I'm new to python and the following piece of code is driving me crazy. It lists the files in a directory and for each file does some stuff. I get a IOError: [Errno2] No such file or directory: my_file_that_is_actually_there!
def loadFile(aFile):
f_gz = gzip.open(aFile, 'rb')
data = f_gz.read()
#do some stuff...
f_gz.close()
return data
def main():
inputFolder = '../myFolder/'
for aFile in os.listdir(inputFolder):
data = loadFile(aFile)
#do some more stuff
The file exists and it's not corrupted. I do not understand how it's possible that python first finds the file when it checks the content of myFolder, and then it cannot find itanymore... This happens on the second iteration of my for loop only with any files.
NOTE: Why does this exception happen ONLY at the second iteration of the loop?? The first file in the folder is found and opened without any issues...
This is because open receives the local name (returned from os.listdir). It doesn't know that you mean that it should look in ../myFolder. So it receives a relative path and applies it to the current dir. To fix it, try:
data = loadFile(os.path.join(inputFolder, aFile))
Related
I have the following code to get messages from group:
getmessage = client.get_messages(dialog, limit=1000)
for message in getmessage:
try:
if message.media == None:
print("message")
continue
else:
print("Media**********")
client.download_media(message)
The code above write the media.
I need to know the file name/file type before I write the file, How can I get it?
You can use a filename of your choosing to ensure that said filename will be used:
filename = 'some-file'
filename = client.download_media(message, filename)
Then filename will be some-file with the correct extension.
Otherwise, Telethon will generate a file name for the file (it will try the original name, and if it exists, append (n) to avoid overwriting existing files), so you can't really know where it will be saved beforehand (but the method does return the final filename).
Based on *.blend file I have to write a script that gets informations about objects and saves them to json. This script can be opened in Blender, or running. The launch should save the json file with the data in the current directory.
So I created this:
import bpy
import json
objects = bpy.context.scene.objects
data = {}
for ob in objects:
item = {}
item['location'] = ob.location
if ob.name == 'Cube':
item['material_name'] = ob.active_material.name
data[ob.name] = item
elif ob.name == 'Camera':
item['camera_type'] = ob.data.type
data[ob.name] = item
elif ob.name == 'Lamp':
item['lamp_type'] = ob.data.type
data[ob.name] = item
with open('scene_objects.json', 'w') as json_file:
json.dump(data, json_file)
However, when I run the script in Blender, I received the following error:
PermissionError: [Errno 13] Permission denied: 'scene_objects.json'
I'm a beginner in using Blender so maybe it's impossible to write to file from Blender? However, if I can do it, I am asking for advice on how?
Your issue isn't with blender, the OS is preventing the creation (or writability) of the file based on file system permissions.
The line -
with open('scene_objects.json', 'w') as json_file:
will create a new file (or open existing) in the current working directory. When running blender that could be one of several options, depending on which OS you are using. It is also possible that starting blender from a GUI can leave you without a valid CWD, or a temporary dir that a user does not have permission to write to.
You can use os.chdir() to change the CWD to one that you know exists and that you can write to. You can also specify a full path instead of just a filename.
I am writing a script in which I want the user to input a file path that will then be parsed by the rest of the script as a way to know which file to work on.
For brevity here is the beginning of the code which is where I have a problem:
### import modules
import pyabf
#### open file and extract basic data
file_path = input('file path?')
abf = pyabf.ABF(file_path)
data = abf.data
If I run the script in this form I get the following error:
File "/Users/XXX/anaconda3/envs/blah_blah/lib/python3.6/site-packages/pyabf/abf.py", line 65, in init
raise ValueError("ABF file does not exist: %s" % self.abfFilePath)
ValueError: ABF file does not exist
And here the portion of the script that gives me this error:
# clean-up file paths and filenames, then open the file
self.abfFilePath = os.path.abspath(abfFilePath)
if not os.path.exists(self.abfFilePath):
raise ValueError("ABF file does not exist: %s" % self.abfFilePath)
self.abfID = os.path.splitext(os.path.basename(self.abfFilePath))[0]
log.debug(self.__repr__())
If I run the lines of the code individually in the console everything works:
abf = pyabf.ABF(file_path) # same path as before works to open the file
data = abf.data # then manages to extract the correct numpy array.
What happens to the path when it is fed into as user input as opposed to being given directly into the argument where it will be used? I have tried to type the path in different ways with or without '' or () but I can't get it to be recognised by the pyabf.ABF script.
I had a look at the os.path info and from what I can understand the os.path.abspath(abfFilePath) line, which is bugging now, should just return an absolute path name. I'm sure it's probably something simple and obvious that I just don't get.
Hopefully someone here can help.
Thanks!
I have this code which along with the rest of it runs on a single file in a folder.
I want to try and run this code on 11 files in the folder. I have to pass parameters to the whole script via an .sh script which I've written.
I've searched on here and found various solutions which have not worked.
def get_m3u_name():
m3u_name = ""
dirs = os.listdir("/tmp/")
for m3u_file in dirs:
if m3u_file.endswith(".m3u") or m3u_file.endswith(".xml"):
m3u_name = m3u_file
return m3u_name
def remove_line(filename, what):
if os.path.isfile(filename):
file_read = open(filename).readlines()
file_write = open(filename, 'w')
for line in file_read:
if what not in line:
file_write.write(line)
file_write.close()
m3ufile = get_m3u_name()
I have tried a different method of deleting the file just processed then looping the script to run again on the next file as I can do this manually but when I use
os.remove(m3ufile)
I get file not found either method of improving my code would be of great help to me. I'm just a newbie at this but pointing me in the right direction would be of great help.
In my code, user uploads file which is saved on server and read using the server path. I'm trying to delete the file from that path after I'm done reading it. But it gives me following error instead:
An error occurred while reading file. [WinError 32] The process cannot access the file because it is being used by another process
I'm reading file using with, and I've tried f.close() and also f.closed but its the same error every time.
This is my code:
f = open(filePath)
with f:
line = f.readline().strip()
tempLst = line.split(fileSeparator)
if(len(lstHeader) != len(tempLst)):
headerErrorMsg = "invalid headers"
hjsonObj["Line No."] = 1
hjsonObj["Error Detail"] = headerErrorMsg
data['lstErrorData'].append(hjsonObj)
data["status"] = True
f.closed
return data
f.closed
after this code I call the remove function:
os.remove(filePath)
Edit: using with open(filePath) as f: and then trying to remove the file gives the same error.
Instead of:
f.closed
You need to say:
f.close()
closed is just a boolean property on the file object to indicate if the file is actually closed.
close() is method on the file object that actually closes the file.
Side note: attempting a file delete after closing a file handle is not 100% reliable. The file might still be getting scanned by the virus scanner or indexer. Or some other system hook is holding on to the file reference, etc... If the delete fails, wait a second and try again.
Use below code:
import os
os.startfile('your_file.py')
To delete after completion:
os.remove('your_file.py')
This
import os
path = 'path/to/file'
with open(path) as f:
for l in f:
print l,
os.remove(path)
should work, with statement will automatically close the file after the nested block of code
if it fails, File could be in use by some external factor. you can use Redo pattern.
while True:
try:
os.remove(path)
break
except:
time.sleep(1)
There is probably an application that is opening the file; check and close the application before executing your code:
os.remove(file_path)
Delete files that are not used by another application.