In python, suppose I have a path like this:
/folderA/folderB/folderC/folderD/
How can I get just the folderD part?
Use os.path.normpath, then os.path.basename:
>>> os.path.basename(os.path.normpath('/folderA/folderB/folderC/folderD/'))
'folderD'
The first strips off any trailing slashes, the second gives you the last part of the path. Using only basename gives everything after the last slash, which in this case is ''.
With python 3 you can use the pathlib module (pathlib.PurePath for example):
>>> import pathlib
>>> path = pathlib.PurePath('/folderA/folderB/folderC/folderD/')
>>> path.name
'folderD'
If you want the last folder name where a file is located:
>>> path = pathlib.PurePath('/folderA/folderB/folderC/folderD/file.py')
>>> path.parent.name
'folderD'
You could do
>>> import os
>>> os.path.basename('/folderA/folderB/folderC/folderD')
UPDATE1: This approach works in case you give it /folderA/folderB/folderC/folderD/xx.py. This gives xx.py as the basename. Which is not what you want I guess. So you could do this -
>>> import os
>>> path = "/folderA/folderB/folderC/folderD"
>>> if os.path.isdir(path):
dirname = os.path.basename(path)
UPDATE2: As lars pointed out, making changes so as to accomodate trailing '/'.
>>> from os.path import normpath, basename
>>> basename(normpath('/folderA/folderB/folderC/folderD/'))
'folderD'
Here is my approach:
>>> import os
>>> print os.path.basename(
os.path.dirname('/folderA/folderB/folderC/folderD/test.py'))
folderD
>>> print os.path.basename(
os.path.dirname('/folderA/folderB/folderC/folderD/'))
folderD
>>> print os.path.basename(
os.path.dirname('/folderA/folderB/folderC/folderD'))
folderC
I was searching for a solution to get the last foldername where the file is located, I just used split two times, to get the right part. It's not the question but google transfered me here.
pathname = "/folderA/folderB/folderC/folderD/filename.py"
head, tail = os.path.split(os.path.split(pathname)[0])
print(head + " " + tail)
I like the parts method of Path for this:
grandparent_directory, parent_directory, filename = Path(export_filename).parts[-3:]
log.info(f'{t: <30}: {num_rows: >7} Rows exported to {grandparent_directory}/{parent_directory}/{filename}')
If you use the native python package pathlib it's really simple.
>>> from pathlib import Path
>>> your_path = Path("/folderA/folderB/folderC/folderD/")
>>> your_path.stem
'folderD'
Suppose you have the path to a file in folderD.
>>> from pathlib import Path
>>> your_path = Path("/folderA/folderB/folderC/folderD/file.txt")
>>> your_path.name
'file.txt'
>>> your_path.parent
'folderD'
During my current projects, I'm often passing rear parts of a path to a function and therefore use the Path module. To get the n-th part in reverse order, I'm using:
from typing import Union
from pathlib import Path
def get_single_subpath_part(base_dir: Union[Path, str], n:int) -> str:
if n ==0:
return Path(base_dir).name
for _ in range(n):
base_dir = Path(base_dir).parent
return getattr(base_dir, "name")
path= "/folderA/folderB/folderC/folderD/"
# for getting the last part:
print(get_single_subpath_part(path, 0))
# yields "folderD"
# for the second last
print(get_single_subpath_part(path, 1))
#yields "folderC"
Furthermore, to pass the n-th part in reverse order of a path containing the remaining path, I use:
from typing import Union
from pathlib import Path
def get_n_last_subparts_path(base_dir: Union[Path, str], n:int) -> Path:
return Path(*Path(base_dir).parts[-n-1:])
path= "/folderA/folderB/folderC/folderD/"
# for getting the last part:
print(get_n_last_subparts_path(path, 0))
# yields a `Path` object of "folderD"
# for second last and last part together
print(get_n_last_subparts_path(path, 1))
# yields a `Path` object of "folderc/folderD"
Note that this function returns a Pathobject which can easily be converted to a string (e.g. str(path))
path = "/folderA/folderB/folderC/folderD/"
last = path.split('/').pop()
str = "/folderA/folderB/folderC/folderD/"
print str.split("/")[-2]
Related
I want to insert a directory name in the middle of a given file path, like this:
directory_name = 'new_dir'
file_path0 = 'dir1/dir2/dir3/dir4/file.txt'
file_path1 = some_func(file_path0, directory_name, position=2)
print(file_path1)
>>> 'dir1/dir2/new_dir/dir3/dir4/file.txt'
I looked through the os.path and pathlib packages, but it looks like they don't have a function that allows for inserting in the middle of a file path. I tried:
import sys,os
from os.path import join
path_ = file_path0.split(os.sep)
path_.insert(2, 'new_dir')
print(join(path_))
but this results in the error
"expected str, bytes or os.PathLike object, not list"
Does anyone know standard python functions that allow such inserting in the middle of a file path? Alternatively - how can I turn path_ to something that can be processed by os.path. I am new to pathlib, so maybe I missed something out there
Edit: Following the answers to the question I can suggest the following solutions:
1.) As Zach Favakeh suggests and as written in this answer just correct my code above to join(*path_) by using the 'splat' operator * and everything is solved.
2.) As suggested by buran you can use the pathlib package, in very short it results in:
from pathlib import PurePath
path_list = list(PurePath(file_path0).parts)
path_list.insert(2, 'new_dir')
file_path1 = PurePath('').joinpath(*path_list)
print(file_path1)
>>> 'dir1/dir2/new_dir/dir3/dir4/file.txt'
Take a look at pathlib.PurePath.parts. It will return separate components of the path and you can insert at desired position and construct the new path
>>> from pathlib import PurePath
>>> file_path0 = 'dir1/dir2/dir3/dir4/file.txt'
>>> p = PurePath(file_path0)
>>> p.parts
('dir1', 'dir2', 'dir3', 'dir4', 'file.txt')
>>> spam = list(p.parts)
>>> spam.insert(2, 'new_dir')
>>> new_path = PurePath('').joinpath(*spam)
>>> new_path
PurePosixPath('dir1/dir2/new_dir/dir3/dir4/file.txt')
This will work with path as a str as well as with pathlib.Path objects
Since you want to use join on a list to produce the pathname, you should do the following using the "splat" operator: Python os.path.join() on a list
Edit: You could also take your np array and concatenate its elements into a string using np.array2string, using '/' as your separator parameter:https://docs.scipy.org/doc/numpy-1.14.0/reference/generated/numpy.array2string.html
Hope this helps.
Solution using regex. The regex will create groups of the following
[^\/]+ - non-'/' characters(i.e. directory names)
\w+\.\w+ - word characters then '.' then word characters (i.e. file name)
import re
directory_name = 'new_dir'
file_path0 = 'dir1/dir2/dir3/dir4/file.txt'
position = 2
regex = re.compile(r'([^\/]+|\w+\.\w+)')
tokens = re.findall(regex, file_path0)
tokens.insert(position, directory_name)
file_path1 = '/'.join(tokens)
Result:
'dir1/dir2/new_dir/dir3/dir4/file.txt'
Your solution has only one flaw. After inserting the new directory in the path list path_.insert(2, 'new_dir')you need to call os.path.join(*path_) to get the new modified path. The error that you get is because you are passing a list as parameter to the join function, but you have to unpack it.
In my case, I knew the portion of path that would precede the insertion point (i.e., "root"). However, the position of the insertion point was not constant due to the possibility of having varying number of path components in the root path. I used Path.relative_to() to break the full path to yield an insertion point for the new_dir.
from pathlib import Path
directory_name = Path('new_dir')
root = Path('dir1/dir2/')
file_path0 = Path('dir1/dir2/dir3/dir4/file.txt')
# non-root component of path
chld = file_path0.relative_to(root)
file_path1 = root / directory_name / chld
print(file_path1)
Result:
'dir1/dir2/new_dir/dir3/dir4/file.txt'
I made a try with your need:
directory_name = '/new_dir'
file_path0 = 'dir1/dir2/dir3/dir4/file.txt'
before_the_newpath = 'dir1/dir2'
position = file_path0.split(before_the_newpath)
new_directory = before_the_newpath + directory_name + position[1]
Hope it helps.
I want to create a list of all the filepath names that match a specific string e.g. "04_DEM" so I can do further processing on the files inside those directories?
e.g.
INPUT
C:\directory\NewZealand\04DEM\DEM_CD23_1232.tif
C:\directory\Australia\04DEM\DEM_CD23_1233.tif
C:\directory\NewZealand\05DSM\DSM_CD23_1232.tif
C:\directory\Australia\05DSM\DSM_CD23_1232.tif
WANTED OUTPUT
C:\directory\NewZealand\04DEM\
C:\directory\Australia\04DEM\
This makes sure that only those files are processed, as some other files in the directories also have the same string "DEM" included in their filename, which I do not want to modify.
This is my bad attempt due to being a rookie with Py code
import os
for dirnames in os.walk('D:\Canterbury_2017Copy'):
print dirnames
if dirnames=='04_DEM' > listofdirectoriestoprocess.txt
print "DONE CHECK TEXT FILE"
You can use os.path for this:
import os
lst = [r'C:\directory\NewZealand\04DEM\DEM_CD23_1232.tif',
r'C:\directory\Australia\04DEM\DEM_CD23_1233.tif',
r'C:\directory\NewZealand\05DSM\DSM_CD23_1232.tif',
r'C:\directory\Australia\05DSM\DSM_CD23_1232.tif']
def filter_paths(lst, x):
return [os.path.split(i)[0] for i in lst if os.path.normpath(i).split(os.sep)[3] == x]
res = list(filter_paths(lst, '04DEM'))
# ['C:\\directory\\NewZealand\\04DEM',
# 'C:\\directory\\Australia\\04DEM']
Use in to check if a required string is in another string.
This is one quick way:
new_list = []
for path in path_list:
if '04DEM' in path:
new_list.append(path)
Demo:
s = 'C:/directory/NewZealand/04DEM/DEM_CD23_1232.tif'
if '04DEM' in s:
print(True)
# True
Make sure you use / or \\ as directory separator instead of \ because the latter escapes characters.
First, you select via regex using re, and then use pathlib:
import re
import pathlib
pattern = re.compile('04DEM')
# You use pattern.search() if s is IN the string
# You use pattern.match() if s COMPLETELY matches the string.
# Apply the correct function to your use case.
files = [s in list_of_files if pattern.search(s)]
all_pruned_paths = set()
for p in files:
total = ""
for d in pathlib.Path(p):
total = os.path.join(total, d)
if pattern.search(s):
break
all_pruned_paths.add(total)
result = list(all_pruned_paths)
This is more robust than using in because you might need to form more complicated queries in the future.
Assume my folder structure to be
+Data
-abc.jpg
-db.jpg
-ap.jpg
Input is 'path/to/Data'
Expected output is ['abc','db','ap']
I saw many similar questions but did not get what exactly I wanted.
I prefer to use os module in python.
simply try this,
l=os.listdir('path')
li=[x.split('.')[0] for x in l]
First list your directory files
split file by. and take first argument.
import os
files_no_ext = [".".join(f.split(".")[:-1]) for f in os.listdir() if os.path.isfile(f)]
print(files_no_ext)
import glob
from Pathlib import Path
for f in glob.glob("*.*"):
print(Path(f).stem)
import os
filenames=next(os.walk(os.getcwd()))[2]
efn=[f.split('.')[0] for f in filenames]
os.getcwd() #for get current directory path
You can use os.listdir which take path as argument and return a list of files and directories in it.
>>> list_ = os.listdir("path/to/Data")
>>> list_
>>> ['abc.jpg', 'dn.jpg', 'ap.jpg']
With that list, you only have to do a comprehension list which split each element on '.' (dot), take all the elements except the last one an join them with '.' (dot) and check if the element is a file using os.path.file().
>>> list_ = ['.'.join(x.split('.')[:-1]) for x in os.listdir("path/to/Data") if os.path.isfile(os.path.join('path/to/Data', x))]
>>> list_
>>> ['abc', 'dn', 'ap']
import os
from os import listdir
from os.path import isfile, join
def filesMinusExtension(path):
# os.path.splitext(f)[0] map with filename without extension with checking if file exists.
files = [os.path.splitext(f)[0] for f in listdir(path) if isfile(join(path, f))];
return files;
Is there something similar to os.path.dirname(path), but in pathlib?
It looks like there is a parents element that contains all the parent directories of a given path. E.g., if you start with:
>>> import pathlib
>>> p = pathlib.Path('/path/to/my/file')
Then p.parents[0] is the directory containing file:
>>> p.parents[0]
PosixPath('/path/to/my')
...and p.parents[1] will be the next directory up:
>>> p.parents[1]
PosixPath('/path/to')
Etc.
p.parent is another way to ask for p.parents[0]. You can convert a Path into a string and get pretty much what you would expect:
>>> str(p.parent)
'/path/to/my'
And also on any Path you can use the .absolute() method to get an absolute path:
>>> os.chdir('/etc')
>>> p = pathlib.Path('../relative/path')
>>> str(p.parent)
'../relative'
>>> str(p.parent.absolute())
'/etc/../relative'
Note that os.path.dirname and pathlib treat paths with a trailing slash differently. The pathlib parent of some/path/ is some:
>>> p = pathlib.Path('some/path/')
>>> p.parent
PosixPath('some')
While os.path.dirname on some/path/ returns some/path:
>>> os.path.dirname('some/path/')
'some/path'
summary
from pathlib import Path
file_path = Path("/Users/yuanz/PycharmProjects/workenv/little_code/code09/sample.csv")
1. get dir path
file_path.parent
# >>> /Users/yuanz/PycharmProjects/workenv/little_code/code09
2. get filename
file_path.name
# >>> sample.csv
3. get file type
file_path.suffix
# >>> .csv
4.add new file in this dir path
file_path.parent.joinpath("dd.png")
I came here looking for something very similar. My solution, based on the above by #larsks, and assuming you want to preserve the entire path except the filename, is to do:
>>> import pathlib
>>> p = pathlib.Path('/path/to/my/file')
>>> pathlib.Path('/'.join(list(p.parts)[1:-1])+'/')
Essentially, list(p.parts)[1:-1] creates a list of Path elements, starting from the second to n-1th, and you join them with a '/' and make a path of the resulting string. Edit The final +'/' adds in the trailing slash - adjust as required.
I am looking for a way to extract a filename and extension from a particular url using Python
lets say a URL looks as follows
picture_page = "http://distilleryimage2.instagram.com/da4ca3509a7b11e19e4a12313813ffc0_7.jpg"
How would I go about getting the following.
filename = "da4ca3509a7b11e19e4a12313813ffc0_7"
file_ext = ".jpg"
try:
# Python 3
from urllib.parse import urlparse
except ImportError:
# Python 2
from urlparse import urlparse
from os.path import splitext, basename
picture_page = "http://distilleryimage2.instagram.com/da4ca3509a7b11e19e4a12313813ffc0_7.jpg"
disassembled = urlparse(picture_page)
filename, file_ext = splitext(basename(disassembled.path))
Only downside with this is that your filename will contain a preceding / which you can always remove yourself.
Try with urlparse.urlsplit to split url, and then os.path.splitext to retrieve filename and extension (use os.path.basename to keep only the last filename) :
import urlparse
import os.path
picture_page = "http://distilleryimage2.instagram.com/da4ca3509a7b11e19e4a12313813ffc0_7.jpg"
print os.path.splitext(os.path.basename(urlparse.urlsplit(picture_page).path))
>>> ('da4ca3509a7b11e19e4a12313813ffc0_7', '.jpg')
filename = picture_page.split('/')[-1].split('.')[0]
file_ext = '.'+picture_page.split('.')[-1]
# Here's your link:
picture_page = "http://distilleryimage2.instagram.com/da4ca3509a7b11e19e4a12313813ffc0_7.jpg"
#Here's your filename and ext:
filename, ext = (picture_page.split('/')[-1].split('.'))
When you do picture_page.split('/'), it will return a list of strings from your url split by a /.
If you know python list indexing well, you'd know that -1 will give you the last element or the first element from the end of the list.
In your case, it will be the filename: da4ca3509a7b11e19e4a12313813ffc0_7.jpg
Splitting that by delimeter ., you get two values:
da4ca3509a7b11e19e4a12313813ffc0_7 and jpg, as expected, because they are separated by a period which you used as a delimeter in your split() call.
Now, since the last split returns two values in the resulting list, you can tuplify it.
Hence, basically, the result would be like:
filename,ext = ('da4ca3509a7b11e19e4a12313813ffc0_7', 'jpg')
os.path.splitext will help you extract the filename and extension once you have extracted the relevant string from the URL using urlparse:
fName, ext = os.path.splitext('yourImage.jpg')
This is the easiest way to find image name and extension using regular expression.
import re
import sys
picture_page = "http://distilleryimage2.instagram.com/da4ca3509a7b11e19e4a12313813ffc0_7.jpg"
regex = re.compile('(.*\/(?P<name>\w+)\.(?P<ext>\w+))')
print regex.search(picture_page).group('name')
print regex.search(picture_page).group('ext')
>>> import re
>>> s = 'picture_page = "http://distilleryimage2.instagram.com/da4ca3509a7b11e19e4a12313813ffc0_7.jpg"'
>>> re.findall(r'\/([a-zA-Z0-9_]*)\.[a-zA-Z]*\"$',s)[0]
'da4ca3509a7b11e19e4a12313813ffc0_7'
>>> re.findall(r'([a-zA-Z]*)\"$',s)[0]
'jpg'