I am splitting a large wordlist by length of the word
i didn't find a different approach for it so i decided to write a script in python for it.
say test.txt has
word
words
i want it to make new text files based on length of line and write the line to it
4.txt
word
5.txt
words
CODE
import os
import sys
basefile = open(sys.argv[1],'rt')
print("Writing.....")
os.mkdir(str(os.path.splitext(sys.argv[1])[0]))
os.chdir(os.path.splitext(sys.argv[1])[0])
#print(basefile)
for line in basefile:
cpyfile=open(str(len(line.strip()))+'.txt',mode = 'a',encoding = 'utf-8')
cpyfile.write(line)
cpyfile.close()
print("Done")
basefile.close()
It works for small files but for larger files it gives out an error after a while
PermissionError: [Errno 13] Permission denied: '10.txt'
or
PermissionError: [Errno 13] Permission denied: '11.txt'
the error file is completely random too and the previous lines written are perfectly okay.
I have tried it on windows using powershell and using gitbash
Any help is appreciated and thanks
I suspect you are running into the issue that Windows does not allow two programs to open the same file at once. I'm not sure what the second program would be. Maybe a virus scanner? Your program works unaltered on Ubuntu using /usr/share/dict/american-english, so I think this may be a Windows thing.
In any case, I think you can solve this by keeping the files open while the program is running.
import os
import sys
basefile = open(sys.argv[1], 'rt')
print("Writing.....")
os.mkdir(str(os.path.splitext(sys.argv[1])[0]))
os.chdir(os.path.splitext(sys.argv[1])[0])
# print(basefile)
files = {}
try:
for line in basefile:
cpyfilename = str(len(line.strip()))+'.txt'
cpyfile = files.get(cpyfilename)
if cpyfile is None:
cpyfile = open(cpyfilename, mode='a', encoding='utf-8')
files[cpyfilename] = cpyfile
cpyfile.write(line)
finally:
for cpyfile in files.values():
# Not strictly necessary because the program is about to end and
# auto-close the files.
cpyfile.close()
print("Done")
basefile.close()
This is my code in pycharm:
#coding=utf-8
import csv
path=r'D:\NLP\cv082_11080.csv'
with open(path) as f: # 采用b的方式处理可以省去很多问题
reader = csv.reader(f)
for row in reader:
print reader
But it always gives this error:
IOError: [Errno 22] invalid mode ('r') or filename: '\xe2\x80\xaaD:\NLP\cv082_11080.csv'
But when I run the code in the Python command line, it works well. So then I lool the setting in pycharm, but the option of file-encoding is right, the project and the IDE coding is both UTF-8.
How do I solve this?
You're getting an error because your path has some invisible characters in it (like the error message suggests):
IOError: [Errno 22] invalid mode ('r') or filename: '\xe2\x80\xaaD:\\NLP\\cv082_11080.csv' in pycharm
^^^^^^^^^^^^
Delete this entire line:
path=r'D:\NLP\cv082_11080.csv'
And type it out by hand again. You probably copy-pasted the path from somewhere and your text editor isn't showing the invisible character.
Those characters are the UTF-8 encoding of \u202a, which is a left-to-right mark.
问题出在编码造成的路径错误。
maybe first line code 没有起到作用
建议#coding=utf-8修改为
# coding:utf-8
注意 # 和 coding 之间有一个空格。
Goal = Open file, encrypt file, write encrypted file.
Trying to use the PyPDF2 module to accomplish this. I have verified theat "input" is a file type object. I have researched this error and it translates to "file not found". I believe that it is linked somehow to the file/file path but am unsure how to debug or troubleshoot. and getting the following error:
Traceback (most recent call last):
File "CommissionSecurity.py", line 52, in <module>
inputStream = PyPDF2.PdfFileReader(input)
File "build\bdist.win-amd64\egg\PyPDF2\pdf.py", line 1065, in __init__
File "build\bdist.win-amd64\egg\PyPDF2\pdf.py", line 1660, in read
IOError: [Errno 22] Invalid argument
Below is the relevant code. I'm not sure how to correct this issue because I'm not really sure what the issue is. Any guidance is appreciated.
for ID in FileDict:
if ID in EmailDict :
path = "C:\\Apps\\CorVu\\DATA\\Reports\\AlliD\\Monthly Commission Reports\\Output\\pdcom1\\"
#print os.listdir(path)
file = os.path.join(path + FileDict[ID])
with open(file, 'rb') as input:
print type(input)
inputStream = PyPDF2.PdfFileReader(input)
output = PyPDF2.PdfFileWriter()
output = inputStream.encrypt(EmailDict[ID][1])
with open(file, 'wb') as outputStream:
output.write(outputStream)
else : continue
I think your problem might be caused by the fact that you use the same filename to both open and write to the file, opening it twice:
with open(file, 'rb') as input :
with open(file, 'wb') as outputStream :
The w mode will truncate the file, thus the second line truncates the input.
I'm not sure what you're intention is, because you can't really try to read from the (beginning) of the file, and at the same time overwrite it. Even if you try to write to the end of the file, you'll have to position the file pointer somewhere.
So create an extra output file that has a different name; you can always rename that output file to your input file after both files are closed, thus overwriting your input file.
Or you could first read the complete file into memory, then write to it:
with open(file, 'rb') as input:
inputStream = PyPDF2.PdfFileReader(input)
output = PyPDF2.PdfFileWriter()
output = input.encrypt(EmailDict[ID][1])
with open(file, 'wb') as outputStream:
output.write(outputStream)
Notes:
you assign inputStream, but never use it
you assign PdfFileWriter() to output, and then assign something else to output in the next line. Hence, you never used the result from the first output = line.
Please check carefully what you're doing, because it feels there are numerous other problems with your code.
Alternatively, here are some other tips that may help:
The documentation suggests that you can also use the filename as first argument to PdfFileReader:
stream – A File object or an object that supports the standard read
and seek methods similar to a File object. Could also be a string
representing a path to a PDF file.
So try:
inputStream = PyPDF2.PdfFileReader(file)
You can also try to set the strict argument to False:
strict (bool) – Determines whether user should be warned of all
problems and also causes some correctable problems to be fatal.
Defaults to True.
For example:
inputStream = PyPDF2.PdfFileReader(file, strict=False)
Using open(file, 'rb') was causing the issue becuase PdfFileReader() does that automagically. I just removed the with statement and that corrected the problem.
with open(file, 'rb') as input:
inputStream = PyPDF2.PdfFileReader(input)
This error raised up because of PDF file is empty.
My PDF file was empty that's why my error was raised up. So First of all i fill my PDF file with some data and Then start reeading it using PyPDF2.PdfFileReader,
And it solved my Problem!!!
Late but, you may be opening an invalid PDF file or an empty file that's named x.pdf and you think it's a PDF file
I am using django. I was trying to compress an icon uploaded by a user to a smaller size using Pythons Image library.
Following is my code:
def resizeImage(icon,ext):
path= os.path.join(settings.SITE_ROOT,'karnadash/static/tempfiles/temp'+ext)
destination = open(path,'wb+')
for chunk in icon.chunks():
destination.write(chunk)
destination.close()
image = Image.open(path)
image= image.resize((50, 50), Image.ANTIALIAS)
image.save(path)
return image
Problem is I am getting an internal server error. The last part of the stack trace is as follows:
line 31, in resizeImage
image.save(path)
File "C:\Python27\lib\site-packages\PIL\Image.py", line 1446, in save
fp = builtins.open(fp, "wb+")
IOError: [Errno 22] invalid mode ('wb') or filename: 'C:/Users/Silent/Documents/Python/karnadash/karnadash/static/tempfiles/temp.jpg'
Can anybody please explain why this is happening?
What solved it for me was switching from backslashes to forward slashes! Who would've thought?!
Similar post: ioerror invalid mode w
Check your file path if it's valid:
C:/Users/Silent/Documents/Python/karnadash/karnadash/static/tempfiles/temp.jpg
Perhaps it contains one karnadash too much.
I had a similar problem when I was trying to save some figures in a fowder. Some figures I could save, but couldn't save others, and I was using the same code. I realized that the name of the figure and the backslash were in conflict with a reserved code.
IOError: [Errno 22] invalid mode ('wb') or filename: '02102016\nDTG.png'
I think "\n" was interpreted as "enter". The problem had been solved when I changed it to forward slash.
I have two zip files, both of them open well with Windows Explorer and 7-zip.
However when i open them with Python's zipfile module [ zipfile.ZipFile("filex.zip") ], one of them gets opened but the other one gives error "BadZipfile: File is not a zip file".
I've made sure that the latter one is a valid Zip File by opening it with 7-Zip and looking at its properties (says 7Zip.ZIP). When I open the file with a text editor, the first two characters are "PK", showing that it is indeed a zip file.
I'm using Python 2.5 and really don't have any clue how to go about for this. I've tried it both with Windows as well as Ubuntu and problem exists on both platforms.
Update: Traceback from Python 2.5.4 on Windows:
Traceback (most recent call last):
File "<module1>", line 5, in <module>
zipfile.ZipFile("c:/temp/test.zip")
File "C:\Python25\lib\zipfile.py", line 346, in init
self._GetContents()
File "C:\Python25\lib\zipfile.py", line 366, in _GetContents
self._RealGetContents()
File "C:\Python25\lib\zipfile.py", line 378, in _RealGetContents
raise BadZipfile, "File is not a zip file"
BadZipfile: File is not a zip file
Basically when the _EndRecData function is called for getting data from End of Central Directory" record, the comment length checkout fails [ endrec[7] == len(comment) ].
The values of locals in the _EndRecData function are as following:
END_BLOCK: 4096,
comment: '\x00',
data: '\xd6\xf6\x03\x00\x88,N8?<e\xf0q\xa8\x1cwK\x87\x0c(\x82a\xee\xc61N\'1qN\x0b\x16K-\x9d\xd57w\x0f\xa31n\xf3dN\x9e\xb1s\xffu\xd1\.....', (truncated)
endrec: ['PK\x05\x06', 0, 0, 4, 4, 268, 199515, 0],
filesize: 199806L,
fpin: <open file 'c:/temp/test.zip', mode 'rb' at 0x045D4F98>,
start: 4073
files named file can confuse python - try naming it something else. if it STILL wont work, try this code:
def fixBadZipfile(zipFile):
f = open(zipFile, 'r+b')
data = f.read()
pos = data.find('\x50\x4b\x05\x06') # End of central directory signature
if (pos > 0):
self._log("Trancating file at location " + str(pos + 22)+ ".")
f.seek(pos + 22) # size of 'ZIP end of central directory record'
f.truncate()
f.close()
else:
# raise error, file is truncated
I run into the same issue. My problem was that it was a gzip instead of a zip file. I switched to the class gzip.GzipFile and it worked like a charm.
astronautlevel's solution works for most cases, but the compressed data and CRCs in the Zip can also contain the same 4 bytes. You should do an rfind (not find), seek to pos+20 and then add write \x00\x00 to the end of the file (tell zip applications that the length of the 'comments' section is 0 bytes long).
# HACK: See http://bugs.python.org/issue10694
# The zip file generated is correct, but because of extra data after the 'central directory' section,
# Some version of python (and some zip applications) can't read the file. By removing the extra data,
# we ensure that all applications can read the zip without issue.
# The ZIP format: http://www.pkware.com/documents/APPNOTE/APPNOTE-6.3.0.TXT
# Finding the end of the central directory:
# http://stackoverflow.com/questions/8593904/how-to-find-the-position-of-central-directory-in-a-zip-file
# http://stackoverflow.com/questions/20276105/why-cant-python-execute-a-zip-archive-passed-via-stdin
# This second link is only losely related, but echos the first, "processing a ZIP archive often requires backwards seeking"
content = zipFileContainer.read()
pos = content.rfind('\x50\x4b\x05\x06') # reverse find: this string of bytes is the end of the zip's central directory.
if pos>0:
zipFileContainer.seek(pos+20) # +20: see secion V.I in 'ZIP format' link above.
zipFileContainer.truncate()
zipFileContainer.write('\x00\x00') # Zip file comment length: 0 byte length; tell zip applications to stop reading.
zipFileContainer.seek(0)
return zipFileContainer
I had the same problem and was able to solve this issue for my files, see my answer at
zipfile cant handle some type of zip data?
I'm very new at python and i was facing the exact same issue, none of the previous methods were working.
Trying to print the 'corrupted' file just before unzipping it returned an empty byte object.
Turned out, I was trying to unzip the file right after writing it to disk, without closing the file handler.
with open(path, 'wb') as outFile:
outFile.write(data)
outFile.close() # was missing this
with zipfile.ZipFile(path, 'r') as zip:
zip.extractall(destination)
Closing the file stream then unzipping the file resolved my issue.
Sometime there are zip file which contain corrupted files and upon unzipping the zip gives badzipfile error. but there are tools like 7zip winrar which ignores these errors and successfully unzip the zip file. you can create a sub process and use this code to unzip your zip file without getting BadZipFile Error.
import subprocess
ziploc = "C:/Program Files/7-Zip/7z.exe" #location where 7zip is installed
cmd = [ziploc, 'e',your_Zip_file.zip ,'-o'+ OutputDirectory ,'-r' ]
sp = subprocess.Popen(cmd, stderr=subprocess.STDOUT, stdout=subprocess.PIPE)
Show the full traceback that you got from Python -- this may give a hint as to what the specific problem is. Unanswered: What software produced the bad file, and on what platform?
Update: Traceback indicates having problem detecting the "End of Central Directory" record in the file -- see function _EndRecData starting at line 128 of C:\Python25\Lib\zipfile.py
Suggestions:
(1) Trace through the above function
(2) Try it on the latest Python
(3) Answer the question above.
(4) Read this and anything else found by google("BadZipfile: File is not a zip file") that appears to be relevant
I faced this problem and was looking for a good and clean solution; But there was no solution until I found this answer. I had the same problem that #marsl (among the answers) had. It was a gzipfile instead of a zipfile in my case.
I could unarchive and decompress my gzipfile with this approach:
with tarfile.open(archive_path, "r:gz") as gzip_file:
gzip_file.extractall()
Have you tried a newer python, or if that is too much trouble, simply a newer zipfile.py? I have successfully used a copy of zipfile.py from Python 2.6.2 (latest at the time) with Python 2.5 in order to open some zip files that weren't supported by Py2.5s zipfile module.
In some cases, you have to confirm if the zip file is actually in gzip format. this was the case for me and i solved it by :
import requests
import tarfile
url = ".tar.gz link"
response = requests.get(url, stream=True)
file = tarfile.open(fileobj=response.raw, mode="r|gz")
file.extractall(path=".")
for this this happened when the file wasn't downloaded fully I think. So I just delete it in my download code.
def download_and_extract(url: str,
path_used_for_zip: Path = Path('~/data/'),
path_used_for_dataset: Path = Path('~/data/tmp/'),
rm_zip_file_after_extraction: bool = True,
force_rewrite_data_from_url_to_file: bool = False,
clean_old_zip_file: bool = False,
gdrive_file_id: Optional[str] = None,
gdrive_filename: Optional[str] = None,
):
"""
Downloads data and tries to extract it according to different protocols/file types.
note:
- to force a download do:
force_rewrite_data_from_url_to_file = True
clean_old_zip_file = True
- to NOT remove file after extraction:
rm_zip_file_after_extraction = False
Tested with:
- zip files, yes!
Later:
- todo: tar, gz, gdrive
force_rewrite_data_from_url_to_file = remvoes the data from url (likely a zip file) and redownloads the zip file.
"""
path_used_for_zip: Path = expanduser(path_used_for_zip)
path_used_for_zip.mkdir(parents=True, exist_ok=True)
path_used_for_dataset: Path = expanduser(path_used_for_dataset)
path_used_for_dataset.mkdir(parents=True, exist_ok=True)
# - download data from url
if gdrive_filename is None: # get data from url, not using gdrive
import ssl
ctx = ssl.create_default_context()
ctx.check_hostname = False
ctx.verify_mode = ssl.CERT_NONE
print("downloading data from url: ", url)
import urllib
import http
response: http.client.HTTPResponse = urllib.request.urlopen(url, context=ctx)
print(f'{type(response)=}')
data = response
# save zipfile like data to path given
filename = url.rpartition('/')[2]
path2file: Path = path_used_for_zip / filename
else: # gdrive case
from torchvision.datasets.utils import download_file_from_google_drive
# if zip not there re-download it or force get the data
path2file: Path = path_used_for_zip / gdrive_filename
if not path2file.exists():
download_file_from_google_drive(gdrive_file_id, path_used_for_zip, gdrive_filename)
filename = gdrive_filename
# -- write downloaded data from the url to a file
print(f'{path2file=}')
print(f'{filename=}')
if clean_old_zip_file:
path2file.unlink(missing_ok=True)
if filename.endswith('.zip') or filename.endswith('.pkl'):
# if path to file does not exist or force to write down the data
if not path2file.exists() or force_rewrite_data_from_url_to_file:
# delete file if there is one if your going to force a rewrite
path2file.unlink(missing_ok=True) if force_rewrite_data_from_url_to_file else None
print(f'about to write downloaded data from url to: {path2file=}')
# wb+ is used sinze the zip file was in bytes, otherwise w+ is fine if the data is a string
with open(path2file, 'wb+') as f:
# with open(path2file, 'w+') as f:
print(f'{f=}')
print(f'{f.name=}')
f.write(data.read())
print(f'done writing downloaded from url to: {path2file=}')
elif filename.endswith('.gz'):
pass # the download of the data doesn't seem to be explicitly handled by me, that is done in the extract step by a magic function tarfile.open
# elif is_tar_file(filename):
# os.system(f'tar -xvzf {path_2_zip_with_filename} -C {path_2_dataset}/')
else:
raise ValueError(f'File type {filename=} not supported.')
# - unzip data written in the file
extract_to = path_used_for_dataset
print(f'about to extract: {path2file=}')
print(f'extract to target: {extract_to=}')
if filename.endswith('.zip'):
import zipfile # this one is for zip files, inspired from l2l
zip_ref = zipfile.ZipFile(path2file, 'r')
zip_ref.extractall(extract_to)
zip_ref.close()
if rm_zip_file_after_extraction:
path2file.unlink(missing_ok=True)
elif filename.endswith('.gz'):
import tarfile
file = tarfile.open(fileobj=response, mode="r|gz")
file.extractall(path=extract_to)
file.close()
elif filename.endswith('.pkl'):
# no need to extract it, but when you use the data make sure you torch.load it or pickle.load it.
print(f'about to test torch.load of: {path2file=}')
data = torch.load(path2file) # just to test
assert data is not None
print(f'{data=}')
pass
else:
raise ValueError(f'File type {filename=} not supported, edit code to support it.')
# path_2_zip_with_filename = path_2_ziplike / filename
# os.system(f'tar -xvzf {path_2_zip_with_filename} -C {path_2_dataset}/')
# if rm_zip_file:
# path_2_zip_with_filename.unlink(missing_ok=True)
# # raise ValueError(f'File type {filename=} not supported.')
print(f'done extracting: {path2file=}')
print(f'extracted at location: {path_used_for_dataset=}')
print(f'-->Succes downloading & extracting dataset at location: {path_used_for_dataset=}')
you can use my code with pip install ultimate-utils for the most up to date version.
In the other case, this warning showing up when the ml/dl model has different format.
For the example:
you want to open pickle, but the model format is .sav
Solution:
you need to change the format to original format
pickle --> .pkl
tensorflow --> .h5
etc.
In my case, the zip file itself was missing from that directory - thus when I tried to unzip it, I got the error "BadZipFile: File is not a zip file". It got resolved after I moved the .zip file to the directory. Please confirm that the file is indeed present in your directory before running the python script.
In my case, the zip file was corrupted. I was trying to download the zip file with urllib.request.urlretrieve but the file wouldn't completely download for some reason.
I connected to a VPN, the file downloaded just fine, and I was able to open the file.