Develop Reference

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'

Path name format - Windows - Python

I am working on a code that takes the path of a file and the reads it.
The code is,
import sys
import os
user_input = input("Enter the path of your file: ")
assert os.path.exists(user_input), "I did not find the file at, "+str(user_input)
However, I am on a Windows machine in which, the path names come with a \, like, C:\your\path\and\file.xlsx
Each time I enter the file name to the prompt, I need to manually replace all \ with / and then run the code.
Is there a solution such that I enter C:\your\path\and\file.xlsx but the code takes it as, C:/your/path/and/file.xlsx
Regards.

Use pythons builtin pathlib like this:
from pathlib import Path
# No matter what slash you use in the input path
# I used Raw-Prefix 'r' here because \t would otherwise be interpreted as tab
path_a = Path(r"C:\Temp\test.txt)
path_b = Path(r"C:\Temp/test.txt)
path_c = Path(r"C:/Temp/test.txt)
path_d = Path(r"C:\\Temp\\test.txt)
print(path_a, path_b, path_c, path_d)
# all Paths will be the same:
# out >> C:\Temp\test.txt C:\Temp\test.txt C:\Temp\test.txt C:\Temp\test.txt
Furthermore you can also easily extend a given Path like this:
path_e = Path("C:\Temp")
extended_path = path_e / "Test"
print(extended_path)
# out >> C:\Temp\Test
So in your case simply do it like this:
import sys
import os
from pathlib import Path
user_input = input("Enter the path of your file: ")
file_path = Path(user_input)
if not os.path.exists(file_input):
print("I did not find the file at, " + str(file_path))

Oddly enough, when I use \ in my path it just worked. But for whatever reason if yours is not, here is a way to change backslash to slash. One with a regex and one using replace:
import sys
import os
import re
user_input = input("Enter the path of your file: ")
#with a regex
user_input_regex = re.sub(r"\\", "/", user_input)
print(user_input_regex)
#using replace
user_input_replace = user_input.replace("\\","/")
print(user_input_replace)
I believe that replace is slightly faster but if you want to change other stuff (or just like using regex) the regex will probably offer more options down the road. The key to this is the \ needs to be escaped with a \ since itself is an escape character.

How to get parent folder name of current directory?

I know there are functions for finding parent directory or path such as.
os.path.dirname(os.path.realpath(__file__))
'C:\Users\jahon\Desktop\Projects\CAA\Result\caa\project_folder'
Is there a function that just returns the parent folder name? In this case it should be project_folder.

You can achieve this easily with os
import os
os.path.basename(os.getcwd())

You can get the last part of any path using basename (from os.path):
>>> from os.path import basename
>>> basename('/path/to/directory')
'directory'
Just to note, if your path ends with / then the last part of the path is empty:
>>> basename('/path/to/directory/')
''

Yes, you can use PurePath.
PurePath(__file__).parent.name == 'parent_dir'

You can use split and os.path.sep to get the list of path elements and then call the last element of the list:
import os
path = 'C:\\Users\\jahon\\Desktop\\Projects\\CAA\\Result\\caa\\project_folder'
if path.split(os.path.sep)[-1]:
parent_folder = path.split(os.path.sep)[-1] # if no backslashes at the end
else:
parent_folder = path.split(os.path.sep)[-2] # with backslashes at the end

How to check if a file exists without looking at its extension name in Python?

A file can have multiple extensions but name of the file will remains same.
I have tried
import os.path
os.path.isfile("Filename")
but this code is looking at the extension of the file also.

This would list all files with same name but different extensions.
import glob
print glob.glob("E:\\Logs\\Filename.*")
You could use this check instead.
import glob
if glob.glob("E:\\Logs\\Filename.*"):
print "Found"
Refer this post.

Try this.
import os
def check_file(dir, prefix):
for s in os.listdir(dir):
if os.path.splitext(s)[0] == prefix and os.path.isfile(os.path.join(dir, s)):
return True
return False
You can call this function like, e.g., check_file("/path/to/dir", "my_file") to search for files of the form /path/to/dir/my_file.*.

You can use fnmatch also,
import os
import fnmatch
print filter(lambda f: fnmatch.fnmatch(f, "Filename.*"), os.listdir(FilePath))
Here, No need to format FilePath. You can simply write like 'C:\Python27\python.*'

This will get all the files with the basename you want (in this case 'tmp') with or without an extension and will exclude things that start with your basename - like tmp_tmp.txt for example:
import re
import os
basename='tmp'
for filename in os.listdir('.'):
if re.match(basename+"(\..*)?$", filename):
print("this file: %s matches my basename"%filename)
Or of course if you prefer them in a list, more succinctly:
[fn for fn in os.listdir('.') if re.match(basename+"(\..*)?$",fn)]

Build the full path filename in 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')