How can I create a full path to a file from parts (e.g. path to the folder, name and extension)?

Question

I need to pass a file path name to a module. How do I build the file path from a directory name, base filename, and a file format string?

The directory may or may not exist at the time of call.

For example:

dir_name='/home/me/dev/my_reports'
base_filename='daily_report'
format = 'pdf'

I need to create a string '/home/me/dev/my_reports/daily_report.pdf'

Concatenating the pieces manually doesn’t seem to be a good way. I tried os.path.join:

join(dir_name,base_filename,format)

but it gives

/home/me/dev/my_reports/daily_report/pdf

Asked By: Damon Julian

||

Source

Answer 1

>>> import os
>>> os.path.join(dir_name, base_filename + "." + format)
'/home/me/dev/my_reports/daily_report.pdf'

Answered By: Mark Longair

Answer 2

This works fine:

os.path.join(dir_name, base_filename + '.' + filename_suffix)

Keep in mind that os.path.join() exists only because different operating systems use different path separator characters. It smooths over that difference so cross-platform code doesn’t have to be cluttered with special cases for each OS. There is no need to do this for file name "extensions" (see footnote) because they are always preceded by a dot character, on every OS that implements them.

If using a function anyway makes you feel better (and you like needlessly complicating your code), you can do this:

os.path.join(dir_name, '.'.join((base_filename, filename_suffix)))

If you prefer to keep your code clean, simply include the dot in the suffix:

suffix = '.pdf'
os.path.join(dir_name, base_filename + suffix)

That approach also happens to be compatible with the suffix conventions in pathlib, which was introduced in python 3.4 a few years after this question was asked. New code that doesn’t require backward compatibility can do this:

suffix = '.pdf'
pathlib.PurePath(dir_name, base_filename + suffix)

You might be tempted to use the shorter Path() instead of PurePath() if you’re only handling paths for the local OS. I would question that choice, given the cross-platform issues behind the original question.

Warning: Do not use pathlib’s with_suffix() for this purpose. That method will corrupt base_filename if it ever contains a dot.

Footnote: Outside of Microsoft operating systems, there is no such thing as a file name "extension". Its presence on Windows comes from MS-DOS and FAT, which borrowed it from CP/M, which has been dead for decades. That dot-plus-three-letters that many of us are accustomed to seeing is just part of the file name on every other modern OS, where it has no built-in meaning.

Answered By: ʇsәɹoɈ

Answer 3

Just use os.path.join to join your path with the filename and extension. Use sys.argv to access arguments passed to the script when executing it:

#!/usr/bin/env python3
# coding: utf-8

# import netCDF4 as nc
import numpy as np
import numpy.ma as ma
import csv as csv

import os.path
import sys

basedir = '/data/reu_data/soil_moisture/'
suffix = 'nc'


def read_fid(filename):
    fid = nc.MFDataset(filename,'r')
    fid.close()
    return fid

def read_var(file, varname):
    fid = nc.Dataset(file, 'r')
    out = fid.variables[varname][:]
    fid.close()
    return out


if __name__ == '__main__':
    if len(sys.argv) < 2:
        print('Please specify a year')

    else:
        filename = os.path.join(basedir, '.'.join((sys.argv[1], suffix)))
        time = read_var(ncf, 'time')
        lat = read_var(ncf, 'lat')
        lon = read_var(ncf, 'lon')
        soil = read_var(ncf, 'soilw')

Simply run the script like:

   # on windows-based systems
   python script.py year

   # on unix-based systems
   ./script.py year

Answered By: albert

Answer 4

In Python 3.4 and above, the pathlib standard library module can be used like so:

>>> from pathlib import Path
>>> dirname = '/home/reports'
>>> filename = 'daily'
>>> suffix = '.pdf'
>>> Path(dirname, filename).with_suffix(suffix)
PosixPath('/home/reports/daily.pdf')

Answered By: Eugene Yarmash

Answer 5

from pathlib import Path

# Build paths inside the project like this: BASE_DIR / 'subdir'.

BASE_DIR = Path(__file__).resolve().parent.parent
TEMPLATE_PATH = Path.joinpath(BASE_DIR,"templates")
print(TEMPLATE_PATH)

Answered By: Ravi

Answer 6

Why not just include the extension in the base filename?

dir_name='/home/me/dev/my_reports/'
base_filename='daily_report.pdf'
os.path.join(dir_name, base_filename)

Answered By: Saltanat Khalyk

Answer 7

import os
def createfile(name, location, extension):
    print(name, extension, location)
    #starting creating a file with some dummy contents
    path = os.path.join(location, name + '.' + extension)
    f = open(path, "a")
    f.write("Your contents!! or whatever you want to put inside this file.")
    f.close()
    print("File creation is successful!!")

def readfile(name, location, extension):
    #open and read the file after the appending:
    path = os.path.join(location, name + '.' + extension)
    f = open(path, "r")
    print(f.read())

#pass the parameters here
createfile('test','./','txt')
readfile('test','./','txt')

Answered By: Nitikesh Pattanayak

How can I create a full path to a file from parts (e.g. path to the folder, name and extension)?

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