir1/sample.txt', so without a leading /.
Using the pathlib library this then becomes very easy
>>> from pathlib import Path
>>> a_path = "dir1/subdir1/sample.txt"
>>> a_path = Path.home() / a_path
>>> print(a_path)
/home/pareto/dir1/subdir1/sample.txt
Again, make sure you are not using absolute paths. Otherwise you would get the following
>>> print(Path.home() / "/dir1/subdir1/sample.txt")
/dir1/subdir1/sample.txt
Related
I am writing a cli-tool that needs some path as input.
I am writing this tool in python and would like to use no python interpreter below 3.6. Using the packagepathlibseems to be the modern way to go when dealing with paths in python. So I would like to leave os and os.path behind if possible.
It seems like pathlib interprets the path ~/test/ as a relative path to the current working directory, the code below shows it
import pathlib
test_path = pathlib.Path('~/test')
absolute_path = test_path.absolute()
print(f"{str(test_path):>31}\n{str(absolute_path):>31}")
# output:
# ~/test
# /home/myUser/~/test
How can I use pathlib to recognize every path starting with ~ as an absolute path and automatically expand ~ into the users home directory?
The answer is easy, use .expanduser() on your Path object instead of .absolute()and it will replace ~ with the home directory of the user running the script, the result is also an absolute path only if ~ is at the beginning:
import pathlib
test_path = pathlib.Path('~/test')
absolute_path = test_path.expanduser()
# If ~ is somewhere in the middle of the path, use .resolve() to get an absolute path.
print(f"{str(test_path):>31}\n{str(absolute_path):>31}")
# output:
# ~/test
# /home/myUser/test
I'm trying to move files from one directory to another using Python - spyder.
My file extension is *.OD which python does not support or read.
I have tried using the wildcard and leaving out the file extension (which does not work). Another file extension cannot be used for this particular file.
Moving python supported extensions such as .txt and .csv works fine.
import shutil
source = '//Original_Filepath/Extract*.od'
target = '//NewFilePath/Extract_*.od'
shutil.copy(source, target)
There are no errors, it just doesn't move/copy the file.
Thanks,
There are a couple of basic mistakes with how you're trying to copy the files. With shutil.copy you should not specify a glob, but instead the exact source and destination.
If instead you want to copy a set of files from one directory to another and (presuming the added underscore isn't a mistake) change the target, then you should try using pathlib in combination with shutil (and re if needed).
pathlib - Object-oriented filesystem paths
Try adapting this:
import pathlib
import shutil
import re
source = pathlib.Path('//Original_Filepath') # pathlib takes care of end slash
source_glob = 'Extract*.od'
target = pathlib.Path('//NewFilePath')
for filename in source.glob(source_glob):
# filename here is a Path object as well
glob_match = re.match(r'Extract(.*)\.od', filename.stem).group(1)
new_filename = "Extract_{}.od".format(glob_match)
shutil.copy(str(filename), str(target / new_filename)) # `/` will create new Path
If you're not interested in editing the target nor using any other advanced feature that pathlib provides then see Xukrao's comment.
Thank you all for your help. Much appreciated! :)
I was also able to copy the file with the below as well (a bit simpler).
I left out the * and used a date string instead.
import shutil
from datetime import datetime
now = datetime.now()
Org_from=os.path.abspath('//Original FilePath')
New_to=os.path.abspath('//New Path')
shutil.copy(os.path.join(org_from, 'File_' + now.strftime("%Y%m%d") + '.od'), os.path.join(New_to, 'File_' + now.strftime("%Y%m%d") + '.od'))
Cheers,
Jen
I'm trying to use relative paths in Python, and I want to put my csv files in a separate folder from my python code.
My python program is in the following folder:
G:\projects\code
I want to read this file which is one level up:
G:\projects\data\sales.csv
How do I specify a path using pathlib that is one level up from my current working folder? I don't want to change the current working folder.
I tried this:
from pathlib import Path
file = Path.cwd() /'..'/'data'/'sales.csv'
But now the 'file' variable equals this:
'G:/projects/code/../data/sales.csv'
I read through the docs and either it isn't explained there or I'm just missing it.
Although it's not a problem that your path includes '..' (you can still use this path to open files, etc. in Python), you can normalize the path using resolve():
from pathlib import Path
path = Path.cwd() / '..' / 'data' / 'sales.csv'
print(path) # WindowsPath('G:/projects/code/../data/sales.csv')
print(path.resolve()) # WindowsPath('G:/projects/data/sales.csv')
NB: I personally would name a variable that contains a path path, not file. So you could later on do file = open(path).
print(
Path(__file__).parent, # the folder
Path(__file__).parent.parent, # the folder's parent
sep='\n'
)
print(
Path(
Path(__file__).parent.parent, 'hello.py'
)
)
results in
C:\Users\isik\Desktop\Python\MessAround\project\module
C:\Users\isik\Desktop\Python\MessAround\project
C:\Users\isik\Desktop\Python\MessAround\project\hello.py
with this file structure
-project
-module
-__init__.py
-hello.py
-__init__.py
while the code is located inside project.module.__init__.py
Do you mean "read my csv files"?
The import keyword has a different meaning in Python (you import only other Python modules).
In any case, in order to read a file located one folder above your Python file, you can use this:
import os
filePath = os.path.dirname(__file__)+'/../'+fileName
fileDesc = open(filePath)
fileData = fileDesc.read()
fileDesc.close()
...
here is an example I used:
import json
from pathlib import Path
def read_files(folder_name, file_name):
base_path = Path.cwd().joinpath('configs','resources')
path = base_path.joinpath(folder_name,file_name)
open_file = open(path,'r')
return json.load(open_file.read())
This is pretty old, but I happened on it looking for something else.
Oddly you never got a direct, obvious, answer -- you want the parent property:
from pathlib import Path
file = Path.cwd().parent / 'data' / 'sales.csv'
Note that some of the answers that say you want __file__ rather than the current working directory may be correct (depending on your use case), in which case it's:
from pathlib import Path
file = Path(__file__).parent.parent / 'data' / 'sales.csv'
(parent of the python file is the code dir, parent of that is the projects dir.
However, It's not great practice to refer to your data by its relative path to your code -- I think using the cwd is a better option -- though what you should do is pass the path to the data in to the script via sys.argv.
I am trying to get the absolute path of a file, but it's not working. My Code is only showing the path to the directory of my running code. This are the two way I have tried:
1)
import os
print(os.path.abspath("More_Types.py"))
2)
import os
print(os.path.realpath("More_Types.py"))
But I continue to get the full path to my running program. How can I get the correct path to file, that is located some where else in my Computer?
PS: I am sorry I can't provide the output because it will reveal all the folders to my running program.
More_Types.py, subdir/More_Types.py and ../More_Types.py are relative paths.
Outcome
If you provide a relative path, realpath and abspath will return an absolute path relative to the current working directory. So essentially, they behave like os.path.join with the current working directory as first argument:
>>> import os.path
>>> os.path.join(os.getcwd(), 'More_Types.py') == os.path.abspath('More_Types.py')
True
This explains, why you get the result you explained.
Explanation
The purpose of abspath is to convert a relative path into an absolute path:
>>> os.path.abspath('More_Types.py')
'/home/meisterluk/More_Types.py'
Unlike abspath, relpath also follows symbolic links. Technically, this is dangerous and can get you caught up in infinite loops (if a link points to any of its parent directories):
>>> os.path.abspath('More_Types.py')
'/net/disk-user/m/meisterluk/More_Types.py'
Proposed solution
However, if you want to retrieve the absolute path relative to some other directory, you can use os.path.join directly:
>>> directory = '/home/foo'
>>> os.path.join(directory, 'More_Types.py')
'/home/foo/More_Types.py'
>>>
I think the best answer to your question is: You need to specify the directory the file is in. Then you can use os.path.join to create an absolute filepath.
You could use this if you want:
import os
a='MoreTypes.py'
for root , dirs , files in os.walk('.') :
for file in files:
if file==a:
print(os.path.join(root,file))
I'm trying to extract tar.gz files which are situated in diffent files named srm01, srm02 and srm03.
The file's name must be in input (a string) to run my code.
I'm trying to do something like this :
import tarfile
import glob
thirdBloc = 'srm01' #Then, that must be 'srm02', or 'srm03'
for f in glob.glob('C://Users//asediri//Downloads/srm/'+thirdBloc+'/'+'*.tar.gz'):
tar = tarfile.open(f)
tar.extractall('C://Users//asediri//Downloads/srm/'+thirdBloc)
I have this error message:
IOError: CRC check failed 0x182518 != 0x7a1780e1L
I want first to be sure that my code find the .tar.gz files. So I tried to just print my paths after glob:
thirdBloc = 'srm01' #Then, that must be 'srm02', or 'srm03'
for f in glob.glob('C://Users//asediri//Downloads/srm/'+thirdBloc+'/'+'*.tar.gz'):
print f
That gives :
C://Users//asediri//Downloads/srm/srm01\20160707000001-server.log.1.tar.gz
C://Users//asediri//Downloads/srm/srm01\20160707003501-server.log.1.tar.gz
The os.path.exists method tell me that my files doesn't exist.
print os.path.exists('C://Users//asediri//Downloads/srm/srm01\20160707000001-server.log.1.tar.gz')
That gives : False
Any way todo properly this work ? What's the best way to have first of all the right paths ?
In order to join paths you have to use os.path.join as follow:
import os
import tarfile
import glob
thirdBloc = 'srm01' #Then, that must be 'srm02', or 'srm03'
for f in glob.glob(os.path.join('C://Users//asediri//Downloads/srm/', thirdBloc, '*.tar.gz'):
tar = tarfile.open(f)
tar.extractall(os.path.join('C://Users//asediri//Downloads/srm/', thirdBloc))
os.path.join will create the correct paths for your filesystem
f = os.path.join('C://Users//asediri//Downloads/srm/', thirdBloc, '*.tar.gz')
C://Users//asediri//Downloads/srm/srm01\20160707000001-server.log.1.tar.gz
Never use \ with python for filepaths, \201 is \x81 character. It results to this:
C://Users//asediri//Downloads/srm/srm01ΓΌ60707000001-server.log.1.tar.gz
this is why os.path.exists does not find it
Or use (r"C:\...")
I would suggest you do this:
import os
os.chdir("C:/Users/asediri/Downloads/srm/srm01")
for f in glob.glob(str(thirdBloc) + ".tar.gz"):
print f