We Keep Coding

Need help to open a file in python

I don't know what's wrong here, all I'm trying to do is to open this file, but it says it can't find such a file or directory, however as I have highlighted on the side, the file is right there. I just want to open it. I have opened files before but never encountered this. I must have missed something, I checked online, and seems like my syntax is correct, but I don't know.
I get the same error when I try with "alphabetical_words" which is just a text file.

When open() function receives a relative path, it looks for that file relative to the current working directory. In other words: relative to the current directory from where the script is run. This can be any arbitrary location.
I guess what you want to do is look for alphabetical.csv file relative to the script location. To do that use the following formula:
from pathlib import Path
# Get directory of this script
THIS_DIR = Path(__file__).absolute().parent
# Get path of CSV file relative to the directory of this script
CSV_PATH = THIS_DIR.joinpath("alphabetical.csv")
# Open the CSV file using a with-block
with CSV_PATH.open(encoding="utf-8") as csvfile:
pass # Do stuff with opened file

You need to import csv. Then you can open the file as a csv file. For example,
with open ("alphabetical.csv", "r") as csvfile:
Then you can use the file.
Make sure the file is in your cwd.

Ignore "temp" files?

I've had a script for a while that has been running without issues however recently had a "hitch" with a temporary file that was within a directory.
The file in question started with '~$' on a windows PC so the script was erroring out on this file as it is not a proper DOCX file. The file in question was not open and occurred after being transferred of a network drive onto an external hard drive. Checking the destination drive (with hidden files on etc) did not show this file either.
I have attempted a quick fix off:
for (dirpath,dirnames,filenames) in os.walk('.'):
for filename in filenames:
if filename.endswith('.docx'):
filesList.append(os.path.join(dirpath,filename))
for file in filesList:
if file.startswith('~$'):
pass
else:
<rest of script>
However the script appears to be ignoring this to proceed then error out again, as the file is not "valid".
Does anyone know either why this isn't working or a quick solution to get it to ignore any files that are like this? I would attempt a if exists, however the file technically does exist so this wouldn't work either.
Sorry if its a bit stupid, but I am a bit stumped as to A. why its there and B. how to code around it.

In the second code block, your variable file contains the whole file path, not just the file name.
Instead skip the "bad" files in your first block instead of appending to the list:
for (dirpath,dirnames,filenames) in os.walk('.'):
for filename in filenames:
if filename.endswith('.docx'):
if not filename.startswith('~$'):
filesList.append(os.path.join(dirpath,filename))
The other option would be to check os.path.basename(file) in your second code block.

Iterating over .wav files in subdirectories of parent directory

Cheers everybody,
I need help with something in python 3.6 exactly. So i have structure of data like this:
|main directory
| |subdirectory's(plural)
| | |.wav files
I'm currently working from a directory where main directory is placed so I don't need to specify paths before that. So firstly I wanna iterate over my main directory and find all subdirectorys. Then in each of them I wanna find the .wav files, and when done with processing them I wanna go to next subdirectory and so on until all of them are opened, and all .wav files are processed. Exactly what I wanna do with those .wav files is input them in my program, process them so i can convert them to numpy arrays, and then I convert that numpy array into some other object (working with tensorflow to be exact, and wanna convert to TF object). I wrote about the whole process if anybody has any fast advices on doing that too so why not.
I tried doing it with for loops like:
for subdirectorys in open(data_path, "r"):
for files in subdirectorys:
#doing some processing stuff with the file
The problem is that it always raises error 13, Permission denied showing on that data_path I gave him but when I go to properties there it seems okay and all permissions are fine.
I tried some other ways like with os.open or i replaced for loop with:
with open(data_path, "r") as data:
and it always raises permission denied error.
os.walk works in some way but it's not what I need, and when i tried to modify it id didn't give errors but it also didnt do anything.
Just to say I'm not any pro programmer in python so I may be missing an obvious thing but ehh, I'm here to ask and learn. I also saw a lot of similiar questions but they mainly focus on .txt files and not specificaly in my case so I need to ask it here.
Anyway thanks for help in advance.

Edit: If you want an example for glob (more sane), here it is:
from pathlib import Path
# The pattern "**" means all subdirectories recursively,
# with "*.wav" meaning all files with any name ending in ".wav".
for file in Path(data_path).glob("**/*.wav"):
if not file.is_file(): # Skip directories
continue
with open(file, "w") as f:
# do stuff
For more info see Path.glob() on the documentation. Glob patterns are a useful thing to know.
Previous answer:
Try using either glob or os.walk(). Here is an example for os.walk().
from os import walk, path
# Recursively walk the directory data_path
for root, _, files in walk(data_path):
# files is a list of files in the current root, so iterate them
for file in files:
# Skip the file if it is not *.wav
if not file.endswith(".wav"):
continue
# os.path.join() will create the path for the file
file = path.join(root, files)
# Do what you need with the file
# You can also use block context to open the files like this
with open(file, "w") as f: # "w" means permission to write. If reading, use "r"
# Do stuff
Note that you may be confused about what open() does. It opens a file for reading, writing, and appending. Directories are not files, and therefore cannot be opened.
I suggest that you Google for documentation and do more reading about the functions used. The documentation will help more than I can.
Another good answer explaining in more detail can be seen here.

import glob
import os
main = '/main_wavs'
wavs = [w for w in glob.glob(os.path.join(main, '*/*.wav')) if os.path.isfile(w)]
In terms of permissions on a path A/B/C... A, B and C must all be accessible. For files that means read permission. For directories, it means read and execute permissions (listing contents).

How to exclude ".DS_Store" path when compressing files in Python

I have a python script that compresses specific files into a zip file. However I have noticed that a file ".DS_Store" is produced within this zip file. Is there a way I can remove this from the zip file or avoid it being created in the first place in my python script. From what I have found online I think on a windows machine this hidden file appears as "macosx" file.
I've tested the zip file with and without the ".DS_Store" hidden file (I manually deleted it). When I remove it, the zip file is able to be processed correctly and when I leave it in, errors are thrown.
This is how I create the zip file in my python script:
#Create zip file of all necessary files
zipf = zipfile.ZipFile(new_path+zip_file_name, 'w', zipfile.ZIP_DEFLATED)
create_zip(new_path,zipf)
zipf.close()
Any advice how to approach removing this hidden file would be appreciated.

Your code uses a function, create_zip, but you haven't shared the code of that function. Presumably, it loops through the contents of a directory and calls the .write method of the ZipFile instance in order to write each file into the archive. If this is the case, just add some logic to that function to exclude any files called .DS_Store.
def create_zip(path, zipfile):
files = os.listdir(path)
for file in files:
if file != '.DS_Store':
zipfile.write(file)

taking data from files which are in folder

How do I get the data from multiple txt files that placed in a specific folder. I started with this could not fix. It gives an error like 'No such file or directory: '.idea' (??)
(Let's say I have an A folder and in that, there are x.txt, y.txt, z.txt and so on. I am trying to get and print the information from all the files x,y,z)
def find_get(folder):
for file in os.listdir(folder):
f = open(file, 'r')
for data in open(file, 'r'):
print data
find_get('filex')
Thanks.

If you just want to print each line:
import glob
import os
def find_get(path):
for f in glob.glob(os.path.join(path,"*.txt")):
with open(os.path.join(path, f)) as data:
for line in data:
print(line)
glob will find only your .txt files in the specified path.
Your error comes from not joining the path to the filename, unless the file was in the same directory you were running the code from python would not be able to find the file without the full path. Another issue is you seem to have a directory .idea which would also give you an error when trying to open it as a file. This also presumes you actually have permissions to read the files in the directory.
If your files were larger I would avoid reading all into memory and/or storing the full content.

First of all make sure you add the folder name to the file name, so you can find the file relative to where the script is executed.
To do so you want to use os.path.join, which as it's name suggests - joins paths. So, using a generator:
def find_get(folder):
for filename in os.listdir(folder):
relative_file_path = os.path.join(folder, filename)
with open(relative_file_path) as f:
# read() gives the entire data from the file
yield f.read()
# this consumes the generator to a list
files_data = list(find_get('filex'))
See what we got in the list that consumed the generator:
print files_data
It may be more convenient to produce tuples which can be used to construct a dict:
def find_get(folder):
for filename in os.listdir(folder):
relative_file_path = os.path.join(folder, filename)
with open(relative_file_path) as f:
# read() gives the entire data from the file
yield (relative_file_path, f.read(), )
# this consumes the generator to a list
files_data = dict(find_get('filex'))
You will now have a mapping from the file's name to it's content.
Also, take a look at the answer by #Padraic Cunningham . He brought up the glob module which is suitable in this case.

The error you're facing is simple: listdir returns filenames, not full pathnames. To turn them into pathnames you can access from your current working directory, you have to join them to the directory path:
for filename in os.listdir(directory):
pathname = os.path.join(directory, filename)
with open(pathname) as f:
# do stuff
So, in your case, there's a file named .idea in the folder directory, but you're trying to open a file named .idea in the current working directory, and there is no such file.
There are at least four other potential problems with your code that you also need to think about and possibly fix after this one:
You don't handle errors. There are many very common reasons you may not be able to open and read a file--it may be a directory, you may not have read access, it may be exclusively locked, it may have been moved since your listdir, etc. And those aren't logic errors in your code or user errors in specifying the wrong directory, they're part of the normal flow of events, so your code should handle them, not just die. Which means you need a try statement.
You don't do anything with the files but print out every line. Basically, this is like running cat folder/* from the shell. Is that what you want? If not, you have to figure out what you want and write the corresponding code.
You open the same file twice in a row, without closing in between. At best this is wasteful, at worst it will mean your code doesn't run on any system where opens are exclusive by default. (Are there such systems? Unless you know the answer to that is "no", you should assume there are.)
You don't close your files. Sure, the garbage collector will get to them eventually--and if you're using CPython and know how it works, you can even prove the maximum number of open file handles that your code can accumulate is fixed and pretty small. But why rely on that? Just use a with statement, or call close.
However, none of those problems are related to your current error. So, while you have to fix them too, don't expect fixing one of them to make the first problem go away.

Full variant:
import os
def find_get(path):
files = {}
for file in os.listdir(path):
if os.path.isfile(os.path.join(path,file)):
with open(os.path.join(path,file), "r") as data:
files[file] = data.read()
return files
print(find_get("filex"))
Output:
{'1.txt': 'dsad', '2.txt': 'fsdfs'}
After the you could generate one file from that content, etc.
Key-thing:
os.listdir return a list of files without full path, so you need to concatenate initial path with fount item to operate.
there could be ideally used dicts :)
os.listdir return files and folders, so you need to check if list item is really file

You should check if the file is actually file and not a folder, since you can't open folders for reading. Also, you can't just open a relative path file, since it is under a folder, so you should get the correct path with os.path.join. Check below:
import os
def find_get(folder):
for file in os.listdir(folder):
if not os.path.isfile(file):
continue # skip other directories
f = open(os.path.join(folder, file), 'r')
for line in f:
print line