pathlib

问题 On Python3.8, I'm trying to use pathlib to concatenate a string to a UNC path that's on a remote computer's C drive. It's weirdly inconsistent. For example: >>> remote = Path("\\\\remote\\", "C$\\Some\\Path") >>> remote WindowsPath('//remote//C$/Some/Path') >>> remote2 = Path(remote, "More") >>> remote2 WindowsPath('/remote/C$/Some/Path/More') Notice how the initial // is turned into / ? Put the initial path in one line though, and everything is fine: >>> remote = Path("\\\\remote\\C$\\Some\

问题 Historically I have always used the following for reading files in python : with open("file", "r") as f: for line in f: # do thing to line Is this still the recommend approach? Are there any drawbacks to using the following: from pathlib import Path path = Path("file") for line in path.open(): # do thing to line Most of the references I found are using the with keyword for opening files for the convenience of not having to explicitly close the file. Is this applicable for the iterator

问题 The the new Path.glob from pathlib seems to behave differently from the old glob.glob when the glob pattern ends in a slash it seems. In [1]: from pathlib import Path In [2]: from glob import glob In [3]: glob('webroot/*/') Out[3]: ['webroot/2017-06-07/'] In [4]: list(Path().glob('webroot/*/')) Out[4]: [PosixPath('webroot/.keep'), PosixPath('webroot/2017-06-07'), PosixPath('webroot/matches.2017-06-07.json')] Is that by design, some compatibility issue I haven’t encountered? And is there a way

问题 Using Python 3.6 on Windows 7 x64, the path "C:" seems identical to an empty path for Path.resolve() : 'Empty' paths are 'current working directory' cwd() : >>> from pathlib import Path >>> Path().resolve() WindowsPath('C:/Users/me') >>> Path(r"").resolve() WindowsPath('C:/Users/me') >>> Path.cwd().resolve() WindowsPath('C:/Users/me') A single letter is interpreted as a folder name: >>> Path(r"C").resolve() WindowsPath('C:/Users/me/C') A full blown drive-letter + colon + backslash points to

问题 Using Python 3.6 on Windows 7 x64, the path "C:" seems identical to an empty path for Path.resolve() : 'Empty' paths are 'current working directory' cwd() : >>> from pathlib import Path >>> Path().resolve() WindowsPath('C:/Users/me') >>> Path(r"").resolve() WindowsPath('C:/Users/me') >>> Path.cwd().resolve() WindowsPath('C:/Users/me') A single letter is interpreted as a folder name: >>> Path(r"C").resolve() WindowsPath('C:/Users/me/C') A full blown drive-letter + colon + backslash points to

问题 Is there a Pathlib equivalent of os.access() ? Without Pathlib the code would look like this: import os os.access('my_folder', os.R_OK) # check if script has read access to folder However, in my code I'm dealing with Pathlib paths, so I would need to do this (this is just an example): # Python 3.5+ from pathlib import Path import os # get path ~/home/github if on Linux my_folder_pathlib = Path.home() / "github" os.access(str(my_folder_pathlib), os.R_OK) The casting to str() is kinda ugly. I

问题 I have a directory I'd like to print out with a trailing slash: my_path = pathlib.Path('abc/def') Is there a nicer way of doing this than os.path.join(str(my_path), '') ? 回答1: No, you didn't miss anything. By design, pathlib strips trailing slashes, and provides no way to display paths with trailing slashes. This has annoyed several people, as mentioned in the bug tracker: pathlib strips trailing slash. A compact way to add slashes in Python 3.6 is to use an f-string, eg f'{some_path}/' or f'

来源： https://stackoverflow.com/questions/54152653/renaming-file-extension-using-pathlib-python-3

问题 When I have a pd.DataFrame with paths, I end up doing a lot of .map(lambda path: Path(path).{method_name} , or apply(axis=1) e.g: ( pd.DataFrame({'base_dir': ['dir_A', 'dir_B'], 'file_name': ['file_0', 'file_1']}) .assign(full_path=lambda df: df.apply(lambda row: Path(row.base_dir) / row.file_name, axis=1)) ) base_dir file_name full_path 0 dir_A file_0 dir_A/file_0 1 dir_B file_1 dir_B/file_1 It seems odd to me especially because pathlib does implement / so that something like df.base_dir /

问题 I'm setting up a script and I need to use some functions from fast-ai package. The fact is that I'm on Windows and when I define my paths, the function from fast-ai named load_learner can't load the model. I've tried to change the function into the package as: state = pickle.load(open(str(path) + '/' + str(fname), 'rb')) instead of: state = pickle.load(open(path/fname, 'rb')) but I obtain this error: File "lib\site-packages\fastai\basic_train.py", line 462, in load_learner state = pickle.load