I'm trying to copy dir1 to dir2. Dir1 contains sub-folders and files. In the moment of copy I'm creating url like this C:/dirA/dir1 and C:/dirB/dir2. As you see all slashes are forwarded. When run I get this error
No such file or directory path C:/dirB/dir2\\folder1\\file.txt
As you see sub-folder and file have backslashes. I really do not know how to change that backslashes because when I create a paths I don't know the names of sub-folders/files. I can't post entire code because it's huge.
To copy I use distutils.dir_util.copy_tree.
It looks like you can use os.path.normpath on parts of your path to normalize them for current OS before you concatenate, on Windows it'll use correct slashes.
Related
I have a path to a file and a path to a directory. The file is supposed to be somewhere (multiple levels) in that directory. I want to compare the beginning of the path to the file with the path to the directory. So what I basically do is:
if file_path.startswith(directory_path):
do_something()
Both paths are strings. Unfortunately, my path to the file includes ".." and ".". So it looks something like this: /home/user/documents/folder/../pictures/house.jpg. As the other path does not contain those dots, the comparison fails, obviously. Is there a way in python to remove those spots from the string? I thought of using path.join() from the os module, which did not work.
Thanks a lot for any help :)
Is there a way in python to remove those spots from the string? I thought of using path.join() from the os module, which did not work. Thanks a lot for any help :)
os.path.abspath will normalise the path and absolutify it. Alternatively, pathlib.Path.resolve().
I'm trying to create an arcpy tool for one of the teams, and I have everything
working.
But when I redirect the path names to be where their data is saved and where they want the outputs to go to, I get errors like ERROR:00732 or ERROR:00210 and its because of the folder directories having spaces in their names.
is there a work around for this?
You Have to put the path between doble quotes.
For example:
Path = “path/to/folder”
I'm writing some python code to generate the relative path. Situation need to be considered:
Under the same folder. I want "." or ".\", both of tham are ok for me.
Other folder. I want like ".\xxx\" and "..\xxx\xxx\"
os.path.relpath() will generate the relative path, but without .\ at the beginning and \ in the end. We can add \ in the end by using os.path.join(dirname, ""). But i can't figure out how to add ".\" at the beginning without impacting the first case when they are under the same folder and "..\xxx\xxx\".
It will give you relative path
import os
dir = os.path.dirname(__file__)
filename = os.path.join(dir,'Path')
The relpath() function will produce the ".." syntax given the appropriate base to start from (second parameter). For instance, supposing you were writing something like a script generator that produces code using relative paths, if the working directory is as the second parameter to relpath() as below indicates, and you want to reference in your code another file in your project under a directory one level up and two deep, you'll get "../blah/blah".. In the case where you want to prefix paths in the same folder, you can simply do a join with ".". That will produce a path with the correct OS specific separator.
print(os.path.relpath("/foo/bar/blah/blah", "/foo/bar/baz"))
>>> ../blah/blah
print(os.path.join('.', 'blah'))
>>> ./blah
I am a very novice coder, and Python is my first (and, practically speaking, only) language. I am charged as part of a research job with manipulating a collection of data analysis scripts, first by getting them to run on my computer. I was able to do this, essentially by removing all lines of coding identifying paths, and running the scripts through a Jupyter terminal opened in the directory where the relevant modules and CSV files live so the script knows where to look (I know that Python defaults to the location of the terminal).
Here are the particular blocks of code whose function I don't understand
import sys
sys.path.append('C:\Users\Ben\Documents\TRACMIP_Project\mymodules/')
import altdata as altdata
I have replaced the pathname in the original code with the path name leading to the directory where the module is; the file containing all the CSV files that end up being referenced here is also in mymodules.
This works depending on where I open the terminal, but the only way I can get it to work consistently is by opening the terminal in mymodules, which is fine for now but won't work when I need to work by accessing the server remotely. I need to understand better precisely what is being done here, and how it relates to the location of the terminal (all the documentation I've found is overly technical for my knowledge level).
Here is another segment I don't understand
import os.path
csvfile = 'csv/' + model +'_' + exp + '.csv'
if os.path.isfile(csvfile): # csv file exists
hcsvfile = open(csvfile )
I get here that it's looking for the CSV file, but I'm not sure how. I'm also not sure why then on some occasions depending on where I open the terminal it's able to find the module but not the CSV files.
I would love an explanation of what I've presented, but more generally I would like information (or a link to information) explaining paths and how they work in scripts in modules, as well as what are ways of manipulating them. Thanks.
sys.path
This is simple list of directories where python will look for modules and packages (.py and dirs with __init__.py file, look at modules tutorial). Extending this list will allow you to load modules (custom libs, etc.) from non default locations (usually you need to change it in runtime, for static dirs you can modify startup script to add needed enviroment variables).
os.path
This module implements some useful functions on pathnames.
... and allows you to find out if file exists, is it link, dir, etc.
Why you failed loading *.csv?
Because sys.path responsible for module loading and only for this. When you use relative path:
csvfile = 'csv/' + model +'_' + exp + '.csv'
open() will look in current working directory
file is either a string or bytes object giving the pathname (absolute or relative to the current working directory)...
You need to use absolute paths by constucting them with os.path module.
I agree with cdarke's comment that you are probably running into an issue with backslashes. Replacing the line with:
sys.path.append(r'C:\Users\Ben\Documents\TRACMIP_Project\mymodules')
will likely solve your problem. Details below.
In general, Python treats paths as if they're relative to the current directory (where your terminal is running). When you feed it an absolute path-- which is a path that includes the root directory, like the C:\ in C:\Users\Ben\Documents\TRACMIP_Project\mymodules-- then Python doesn't care about the working directory anymore, it just looks where you tell it to look.
Backslashes are used to make special characters within strings, such as line breaks (\n) and tabs (\t). The snag you've hit is that Python paths are strings first, paths second. So the \U, \B, \D, \T and \m in your path are getting misinterpreted as special characters and messing up Python's path interpretation. If you prefix the string with 'r', Python will ignore the special characters meaning of the backslash and just interpret it as a literal backslash (what you want).
The reason it still works if you run the script from the mymodules directory is because Python automatically looks in the working directory for files when asked. sys.path.append(path) is telling the computer to include that directory when it looks for commands, so that you can use files in that directory no matter where you're running the script. The faulty path will still get added, but its meaningless. There is no directory where you point it, so there's nothing to find there.
As for path manipulation in general, the "safest" way is to use the function in os.path, which are platform-independent and will give the correct output whether you're working in a Windows or a Unix environment (usually).
EDIT: Forgot to cover the second part. Since Python paths are strings, you can build them using string operations. That's what is happening with the line
csvfile = 'csv/' + model +'_' + exp + '.csv'
Presumably model and exp are strings that appear in the filenames in the csv/ folder. With model = "foo" and exp = "bar", you'd get csv/foo_bar.csv which is a relative path to a file (that is, relative to your working directory). The code makes sure a file actually exists at that path and then opens it. Assuming the csv/ folder is in the same path as you added in sys.path.append, this path should work regardless of where you run the file, but I'm not 100% certain on that. EDIT: outoftime pointed out that sys.path.append only works for modules, not opening files, so you'll need to either expand csv/ into an absolute path or always run in its parent directory.
Also, I think Python is smart enough to not care about the direction of slashes in paths, but you should probably not mix them. All backslashes or all forward slashes only. os.path.join will normalize them for you. I'd probably change the line to
csvfile = os.path.join('csv\', model + '_' + exp + '.csv')
for consistency's sake.
I have some homework that I am trying to complete. I don't want the answer. I'm just having trouble in starting. The work I have tried is not working at all... Can someone please just provide a push in the right direction. I am trying to learn but after trying and trying I need some help.
I know I can you os.path.basename() to get the basename and then add it to the file name but I can't get it together.
Here is the assignment
In this project, write a function that takes a directory path and creates an archive of the directory only. For example, if the same path were used as in the example ("c:\\xxxx\\Archives\\archive_me"), the zipfile would contain archive_me\\groucho, archive_me\\harpo and archive_me\\chico.
The base directory (archive_me in the example above) is the final element of the input, and all paths recorded in the zipfile should start with the base directory.
If the directory contains sub-directories, the sub-directory names and any files in the sub-directories should not be included. (Hint: You can use isfile() to determine if a filename represents a regular file and not a directory.)
Thanks again any direction would be great.
It would help to know what you tried yourself, so I'm only giving a few pointers to methods in the standard libraries:
os.listdir to get the a list of files and folders under a given directory (beware, it returns only the file/folder name, not the full path!)
os.path.isfile as mentioned in the assignment to check if a given path represents a file or a folder
os.path.isdir, the opposite of os.path.isfile (thanks inspectorG4adget)
os.path.join to join a filename with the basedir without having to worry about slashes and delimiters
ZipFile for handling, well, zip files
zipFile.write to write the files found to the zip
I'm not sure you'll need all of those, but it doesn't hurt knowing they exist.