I want to open a file to write to.
with open(oname.text , 'w') as f:
and now I want to write the files in a folder "Playlist"
I know that I have to use os.path But I do not know how to use it
ty all
path = os.path.join('Playlist', oname.text)
with open(path, 'w') as f:
...
If you're not sure if the 'Playlist' subdir of the current directory already exists, prefix that with:
if not os.path.isdir('Playlist'):
if os.path.exists('Playlist'):
raise RuntimeError('Playlist exists and is a file, now what?!')
os.mkdir('Playlist')
This raises an exception if 'Playlist' does exist but as a file, not a directory -- handle this anomalous case as you wish, but unless you remove or rename the file, you're not going to be able to have it as a directory as well!
Use os.makedirs instead of os.mkdir if the path you desire has multiple levels of directories, e.g Play/List/Whatever (you could use it anyway just in case).
You could change the current working directory using os.chdir function.
os.chdir('Playlist')
with open(oname.text , 'w') as f:
...
Use with statement and os.path.join method
dir_path = "/home/Playlist"
file_path = os.path.join('dir_path, "oname.txt")
content = """ Some content..."""
with open(file_path, 'wb') as fp:
fp.write(content)
OR
fp = open(file_path, "wb"):
fp.write(content)
fp.close()
Related
I tried using open but it gives an error that the folder doesn't exist, which makes no sense since this is a command to create a folder, not read one. I saw Automatically creating directories with file output, but there is an error saying this is a Errno 30: Read only system: "/folder". Does anyone know how to avoid Error 30?
My code so far:
import os
filename = "/folder/y.txt"
os.makedirs(os.path.dirname(filename), exist_ok=True)
with open(filename, "w") as f:
f.write("FOOBAR")
I figured i just shouldn't put the slash behind the "folder"filename = "folder/y.txt" os.makedirs(os.path.dirname(filename), exist_ok=True) with open(filename, "w") as f: f.write("FOOBAR")
I want to do the following:
1) Ask the user for input for a file path they wish a directory listing for.
2) Take this file path and enter the results, in a list, in a text file in the directory they input NOT the current directory.
I am very nearly there but the last step is that I can't seem to save the file to the directory the user has input only the current directory. I have set out the current code below (which works for the current directory). I have tried various variations to try and save it to the directory input by the user but to no avail - any help would be much appreciated.
CODE BELOW
import os
filenames = os.path.join(input('Please enter your file path: '))
with open ("files.txt", "w") as a:
for path, subdirs, files in os.walk(str(filenames)):
for filename in files:
f = os.path.join(path, filename)
a.write(str(f) + os.linesep)
I came across this link https://cmdlinetips.com/2012/09/three-ways-to-write-text-to-a-file-in-python/. I think your issue has something to do with you needing to provide the full path name and or the way you are using the close() method.
with open(out_filename, 'w') as out_file:
..
..
.. parsed_line
out_file.write(parsed_line)
You have to alter the with open ("files.txt", "w") as a: statement to not only include the filename, but also the path. This is where you should use os.path.join(). Id could be handy to first check the user input for existence with os.path.exists(filepath).
os.path.join(input(...)) does not really make sense for the input, since it returns a single str, so there are no separate things to be joined.
import os
filepath = input('Please enter your file path: ')
if os.path.exists(filepath):
with open (os.path.join(filepath, "files.txt"), "w") as a:
for path, subdirs, files in os.walk(filepath):
for filename in files:
f = os.path.join(path, filename)
a.write(f + os.linesep)
Notice that your file listing will always include a files.txt-entry, since the file is created before os.walk() gets the file list.
As ShadowRanger kindly points out, this LBYL (look before you leap) approach is unsafe, since the existence check could pass, although the file system is changed later while the process is running, leading to an exception.
The mentioned EAFP (it's easier to ask for forgiveness than permission) approach would use a try... except block to handle all errors.
This approach could look like this:
import os
filepath = input('Please enter your file path: ')
try:
with open (os.path.join(filepath, "files.txt"), "w") as a:
for path, subdirs, files in os.walk(filepath):
for filename in files:
f = os.path.join(path, filename)
a.write(f + os.linesep)
except:
print("Could not generate directory listing file.")
You should further refine it by catching specific exceptions. The more code is in the try block, the more errors unrelated to the directory reading and file writing are also caught and suppressed.
Move to the selected directory then do things.
Extra tip: In python 2 use raw_input to avoid special chars error like : or \ ( just use input in python 3 )
import os
filenames = raw_input('Please enter your file path: ')
if not os.path.exists(filenames):
print 'BAD PATH'
return
os.chdir(filenames)
with open ("files.txt", "w") as a:
for path, subdirs, files in os.walk('.'):
for filename in files:
f = os.path.join(path, filename)
a.write(str(f) + os.linesep)
I want to open any .txt file in the same directory.
In ruby I can do
File.open("*.txt").each do |line|
puts line
end
In python I can't do this it will give an error
file = open("*.txt","r")
print(file.read())
file.close()
It gives an error invalid argument.
So is there any way around it?
You can directly use the glob module for this
import glob
for file in glob.glob('*.txt'):
with open(file, 'r') as f:
print(f.read())
Use os.listdir to list all files in the current directory.
all_files = os.listdir()
Then, filter the ones which have the extension you are looking for and open each one of them in a loop.
for filename in all_files:
if filename.lower().endswith('.txt'):
with open(filename, 'rt') as f:
f.read()
I am trying to open the file from folder and read it but it's not locating it. I am using Python3
Here is my code:
import os
import glob
prefix_path = "C:/Users/mpotd/Documents/GitHub/Python-Sample-
codes/Mayur_Python_code/Question/wx_data/"
target_path = open('MissingPrcpData.txt', 'w')
file_array = [os.path.abspath(f) for f in os.listdir(prefix_path) if
f.endswith('.txt')]
file_array.sort() # file is sorted list
for f_obj in range(len(file_array)):
file = os.path.abspath(file_array[f_obj])
join_file = os.path.join(prefix_path, file) #whole file path
for filename in file_array:
log = open(filename, 'r')#<---- Error is here
Error: FileNotFoundError: [Errno 2] No such file or directory: 'USC00110072.txt'
You are not giving the full path to a file to the open(), just its name - a relative path.
Non-absolute paths specify locations in relation to current working directory (CWD, see os.getcwd).
You would have to either os.path.join() correct directory path to it, or os.chdir() to the directory that the files reside in.
Also, remember that os.path.abspath() can't deduce the full path to a file just by it's name. It will only prefix its input with the path of the current working directory, if the given path is relative.
Looks like you are forgetting to modify the the file_array list. To fix this, change the first loop to this:
file_array = [os.path.join(prefix_path, name) for name in file_array]
Let me reiterate.
This line in your code:
file_array = [os.path.abspath(f) for f in os.listdir(prefix_path) if f.endswith('.txt')]
is wrong. It will not give you a list with correct absolute paths. What you should've done is:
import os
import glob
prefix_path = ("C:/Users/mpotd/Documents/GitHub/Python-Sample-"
"codes/Mayur_Python_code/Question/wx_data/")
target_path = open('MissingPrcpData.txt', 'w')
file_array = [f for f in os.listdir(prefix_path) if f.endswith('.txt')]
file_array.sort() # file is sorted list
file_array = [os.path.join(prefix_path, name) for name in file_array]
for filename in file_array:
log = open(filename, 'r')
You are using relative path where you should be using an absolute one. It's a good idea to use os.path to work with file paths. Easy fix for your code is:
prefix = os.path.abspath(prefix_path)
file_list = [os.path.join(prefix, f) for f in os.listdir(prefix) if f.endswith('.txt')]
Note that there are some other issues with your code:
In python you can do for thing in things. You did for thing in range(len(things)) it's much less readable and unnecessary.
You should use context managers when you open a file. Read more here.
Trying to extract all the zip files and giving the same name to the folder where all the files are gonna be.
Looping through all the files in the folder and then looping through the lines within those files to write on a different text file.
This is my code so far:
#!usr/bin/env python3
import glob
import os
import zipfile
zip_files = glob.glob('*.zip')
for zip_filename in zip_files:
dir_name = os.path.splitext(zip_filename)[0]
os.mkdir(dir_name)
zip_handler = zipfile.ZipFile(zip_filename, "r")
zip_handler.extractall(dir_name)
path = dir_name
fOut = open("Output.txt", "w")
for filename in os.listdir(path):
for line in filename.read().splitlines():
print(line)
fOut.write(line + "\n")
fOut.close()
This is the error that I encounter:
for line in filename.read().splitlines():
AttributeError: 'str' object has no attribute 'read'
You need to open the file and also join the path to the file, also using splitlines and then adding a newline to each line is a bit redundant:
path = dir_name
with open("Output.txt", "w") as fOut:
for filename in os.listdir(path):
# join filename to path to avoid file not being found
with open(os.path.join(path, filename)):
for line in filename:
fOut.write(line)
You should always use with to open your files as it will close them automatically. If the files are not large you can simply fOut.write(f.read()) and remove the loop.
You also set path = dir_name which means path will be set to whatever the last value of dir_name was in your first loop which may or may not be what you want. You can also use iglob to avoid creating a full list zip_files = glob.iglob('*.zip').