Python use lxml with fileinput - python

Having a simple xml
<?xml version="1.0" encoding="UTF-8" ?>
<root>
<child>abc</child>
</root>
I wanted to parse it from file and this works well:
with open('tst.xml') as test_xml:
for _, element in lxml.etree.iterparse(test_xml, tag='child'):
print element.text # prints abc as expected
However, I tried to modify script then to allow it parse xml either from file or from stdin and did not succeed:
fi = fileinput.input('tst.xml')
for _, element in lxml.etree.iterparse(fi, tag='child'):
print element.text
# File "iterparse.pxi", line 371, in lxml.etree.iterparse.__init__ (src/lxml/lxml.etree.c:97283)
# File "apihelpers.pxi", line 1411, in lxml.etree._encodeFilename (src/lxml/lxml.etree.c:22515)
# TypeError: Argument must be string or unicode.
I'm not sure in what I'm doing wrong. Is the FileInput object not a file-like object in python?

Without deep investigation, it seems that the reason of an exception is that FileInput class does not provide read method.
To achieve my goal, I ended up writing my own wrapper for now:
class FileInput(object):
def __init__(self, filename=None, *args, **kwargs):
self.file = open(filename, *args, **kwargs) if filename and filename != "-" else sys.stdin
def __enter__(self):
return self.file
def __exit__(self, type, value, traceback):
if self.file is not sys.stdin:
self.file.close()
def __getattr__(self, name):
return getattr(self.file, name)
I'll be waiting for a better answer though.

You should not try to use the fileinput module, but directly do it like that:
if filename == '-': # or, if we don't have a filename argument
f = sys.stdin
else:
f = open(filename, 'r')

Related

How to update a file inside a folder in a zipfile without unzipping the zip in Python? [duplicate]

I have archive.zip with two files: hello.txt and world.txt
I want to overwrite hello.txt file with new one with that code:
import zipfile
z = zipfile.ZipFile('archive.zip','a')
z.write('hello.txt')
z.close()
but it won't overwrite file, somehow it creates another instance of hello.txt — take a look at winzip screenshot:
Since there is no smth like zipfile.remove(), what's the best way to handle this problem?
There's no way to do that with python zipfile module. You have to create a new zip file and recompress everything again from the first file, plus the new modified file.
Below is some code to do just that. But note that it isn't efficient, since it decompresses and then recompresses all data.
import tempfile
import zipfile
import shutil
import os
def remove_from_zip(zipfname, *filenames):
tempdir = tempfile.mkdtemp()
try:
tempname = os.path.join(tempdir, 'new.zip')
with zipfile.ZipFile(zipfname, 'r') as zipread:
with zipfile.ZipFile(tempname, 'w') as zipwrite:
for item in zipread.infolist():
if item.filename not in filenames:
data = zipread.read(item.filename)
zipwrite.writestr(item, data)
shutil.move(tempname, zipfname)
finally:
shutil.rmtree(tempdir)
Usage:
remove_from_zip('archive.zip', 'hello.txt')
with zipfile.ZipFile('archive.zip', 'a') as z:
z.write('hello.txt')
Building on nosklo's answer.
UpdateableZipFile A class that inherits from ZipFile, maintians the same interface but adds the ability to overwrite files (via writestr or write) and removing files.
import os
import shutil
import tempfile
from zipfile import ZipFile, ZIP_STORED, ZipInfo
class UpdateableZipFile(ZipFile):
"""
Add delete (via remove_file) and update (via writestr and write methods)
To enable update features use UpdateableZipFile with the 'with statement',
Upon __exit__ (if updates were applied) a new zip file will override the exiting one with the updates
"""
class DeleteMarker(object):
pass
def __init__(self, file, mode="r", compression=ZIP_STORED, allowZip64=False):
# Init base
super(UpdateableZipFile, self).__init__(file, mode=mode,
compression=compression,
allowZip64=allowZip64)
# track file to override in zip
self._replace = {}
# Whether the with statement was called
self._allow_updates = False
def writestr(self, zinfo_or_arcname, bytes, compress_type=None):
if isinstance(zinfo_or_arcname, ZipInfo):
name = zinfo_or_arcname.filename
else:
name = zinfo_or_arcname
# If the file exits, and needs to be overridden,
# mark the entry, and create a temp-file for it
# we allow this only if the with statement is used
if self._allow_updates and name in self.namelist():
temp_file = self._replace[name] = self._replace.get(name,
tempfile.TemporaryFile())
temp_file.write(bytes)
# Otherwise just act normally
else:
super(UpdateableZipFile, self).writestr(zinfo_or_arcname,
bytes, compress_type=compress_type)
def write(self, filename, arcname=None, compress_type=None):
arcname = arcname or filename
# If the file exits, and needs to be overridden,
# mark the entry, and create a temp-file for it
# we allow this only if the with statement is used
if self._allow_updates and arcname in self.namelist():
temp_file = self._replace[arcname] = self._replace.get(arcname,
tempfile.TemporaryFile())
with open(filename, "rb") as source:
shutil.copyfileobj(source, temp_file)
# Otherwise just act normally
else:
super(UpdateableZipFile, self).write(filename,
arcname=arcname, compress_type=compress_type)
def __enter__(self):
# Allow updates
self._allow_updates = True
return self
def __exit__(self, exc_type, exc_val, exc_tb):
# call base to close zip file, organically
try:
super(UpdateableZipFile, self).__exit__(exc_type, exc_val, exc_tb)
if len(self._replace) > 0:
self._rebuild_zip()
finally:
# In case rebuild zip failed,
# be sure to still release all the temp files
self._close_all_temp_files()
self._allow_updates = False
def _close_all_temp_files(self):
for temp_file in self._replace.itervalues():
if hasattr(temp_file, 'close'):
temp_file.close()
def remove_file(self, path):
self._replace[path] = self.DeleteMarker()
def _rebuild_zip(self):
tempdir = tempfile.mkdtemp()
try:
temp_zip_path = os.path.join(tempdir, 'new.zip')
with ZipFile(self.filename, 'r') as zip_read:
# Create new zip with assigned properties
with ZipFile(temp_zip_path, 'w', compression=self.compression,
allowZip64=self._allowZip64) as zip_write:
for item in zip_read.infolist():
# Check if the file should be replaced / or deleted
replacement = self._replace.get(item.filename, None)
# If marked for deletion, do not copy file to new zipfile
if isinstance(replacement, self.DeleteMarker):
del self._replace[item.filename]
continue
# If marked for replacement, copy temp_file, instead of old file
elif replacement is not None:
del self._replace[item.filename]
# Write replacement to archive,
# and then close it (deleting the temp file)
replacement.seek(0)
data = replacement.read()
replacement.close()
else:
data = zip_read.read(item.filename)
zip_write.writestr(item, data)
# Override the archive with the updated one
shutil.move(temp_zip_path, self.filename)
finally:
shutil.rmtree(tempdir)
usage example:
with UpdateableZipFile("C:\Temp\Test2.docx", "a") as o:
# Overwrite a file with a string
o.writestr("word/document.xml", "Some data")
# exclude an exiting file from the zip
o.remove_file("word/fontTable.xml")
# Write a new file (with no conflict) to the zp
o.writestr("new_file", "more data")
# Overwrite a file with a file
o.write(r"C:\Temp\example.png", "word/settings.xml")
Based on this answer here's a quick and dirty way to monkey patch stock zipfile to support file deletion (while we waiting for it being accepted for python:main):
from zipfile import ZipFile, ZipInfo
from operator import attrgetter
import functools
def enable_zip_remove(func):
def _zipfile_remove_member(self, member):
# get a sorted filelist by header offset, in case the dir order
# doesn't match the actual entry order
fp = self.fp
entry_offset = 0
filelist = sorted(self.filelist, key=attrgetter('header_offset'))
for i in range(len(filelist)):
info = filelist[i]
# find the target member
if info.header_offset < member.header_offset:
continue
# get the total size of the entry
entry_size = None
if i == len(filelist) - 1:
entry_size = self.start_dir - info.header_offset
else:
entry_size = filelist[i + 1].header_offset - info.header_offset
# found the member, set the entry offset
if member == info:
entry_offset = entry_size
continue
# Move entry
# read the actual entry data
fp.seek(info.header_offset)
entry_data = fp.read(entry_size)
# update the header
info.header_offset -= entry_offset
# write the entry to the new position
fp.seek(info.header_offset)
fp.write(entry_data)
fp.flush()
# update state
self.start_dir -= entry_offset
self.filelist.remove(member)
del self.NameToInfo[member.filename]
self._didModify = True
# seek to the start of the central dir
fp.seek(self.start_dir)
def zipfile_remove(self, member):
"""Remove a file from the archive. The archive must be open with mode 'a'"""
if self.mode != 'a':
raise RuntimeError("remove() requires mode 'a'")
if not self.fp:
raise ValueError(
"Attempt to write to ZIP archive that was already closed")
if self._writing:
raise ValueError(
"Can't write to ZIP archive while an open writing handle exists."
)
# Make sure we have an info object
if isinstance(member, ZipInfo):
# 'member' is already an info object
zinfo = member
else:
# get the info object
zinfo = self.getinfo(member)
return self._zipfile_remove_member(zinfo)
#functools.wraps(func)
def wrapper(*args, **kwargs):
if not hasattr(ZipFile, "remove"):
setattr(ZipFile, "_zipfile_remove_member", _zipfile_remove_member)
setattr(ZipFile, "remove", zipfile_remove)
return func(*args, **kwargs)
return wrapper
Usage:
#enable_zip_remove
def replace_zip_file():
with ZipFile("archive.zip", "a") as z:
z.remove("hello.txt")
z.write("hello.txt")
P.S. NSFW
My solution is similar to the other answers but uses SQLite to manage the intermediate files and provides __getitem__, __setitem__ and __delitem__ for an easy interface.
By default the db is in-memory but you can provide a temp file path if you have a zip larger than available memory.
And of course SQLite is built into Python and faster than the file system
import sqlite3
import subprocess
import zipfile
from pathlib import Path
from sql import CREATE_TABLE, DELETE_FILE, INSERT_FILE, SELECT_CONTENT
class EditableZip:
"""Intended to make editing files inside zip archive easy, this class is capable of loading files
from a zip file into a sqlite database, facilitates editing/removing/adding files, and saving
to a zip.
The database can be in-memory (default) or in a temporary on disk file if
temp_db_path is provided.
If an on-disk file is used, EditableZip.close can be called to remove the file or EditableZip
can be used as a context manager.
If auto_save is set to True and an initial zip_path was provided then the file will
be overwritten when EditableZip closes. If you wish to save to a different file,
or no zip_path is used in instantiation, auto_save can take a file path.
Files can be added by item assignment
with EditableZip(auto_save="example.zip") as ez:
ez["thing.txt"] = "stuff"
# empty dir
ez["empty/"] = None
Assignment accepts Non-text files as bytes.
EditableZip is subscriptable. If the subscript is a path in the db, the data will be returned.
EditableZip.files can be used to iterate over files in the db.
"""
def __init__(
self,
zip_path: None | str | Path = None,
temp_db_path: None | Path = None,
auto_save: bool | str | Path = False,
):
self.temp_db_path, self.auto_save, self.file_path = (
temp_db_path,
auto_save,
zip_path,
)
self.db = sqlite3.connect(
str(temp_db_path if temp_db_path is not None else ":memory:")
)
self.db.execute(CREATE_TABLE)
if self.file_path:
self.load(self.file_path)
#property
def files(self):
"Returns a generator of all file paths in the database."
try:
return (
i[0] for i in self.db.execute("SELECT file_path FROM files").fetchall()
)
except TypeError:
return None
def load(self, zip_path: str | Path) -> None:
"Add all files from zip at zip_path to db."
with zipfile.ZipFile(zip_path, mode="r") as archive:
for item in archive.infolist():
self[item.filename] = (
None if item.filename[-1] == "/" else archive.read(item)
)
def save(self, zip_path: None | str | Path) -> Path:
"Save all files from db to zip at zip_path."
zip_path = self.file_path if zip_path is None else zip_path
with zipfile.ZipFile(zip_path, "w") as archive:
for file in self.files:
if file_data := self.fetch(file):
archive.writestr(file, file_data)
else:
archive.writestr(zipfile.ZipInfo(file), "")
return zip_path
def close(self):
"Auto save if applicable and close + remove db."
if self.auto_save:
self.save(
zip_path=self.auto_save
if isinstance(self.auto_save, (str, Path))
else None
)
self.db.close()
if isinstance(self.temp_db_path, Path):
self.temp_db_path.unlink(missing_ok=True)
def fetch(self, file_path: str) -> bytes:
"Get content of db file for file_path."
try:
return self.db.execute(SELECT_CONTENT, {"file_path": file_path}).fetchone()[
0
]
except TypeError:
return None
def __getitem__(self, key):
result = self.fetch(key)
try:
return result.decode("utf-8")
except AttributeError:
return result
def __setitem__(self, file_path, content: str | bytes):
if isinstance(content, str):
content = content.encode("utf-8")
self.db.execute(
INSERT_FILE,
{"file_path": file_path, "file_content": content},
)
def __delitem__(self, file_path):
self.db.execute(DELETE_FILE, {"file_path": file_path})
def __enter__(self):
return self
def __exit__(self, exc_type, exc_value, traceback):
self.close()
if __name__ == "__main__":
# A use case: editing epub files.
# File source:
# https://archiveofourown.org/downloads/13795605/Victoria%20Potter%20and%20the.epub?updated_at=1650231615
file_path = Path("Victoria Potter and the.epub")
new_file = (file_path.parent / (file_path.stem + "- lowercase")).with_suffix(
file_path.suffix
)
# Create a copy of the epub with all letters lowercase
with EditableZip(zip_path=file_path, auto_save=new_file) as ez:
for file in ez.files:
if Path(file).suffix in [".html", ".xhtml"]:
ez[file] = ez[file].lower()
Reference: Delete file from zipfile with the ZipFile Module
In short,
You can take the code from https://github.com/python/cpython/blob/659eb048cc9cac73c46349eb29845bc5cd630f09/Lib/zipfile.py and create a separate file from it. After that just reference it from your project instead of built-in python library: import myproject.zipfile as zipfile.
Usage:
with zipfile.ZipFile(f"archive.zip", "a") as z:
z.remove(f"firstfile.txt")

python class file context manager

I am trying to open a file in a class and close it on exit in this manner.
class PlanetaryImage(object):
#classmethod
def open(cls, filename):
with open(filename, 'rb') as fp:
return cls(fp, filename)
def __init__(self, stream, filename=None, memory_layout='DISK'):
self.filename = filename
self._parse_data(stream)
def _parse_data(self, stream):
data_stream = stream
try:
if self.data_filename is not None:
dirpath = os.path.dirname(self.filename)
data_file = os.path.abspath(
os.path.join(dirpath, self.data_filename))
data_stream = open(data_file, 'rb')
data_stream.seek(self.start_byte)
if self.format in self.BAND_STORAGE_TYPE:
return getattr(self, self.BAND_STORAGE_TYPE[self.format])(data_stream)
raise Exception('Unkown format (%s)' % self.format)
finally:
data_stream.close()
There are certain cases where I am having to use open one more file in _parse_data function. I wanted to use with but the if statements make it difficult. Any suggestions on how to make the try section more pythonic.
There's no reason for _parse_data to try to open a file. It should be the caller's responsibility to either use PlanetaryImage.open with a file name or to provide an open file handle to __init__. _parse_data should do just one thing: parse the data from its stream argument.
class PlanetaryImage(object):
#classmethod
def open(cls, filename):
with open(filename, 'rb') as fp:
return cls(fp, filename)
def __init__(self, stream, memory_layout='DISK'):
self._parse_data(stream)
def _parse_data(self, data_stream):
try:
data_stream.seek(self.start_byte)
if self.format in self.BAND_STORAGE_TYPE:
return getattr(self, self.BAND_STORAGE_TYPE[self.format])(data_stream)
raise Exception('Unkown format (%s)' % self.format)
finally:
data_stream.close()
Now, there are simply two options for using the class:
with open(filename, 'rb') as fp:
x = PlanetaryImage(fp)
...
or
x = PlanetaryImage(filename)
....

python: prevent creating empty files

For a bulk task I create a couple of instances of the ProgressLog object which will each create an empty log-file no matter if there actually will be any errors. what is the best way to prevent this?
class ProgressLog(object):
"""Write a message to log + progress indicator.
"""
total = 0
def __init__(self, name):
source_path, file_name = os.path.split(name)
self.name = file_name
self.source_path = source_path
self.log_dir_name = r'log'
self.make_log_dir(self.source_path, self.log_dir_name)
self.reset()
log_file = self._logfilename()
try:
self.f = open(log_file, 'w')
print('\n***logging errors to {0}***\n'.format(log_file))
except IOError, err:
msg = 'Cannot open logfile {0}. Traceback is: {1}'.format(
log_file, err)
raise msg
def _logfilename(self):
## hms_ddmmyyyy format
log_name = r'{1}_{0}{2}_errors.csv'.format(
time.strftime("%I%M%S"),
time.strftime("%d%m%Y"),
self.name)
return os.path.join(self.source_path, self.log_dir_name, log_name)
There is no "magical" way to do it, you simply need to refactor the code to open the log file only on first actual call to log.
To achieve this, extract the part of __init__ that opens the log file into a separate _open_log method. In __init__ initialize self.f to None. Then, your actual logging method can begin with:
if self.f is None:
self._open_log()

Copying Python console output to file and adding timestamp

I am trying to copy my console log to a file when running a script utilizing a snippet found on this link. I tried to customize it by adding a timestamp utilizing strftime from the time module, but the snippet now adds a timestamp to both the start of a new row and the end:
2014-12-10 12:15:35: Working on local page 12014-12-10 12:15:35:
What did I do wrong? How would I fix this so that the timestamp is only shown at the start of a newline?
from time import strftime
class copyConsoleToFile(object):
""" Enables logging of console output to a file, use
>> tlogger = copyConsoleToFile('logfile.txt', 'w')
at the start of the code to start logging.
"""
def __init__(self, name, mode):
self.file = open(name, mode)
self.stdout = sys.stdout
sys.stdout = self
def close(self):
if self.stdout is not None:
sys.stdout = self.stdout
self.stdout = None
if self.file is not None:
self.file.close()
self.file = None
def write(self, data):
self.file.write(strftime("%Y-%m-%d %H:%M:%S") + ': ' + data)
self.stdout.write(data)
def flush(self):
self.file.flush()
self.stdout.flush()
def __del__(self):
self.close()
As stated in the comments, it seems that a sub-process adds an extra write to each line. I suggest to remove the non-useful extra characters:
with open('logfile.txt','r+') as fopen:
string = ""
for line in fopen.readlines():
string = string + line[:-23] + "\n"
with open('logfile.txt','w') as fopen:
fopen.write(string)
The code was adapted from: https://stackoverflow.com/a/45594783/1751393

AttributeError: 'str' object has no attribute 'filename' in Python

I'm new to Python and currently learning, I had a task to do some reading and writing to files with a python script. The reading part of my script seems to work as expected however the write section is throwing an error. It's probably something trivial I have done but here is my code:
class LogMessage():
def __init__(self, filename):
self.filename = filename
def read(self):
inputFile = open(self.filename)
for line in inputFile:
print(line, end='')
def write(self):
outputFile = open(self.filename)
#writeInput = input('What data do you wish to write?:\n')
for line in writeInput:
print(line,file = outputFile, end='')
filename = LogMessage('new.txt')
filename.read()
writeInput = input('What data do you wish to write?:\n')
LogMessage.write(writeInput)
The read part works but taking user data and writing it to the file and gives this error:
Traceback (most recent call last):
File "/home/alex/workspace/Python/Learn Python/labEx9.py", line 22, in <module>
LogMessage.write(writeInput)
File "/home/alex/workspace/Python/Learn Python/labEx9.py", line 11, in write
outputFile = open(self.filename)
AttributeError: 'str' object has no attribute 'filename'
can anyone help me, thanks a lot.
Alex
You must call 'write' on 'filename', which is an instance of LogMessage, not on the LogMessage class.
Apart from this, there are other issues (e.g. 'writeInput' is not defined in method 'write')
If you get such errors while using flask check your html code( your_form.) and add this to your html :
<form method="POST" action="" enctype="multipart/form-data">
enctype="multipart/form-data" would help.
class LogMessage():
def __init__(self, filename):
self.filename = filename
def read(self):
inputFile = open(self.filename)
for line in inputFile:
print(line, end='')
def write(self):
writeInput = input('What data do you wish to write?:\n')
outputFile = open(self.filename, 'w')
for line in writeInput:
print(line, file = outputFile, end='')
filename = LogMessage('new.txt')
filename.write()
filename.read()

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