Build the full path filename in Python

Build the full path filename in Python - python

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

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.
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.

If you are fortunate enough to be running Python 3.4+, you can use pathlib:
>>> from pathlib import Path
>>> dirname = '/home/reports'
>>> filename = 'daily'
>>> suffix = '.pdf'
>>> Path(dirname, filename).with_suffix(suffix)
PosixPath('/home/reports/daily.pdf')

Um, why not just:
>>> import os
>>> os.path.join(dir_name, base_filename + "." + format)
'/home/me/dev/my_reports/daily_report.pdf'

Is not it better to add the format in the base filename?
dir_name='/home/me/dev/my_reports/'
base_filename='daily_report.pdf'
os.path.join(dir_name, base_filename)

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

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)

Adding code below for better understanding:
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')

Related

how can i get the path plus the name of the first file in a folder and so on in python

I got this code and i need to take the first path of a file and the files name an have put it as a string
from pathlib import Path
from os import walk
import os
from posixpath import dirname
f = []
jhon = r'C:\Users\ioshu\Desktop\you'
for (dirpath, dirnames, filenames) in walk(jhon):
f.extend(filenames)
f.extend(dirnames)
break
Ben1= filenames[:1]
Ben2= dirpath[:2]
dataFolder = Path(r'C:\Users\ioshu\Desktop\you')
print(Ben1 , dataFolder)
print(dataFolder)
The print (ben1, dataFolder)
the output" of that file is
C:\Users\ioshu\Desktop\you ['07a5iya4vfm91-DASH_720.mp4']
The problem is that i need the out put to be like this C:\Users\ioshu\Desktop\you\0q74nqluycm91-DASH_720

Using walk will walk the whole tree, which is overkill for your needs. You can simply
first_file_name = os.listdir('/etc')[0]
if you are sure there are only files, or:
import os
path = '/etc' # any path you want
first_file = None
for i in os.listdir(path):
file_path = os.path.join(path, i)
if os.path.isfile(file_path):
first_file = file_path
break # assuming you don't need to sort the names
Always use os.path.join to join paths, works on Linux, Windows, MacOS and any other supported platform.
PS: Ben1 = filenames[:1] returns a list with one element, not the element. If you need the element then: Ben1 = filenames[0].
PS2: If you want to use pathlib then dataFolder / filenames[0] or something will help.

How to include a variable in 'r' raw string

I've created a new variable for user path, but not sure how to add in to the following.
import os, pwd
path=os.getcwd()
#crif0 = r '/abc/crif/gpio_mem_0_crif.xml' - original
crif0 = r (path+ '/crif/gpio_mem_0_crif.xml') - I tried with this but doesn't work

As mentioned in the comments, r is a literal prefix which you cannot apply to anything but string literals, so path + r'/crif/...' is enough. However, in this particular case when you need to compose a file path, I'd use the standard library, which makes the code more portable:
import os
path = os.getcwd()
crif0 = os.path.join(path, 'crif', 'gpio_mem_0_crif.xml')
or, in a more modern way using path objects rather than strings:
from pathlib import Path
crif0 = Path.cwd() / 'crif' / 'gpio_mem0_crif.xml'

Creating Modules and using them

I am playing around with creating modules.I have two python scripts.The first (the module) has:
def abspath(relpath):
import os
absdir = os.path.realpath('__file__')
absdir = absdir.split('_')[0].replace('\\', '/')
filename = str(absdir + relpath )
print (filename)
return filename;
The second file (main) has:
import file_tools as ft
filename = ft.abspath('some/path/')
When I run Main, filename appears empty (Type:None). I have run the filename = abspath(etc) within the 'module', and it works. Clearly, I am missing something here!!
and doing this, so any help would be useful.
Thank's all.
MT

The problem lies in how you're finding the working directory; the preferred method being os.getcwd() (or os.getcwdb for Unix compatibility). Using that, we can see that your function boils down to:
def abspath(relpath):
return os.path.join(os.getcwd(), relpath)

Changing file extension in Python

Suppose from index.py with CGI, I have post file foo.fasta to display file. I want to change foo.fasta's file extension to be foo.aln in display file. How can I do it?

An elegant way using pathlib.Path:
from pathlib import Path
p = Path('mysequence.fasta')
p.rename(p.with_suffix('.aln'))

os.path.splitext(), os.rename()
for example:
# renamee is the file getting renamed, pre is the part of file name before extension and ext is current extension
pre, ext = os.path.splitext(renamee)
os.rename(renamee, pre + new_extension)

import os
thisFile = "mysequence.fasta"
base = os.path.splitext(thisFile)[0]
os.rename(thisFile, base + ".aln")
Where thisFile = the absolute path of the file you are changing

Starting from Python 3.4 there's pathlib built-in library. So the code could be something like:
from pathlib import Path
filename = "mysequence.fasta"
new_filename = Path(filename).stem + ".aln"
https://docs.python.org/3.4/library/pathlib.html#pathlib.PurePath.stem
I love pathlib :)

Use this:
os.path.splitext("name.fasta")[0]+".aln"
And here is how the above works:
The splitext method separates the name from the extension creating a tuple:
os.path.splitext("name.fasta")
the created tuple now contains the strings "name" and "fasta".
Then you need to access only the string "name" which is the first element of the tuple:
os.path.splitext("name.fasta")[0]
And then you want to add a new extension to that name:
os.path.splitext("name.fasta")[0]+".aln"

As AnaPana mentioned pathlib is more new and easier in python 3.4 and there is new with_suffix method that can handle this problem easily:
from pathlib import Path
new_filename = Path(mysequence.fasta).with_suffix('.aln')

Using pathlib and preserving full path:
from pathlib import Path
p = Path('/User/my/path')
new_p = Path(p.parent.as_posix() + '/' + p.stem + '.aln')

Sadly, I experienced a case of multiple dots on file name that splittext does not worked well... my work around:
file = r'C:\Docs\file.2020.1.1.xls'
ext = '.'+ os.path.realpath(file).split('.')[-1:][0]
filefinal = file.replace(ext,'')
filefinal = file + '.zip'
os.rename(file ,filefinal)

>> file = r'C:\Docs\file.2020.1.1.xls'
>> ext = '.'+ os.path.realpath(file).split('.')[-1:][0]
>> filefinal = file.replace(ext,'.zip')
>> os.rename(file ,filefinal)
Bad logic for repeating extension, sample: 'C:\Docs\.xls_aaa.xls.xls'

How to rename a file using Python

I want to change a.txt to b.kml.

Use os.rename:
import os
os.rename('a.txt', 'b.kml')
Usage:
os.rename('from.extension.whatever','to.another.extension')

File may be inside a directory, in that case specify the path:
import os
old_file = os.path.join("directory", "a.txt")
new_file = os.path.join("directory", "b.kml")
os.rename(old_file, new_file)

As of Python 3.4 one can use the pathlib module to solve this.
If you happen to be on an older version, you can use the backported version found here
Let's assume you are not in the root path (just to add a bit of difficulty to it) you want to rename, and have to provide a full path, we can look at this:
some_path = 'a/b/c/the_file.extension'
So, you can take your path and create a Path object out of it:
from pathlib import Path
p = Path(some_path)
Just to provide some information around this object we have now, we can extract things out of it. For example, if for whatever reason we want to rename the file by modifying the filename from the_file to the_file_1, then we can get the filename part:
name_without_extension = p.stem
And still hold the extension in hand as well:
ext = p.suffix
We can perform our modification with a simple string manipulation:
Python 3.6 and greater make use of f-strings!
new_file_name = f"{name_without_extension}_1"
Otherwise:
new_file_name = "{}_{}".format(name_without_extension, 1)
And now we can perform our rename by calling the rename method on the path object we created and appending the ext to complete the proper rename structure we want:
p.rename(Path(p.parent, new_file_name + ext))
More shortly to showcase its simplicity:
Python 3.6+:
from pathlib import Path
p = Path(some_path)
p.rename(Path(p.parent, f"{p.stem}_1_{p.suffix}"))
Versions less than Python 3.6 use the string format method instead:
from pathlib import Path
p = Path(some_path)
p.rename(Path(p.parent, "{}_{}_{}".format(p.stem, 1, p.suffix))

import shutil
shutil.move('a.txt', 'b.kml')
This will work to rename or move a file.

os.rename(old, new)
This is found in the Python docs: http://docs.python.org/library/os.html

As of Python version 3.3 and later, it is generally preferred to use os.replace instead of os.rename so FileExistsError is not raised if the destination file already exists.
assert os.path.isfile('old.txt')
assert os.path.isfile('new.txt')
os.rename('old.txt', 'new.txt')
# Raises FileExistsError
os.replace('old.txt', 'new.txt')
# Does not raise exception
assert not os.path.isfile('old.txt')
assert os.path.isfile('new.txt')
See the documentation.

Use os.rename. But you have to pass full path of both files to the function. If I have a file a.txt on my desktop so I will do and also I have to give full of renamed file too.
os.rename('C:\\Users\\Desktop\\a.txt', 'C:\\Users\\Desktop\\b.kml')

One important point to note here, we should check if any files exists with the new filename.
suppose if b.kml file exists then renaming other file with the same filename leads to deletion of existing b.kml.
import os
if not os.path.exists('b.kml'):
os.rename('a.txt','b.kml')

import os
# Set the path
path = 'a\\b\\c'
# save current working directory
saved_cwd = os.getcwd()
# change your cwd to the directory which contains files
os.chdir(path)
os.rename('a.txt', 'b.klm')
# moving back to the directory you were in
os.chdir(saved_cwd)

Using the Pathlib library's Path.rename instead of os.rename:
import pathlib
original_path = pathlib.Path('a.txt')
new_path = original_path.rename('b.kml')

Here is an example using pathlib only without touching os which changes the names of all files in a directory, based on a string replace operation without using also string concatenation:
from pathlib import Path
path = Path('/talend/studio/plugins/org.talend.designer.components.bigdata_7.3.1.20200214_1052\components/tMongoDB44Connection')
for p in path.glob("tMongoDBConnection*"):
new_name = p.name.replace("tMongoDBConnection", "tMongoDB44Connection")
new_name = p.parent/new_name
p.rename(new_name)

import shutil
import os
files = os.listdir("./pics/")
for key in range(0, len(files)):
print files[key]
shutil.move("./pics/" + files[key],"./pics/img" + str(key) + ".jpeg")
This should do it. python 3+

How to change the first letter of filename in a directory:
import os
path = "/"
for file in os.listdir(path):
os.rename(path + file, path + file.lower().capitalize())
then = os.listdir(path)
print(then)

If you are Using Windows and you want to rename your 1000s of files in a folder then:
You can use the below code. (Python3)
import os
path = os.chdir(input("Enter the path of the Your Image Folder : ")) #Here put the path of your folder where your images are stored
image_name = input("Enter your Image name : ") #Here, enter the name you want your images to have
i = 0
for file in os.listdir(path):
new_file_name = image_name+"_" + str(i) + ".jpg" #here you can change the extention of your renmamed file.
os.rename(file,new_file_name)
i = i + 1
input("Renamed all Images!!")

os.chdir(r"D:\Folder1\Folder2")
os.rename(src,dst)
#src and dst should be inside Folder2

import os
import re
from pathlib import Path
for f in os.listdir(training_data_dir2):
for file in os.listdir( training_data_dir2 + '/' + f):
oldfile= Path(training_data_dir2 + '/' + f + '/' + file)
newfile = Path(training_data_dir2 + '/' + f + '/' + file[49:])
p=oldfile
p.rename(newfile)

You can use os.system to invoke terminal to accomplish the task:
os.system('mv oldfile newfile')

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