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Looping for files in a directory, but cannot open the files in the directory because it doesn't exist

I was trying to read data from a bunch of textfiles in a directory, but getting an error while opening the file
import os
fileList = os.listdir("Desktop/SLUI")
for txtName in fileList:
#Open the textfile
UIname=str(txtName)
userDTL=open(UIname,'r')
if userDTL.mode=='r':
line=userDTL.readlines()
string1=line[0]
string2=line[1]
string3=line[2]
UserDTL.close()
print(string1)
Here is the error when I try to run this code via cmd.exe
File "C:\Users\*****\Desktop\programName.py", line 24, in <module>
userDTL=open(UIname,'r')
FileNotFoundError: [Errno 2] No such file or directory: 'file1.txt'

It's because os.listdir only displays the name of the files, and to open them you need the whole path.
You need to redefine UIname as such:
UIname=os.path.join("Desktop/SLUI",txtName) # I don't think you need the string conversion.
os.path.join() will properly join two (or more) bits of paths, whichever OS you use.
I'm not really familiar with how Python works on Windows, so you might need to replace "Desktop/SLUI" by the appropriate path to your desktop (C:\Users*****\Desktop).

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 input multiple files from a directory

First and foremost, I am recently new to Unix and I have tried to find a solution to my question online, but I could not find a solution.
So I am running Python through my Unix terminal, and I have a program that parses xml files and inputs the results into a .dat file.
My program works, but I have to input every single xml file (which number over 50) individually.
For example:
clamshell: python3 my_parser2.py 'items-0.xml' 'items-1.xml' 'items-2.xml' 'items-3.xml' .....`
So I was wondering if it is possible to read from the directory, which contains all of my files into my program? Rather than typing all the xml file names individually and running the program that way.
Any help on this is greatly appreciated.

import glob
listOffiles = glob.glob('directory/*.xml')

The shell itself can expand wildcards so, if you don't care about the order of the input files, just use:
python3 my_parser2.py items-*.xml
If the numeric order is important (you want 0..9, 10-99 and so on in that order, you may have to adjust the wildcard arguments slightly to guarantee this, such as with:
python3 my_parser2.py items-[0-9].xml items-[1-9][0-9].xml items-[1-9][0-9][0-9].xml

python3 my_parser2.py *.xml should work.

Other than the command line option, you could just use glob from within your script and bypass the need for command arguments:
import glob
filenames = glob.glob("*.xml")
This will return all .xml files (as filenames) in the directory from which you are running the script.
Then, if needed you can simply iterate through all the files with a basic loop:
for file in filenames:
with open(file, 'r') as f:
# do stuff to f.

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

simple python file search and record to csv

import os, csv
f=open("C:\\tempa\\file.csv", 'wb') #write to an existing blank csv file
w=csv.writer(f)
for path, dirs, files, in os.walk("C:\\tempa"):
for filename in files:
w.writerow([filename])
running win7 64bit latest python, using anaconda spyder, pyscripter issue persists regardless of the ide.
I have some media in folders in tempa jpg, pdf and mov... and I wanted to get a file list of all of them, and the code works but it stops without any issue at row 113, nothing special with the file it stops on, no weird characters.
I could have 3 blocks of code one for each folder to work around this weird bug. but it shouldnt have an issue.. the folders are all in the root folder without going too deep in sub folders:
C:\
-tempa
-jpg
-pdf
-mov
I have heard there are issues with os.walk but I didn't expext anything weird like this.
Maybe I need an f=close?

You were examining the file before it was fully closed. (f won't be closed until, at least, it is no longer referenced by any in-scope variable name.) If you examine a file before it is closed, you may not see the final, partial, data buffer.
Use the file object's context manager to ensure that the file is flushed and closed in all cases:
import os, csv
with open("C:\\tempa\\file.csv", 'wb') as f: #write to an existing blank csv file
w=csv.writer(f)
for path, dirs, files, in os.walk("C:\\tempa"):
for filename in files:
w.writerow([filename])
# Now no need for f.close()