I am trying to get the rest of the directory after a specific folder. Say, for example:
/heres/some/folders/with/files
And I want everything after /folders/:
/with/files
This is what I have:
Tk().withdraw()
directory = os.path.basename(filedialog.askdirectory())
print(directory)
I know that basename only prints the current folder, but I could not find anything on this to even try and code it.
Related
I'm making a program to back up files and folders to a destination.
The problem I'm currently facing is if I have a folder inside a folder and so on, with files in between them, I can't Sync them at the destination.
e.g.:
The source contains folder 1 and file 2. Folder 1 contains folder 2, folder 2 contains folder 3 and files etc...
The backup only contains folder 1 and file 2.
If the backup doesn't exist I simply use: shutil.copytree(path, path_backup), but in the case, I need to sync I can't get the files and folders or at least I'm not seeing a way to do it. I have walked the directory with for path, dir, files in os.walk(directory) and even used what someone suggest in another post:
def walk_folder(target_path, path_backup):
for files in os.scandir(target_path):
if os.path.isfile(files):
file_name = os.path.abspath(files)
print(file_name)
os.makedirs(path_backup)
elif os.path.isdir(files):
walk_folder(files, path_backup)
Is there a way to make the directories in the backup folder from the ground up and then add the info alongside or is the only way to just delete the whole folder and use shutil.copytree(path, path_backup).
With makedirs, all it does is say it can't create because the folder already exists, this is understandable as it's trying to write in the Source folder and not in the backup. Is there a way to make the path to replace Source for backup?
If any more code is needed feel free to ask!
I have a code that locates files in a folder for me by name and moves them to another set path.
import os, shutil
files = os.listdir("D:/Python/Test")
one = "one"
two = "two"
oney = "fold1"
twoy="fold2"
def findfile(x,y):
for file in files:
if x in file.lower():
while x in file.lower():
src = ('D:/Python/Test/'+''.join(file))
dest = ('D:/Python/Test/'+y)
if not os.path.exists(dest):
os.makedirs(dest)
shutil.move(src,dest)
break
findfile(one,oney)
findfile(two,twoy)
In this case, this program moves all the files in the Test folder to another path depending on the name, let's say one as an example:
If there is a .png named one, it will move it to the fold1 folder. The problem is that my code does not distinguish between types of files and what I would like is that it excludes the folders from the search.
If there is a folder called one, don't move it to the fold1 folder, only move it if it is a folder! The other files if you have to move them.
The files in the folder contain the string one, they are not called exactly that.
I don't know if I have explained myself very well, if something is not clear leave me a comment and I will try to explain it better!
Thanks in advance for your help!
os.path.isdir(path)
os.path.isdir() method in Python is used to check whether the specified path is an existing directory or not. This method follows symbolic link, that means if the specified path is a symbolic link pointing to a directory then the method will return True.
Check with that function before.
Home this helps :)
I am trying to loop through sub folders in a directory, but my for loop is skipping over some folders for a reason that is not clear to me, despite having the same name structure and properties as the other folders in the directory. I should add that the directory is a mapped network drive, but I'm not sure why that would matter. I've tried both os.walk and listdir to pick up the names of the folders:
root = r"mydir"
for directory, subdirectory, files in os.walk(mydir):
print(directory)
def get_immediate_subdirectories(a_dir):
return [name for name in os.listdir(a_dir)
if os.path.isdir(os.path.join(a_dir, name))]
Is there a way to force the code to pick up everything in my main directory other than what I'm doing here?
Expected Output:
['11111-Apples',
'11112-Bananas',
'11113-Grapes',
'11114-Oranges',
'11115-Pears']
Actual Output:
['11112-Bananas',
'11113-Grapes',
'11115-Pears']
I am so new with python and pycharm and i got confuse!!
When I run my project in pycharm it gives me an error about not finding the path of my file. The physical file path is:
'../Project/BC/RequiredFiles/resources/a_reqs.csv'
My project working directory is "Project/BC" and the project running file (startApp.sh) is there too. but the .py file that wants to work with a_req.csv is inside the "RequiredFiles" folder. There is the following code in the .py file:
reqsfile = os.getcwd() + "/resources/a_reqs.csv"
it returns: '../Project/BC/resources/a_reqs.csv'
instead of: '../Project/BC/resources/RequiredFiles/a_reqs.csv'
while the .py file is in "RequiredFiles" the os.getcwd() must include it too. but it does not.
The problem is that i can not change the addressing code. because this code works in another IDE and other people who work with the code in other platform or OS do not have any problem. I am working in mac OS and if i am not mistaken the code works with windows!!
So, how can i tell Pycharm (in mac) to see and load "RequiredFiles" folder as the subfolder of my working directory!!!
os.getcwd returns the current working directory of the process (which may be the directory where startApp.sh is located or another one, depending on the PyCharm's run configuration setting, or, if you start the program from the command line, the directory in which you execute the command).
To make a path independent on the current working directory, you can take the directory where your Python file is located and build the path from it:
os.path.dirname(__file__) + "/resources/a_reqs.csv"
From your question what I see is:
reqsfile = os.getcwd() + "/resources/a_reqs.csv"
Which produces: "../Project/BC/resources/a_reqs.csv", whereas your desired output is
"../Project/BC/resources/RequiredFiles/a_reqs.csv". Since we know os.getcwd is returning "/Project/BC/", then to get your desired result you should be doing:
reqsfile = os.getcwd() + "/resources/RequiredFiles/a_reqs.csv"
But since you want the solution to work with or without the RequiredFiles subdirectory you could apply a conditional solution, ie something like:
import os.path
if os.path.exists(os.getcwd() + "/resources/RequiredFiles/a_reqs.csv"):
reqsfile = os.getcwd() + "/resources/RequiredFiles/a_reqs.csv"
else:
reqsfile = os.getcwd() + "/resources/a_reqs.csv"
This solution will set the reqsfile to the csv in the RequiredFiles directory if the directory exists, and thus will work for you. On the other-hand, if the RequiredFiles directory doesn't exist, it will default to the csv in /resources/.
Typically when groups collaborate on projects, the maintain the same file hierarchy so that these types of issues are avoided, so you might want to consider moving the csv from /RequiredFiles/ to /resources/.
I am new to python and have to play with directories.
Can someone help me with it?
I have to know a path with newly created folder. A new temp folder is generated by my code in which some files are copied. I have to know the path of this folder , but folder name is random , so I want to fetch the folder name of latest created folder.
What shall i do?
You can do something like this:
import os
directory = '.' # current dir
folders = os.walk(directory).next()[1]
creation_times = [(folder, os.path.getctime(folder)) for folder in folders]
creation_times.sort(key=lambda x: x[1]) # sort by creation time
Then you can pick the last element of this list:
most_recent = creation_times[-1][0]
Its a bad practice to fetch the latest folder created because some other application may create a folder right before you fetch the latest folder, but if you really want to try it out the code what elyase has demonstrated is fine.