I am creating a zip file using the zipfile module. It works like a charm. but that's file, saved in the executed script place.
my script path is a:
[b]c:/User/Administrator/[/b]script.py
and the zipfile saved in:
[b]c:/User/Administrator/[/b]backup.zip
but I want, [b]creating a zipfile, in another path[/b], like this:
[b]d:/backups/[/b]backup.zip
my code like this:
import zipfile
zip_file = zipfile.ZipFile("backup.zip", 'w')
with zip_file:
for file in filePaths:
zip_file.write(file)
my question is a how can I adding custom path for saving the zip_file. because I have not an enough space in C:
tnx a lot.
Give the path you want to ZipFile function.
When you give only the name of the file, it will save the file in the current directory which the program is running.
Do this instead:
import zipfile
# For example you want to save it in drive 'D'
path = "D:\\PathToYourDir\\backup.zip"
zip_file = zipfile.ZipFile(path, 'w')
with zip_file:
for file in filePaths:
zip_file.write(file)
Related
I know how to open a file in a zip file without unzipping. It can be done as follows:
import zipfile
archive = zipfile.ZipFile(path_to_zipfile, 'r')
my_zipfile = archive.open('file_01.cdf') # "file_01.cdf" is contained in the zipfile
My problem is: the file "file_01.cdf" contained in the zip file is in CDF format. Normally, a CDF format file can be read as follows:
from spacepy import pycdf
cdf_file = pycdf.CDF(path_to_file)
data = cdf_file.copy()
The problem is: function pycdf.CDF accept argument in string format. However, I want to pass an zipfile object as an argument to pycdf.CDF function. (Because I don't unzip the zip file, I don't have path to file as a string). Any idea to overcome this problem? Thank you very much!
I know you specifically asked about doing it without unzipping but I think that's not possible (someone will correct me if I'm wrong, I hope).
You could do something like:
import zipfile
import os
import tempfile
import shutil
path = r'C:\Users\XXX\Desktop\test.zip' # my zip file
archive = zipfile.ZipFile(path, 'r')
tmpdir = tempfile.mkdtemp(dir=os.getcwd()) # create a temp folder in the working directory
archive.extract('test.txt', path=tmpdir) # extract your file to the temp folder
path_to_my_file = os.path.join(tmpdir, 'test.txt') # this is the path to the extracted file
# -------
# do stuff with your file path
# -------
print(path_to_my_file)
shutil.rmtree(tmpdir) # delete the temp folder
I am trying to zip a single file in python. For whatever reason, I'm having a hard time getting down the syntax. What I am trying to do is keep the original file and create a new zipped file of the original (like what a Mac or Windows would do if you archive a file).
Here is what I have so far:
import zipfile
myfilepath = '/tmp/%s' % self.file_name
myzippath = myfilepath.replace('.xml', '.zip')
zipfile.ZipFile(myzippath, 'w').write(open(myfilepath).read()) # does not zip the file properly
The correct way to zip file is:
zipfile.ZipFile('hello.zip', mode='w').write("hello.csv")
# assume your xxx.py under the same dir with hello.csv
The python official doc says:
ZipFile.write(filename, arcname=None, compress_type=None)
Write the file named filename to the archive, giving it the archive name arcname
You pass open(filename).read() into write(). open(filename).read() is a single string that contains the whole content of file filename, it would throw FileNotFoundError because it is trying to find a file named with the string content.
If the file to be zipped (filename) is in a different directory called pathname, you should use the arcname parameter. Otherwise, it will recreate the full folder hierarchy to the file folder.
from zipfile import ZipFile
import os
with ZipFile(zip_file, 'w') as zipf:
zipf.write(os.path.join(pathname,filename), arcname=filename)
Try calling zipfile.close() afterwards?
from zipfile import ZipFile
zipf = ZipFile("main.zip","w", zipfile.ZIP_DEFLATED)
zipf.write("main.json")
zipf.close()
Since you also want to specify the directory try using os.chdir:
#!/usr/bin/python
from zipfile import ZipFile
import os
os.chdir('/path/of/target/and/destination')
ZipFile('archive.zip', 'w').write('original_file.txt')
Python zipfile : Work with Zip archives
Python Miscellaneous operating system interfaces
I want to compress one text file using shutil.make_archive command. I am using the following command:
shutil.make_archive('gzipped'+fname, 'gztar', os.path.join(os.getcwd(), fname))
OSError: [Errno 20] Not a directory: '/home/user/file.txt'
I tried several variants but it keeps trying to compress the whole folders. How to do it correctly?
Actually shutil.make_archive can make one-file archive! Just pass path to target directory as root_dir and target filename as base_dir.
Try this:
import shutil
file_to_zip = 'test.txt' # file to zip
target_path = 'C:\\test_yard\\' # dir, where file is
try:
shutil.make_archive(target_path + 'archive', 'zip', target_path, file_to_zip)
except OSError:
pass
shutil can't create an archive from one file. You can use tarfile, instead:
tar = tarfile.open(fname + ".tar.gz", 'w:qz')
os.chdir('/home/user')
tar.add("file.txt")
tar.close()
or
tar = tarfile.open(fname + ".tar.gz", 'w:qz')
tar.addfile(tarfile.TarInfo("/home/user/file.txt"), "/home/user/file.txt")
tar.close()
Try this and Check shutil
copy your file to a directory.
cd directory
shutil.make_archive('gzipped', 'gztar', os.getcwd())
#CommonSense had a good answer, but the file will always be created zipped inside its parent directories. If you need to create a zipfile without the extra directories, just use the zipfile module directly
import os, zipfile
inpath = "test.txt"
outpath = "test.zip"
with zipfile.ZipFile(outpath, "w", compression=zipfile.ZIP_DEFLATED) as zf:
zf.write(inpath, os.path.basename(inpath))
If you don't mind doing a file copy op:
def single_file_to_archive(full_path, archive_name_no_ext):
tmp_dir = tempfile.mkdtemp()
shutil.copy2(full_path, tmp_dir)
shutil.make_archive(archive_name_no_ext, "zip", tmp_dir, '.')
shutil.rmtree(tmp_dir)
Archiving a directory to another destination was a pickle for me but shutil.make_archive not zipping to correct destination helped a lot.
from shutil import make_archive
make_archive(
base_name=path_to_directory_to_archive},
format="gztar",
root_dir=destination_path,
base_dir=destination_path)
Normally, executing the following code will pickle an object to a file in my current directory:
fp = open('somefile.txt', 'wb')
pickle.dump(object, fp)
How do I re-direct the output from pickle.dump to a different directory?
with open('/full/path/to/file', 'wb') as f:
pickle.dump(object, f)
How about combination of pathlib and with, I think it more flexible and safer.
# from python 3.4
from pathlib import Path
my_path = Path("{path to you want to set root like . or ~}") / "path" / "to" / "file"
with my_path.open('wb') as fp:
pickle.dump(object, fp)
If you wish to save the file to a sub-folder located inside the folder containing your code you can use the pathlib module. This will allow the code to work even if its location is moved on your computer, or you code is added to a different machine.
import the module:
from pathlib import Path
Set root equal to your current folder:
root = Path(".")
Create a path to your sub-folder and file name:
my_path = root / "my_sub_folder" / "my_file_name"
Open your file, dump data to your file, close your file:
my_file = open(my_path, 'wb')
my_file = pickle.dump("data_to_save", my_file)
my_file.close()
Note if your my_file doesn't currently exist you will want to create it before running this code. *
You can just try
import pickle
filepath = r'<FolderName>/<FileName>.pkl'
pickle.dump(<model>, open(filepath, 'wb'))
so final code look likes
If your Folder name is Models
and you wanted to pickle randomForest Model
import pickle
filepath = r'Models/rfPickle.pkl'
pickle.dump(randomForest, open(filepath, 'wb'))
Actually i am writting a script which writes two files into a desktop, let it be as "a.txt" and "b.txt"....... so after writing into a desktop i have to read this files and zip into a folder....
can anyone help on this....i know how to zip a folder but dono how to add two files in to a zip
Reading from folder i know its like this
def zipdir(basedir, archivename):
assert os.path.isdir(basedir)
with closing(ZipFile(archivename, "w", ZIP_DEFLATED)) as z:
for root, dirs, files in os.walk(basedir):
for fn in files:
absfn = os.path.join(root, fn)
zfn = absfn[len(basedir)+len(os.sep):]
z.write(absfn, zfn)
if __name__ == '__main__':
import sys
basedir = sys.argv[1]
archivename = sys.argv[2]
zipdir(basedir, archivename)
The code which now i using is
import zipfile
zip = zipfile.ZipFile('Python.zip', 'a')
zip.write('fields.txt')
zip.write('grp.txt')
zip.close()
This is creating file of those two plus some extra folder which contains all files.......
you need to open the zip file with "a" -append parameter. Then you can use the write parameter without overwriting the file.
source: 12.4.1
EDIT:
zip.write('file.pdf','/folder/file.pdf')
The easiest wayh is to use shutil library. put all the files you want to zip in a single directoty(folder)
import shutil
shutil.make_archive(output_filename_dont_add_.zip, 'zip', directory_to_download)
Remember if you work with ipython you can use relative address for directory_to_download