如何在Python中创建目录的zip存档?


489

如何在Python中创建目录结构的zip存档?


21
不要使用已接受答案中建议的解决方案,而要使用make_archivefrom来进一步解决shutil(如果您要递归压缩单个目录)。
马拉纳

Answers:


526

正如其他人指出的那样,您应该使用zipfile。该文档告诉您可用的功能,但并未真正说明如何使用它们来压缩整个目录。我认为用一些示例代码来解释是最简单的:

#!/usr/bin/env python
import os
import zipfile

def zipdir(path, ziph):
    # ziph is zipfile handle
    for root, dirs, files in os.walk(path):
        for file in files:
            ziph.write(os.path.join(root, file))

if __name__ == '__main__':
    zipf = zipfile.ZipFile('Python.zip', 'w', zipfile.ZIP_DEFLATED)
    zipdir('tmp/', zipf)
    zipf.close()

改编自:http : //www.devshed.com/c/a/Python/Python-UnZipped/


129
我将在write调用中添加第二个参数通过os.path.relpath(os.path.join(root, file), os.path.join(path, '..'))。这样一来,您就可以从任何工作目录中压缩目录,而无需在存档中获取完整的绝对路径。
Reimund 2013年

8
当我尝试压缩文件夹并将输出的zip输出到同一文件夹时,发生了一个有趣的递归。:-)
Sibbs赌博

13
shutil只需一行就可以轻松完成。请检查下面的答案
。– droidlabour

7
你可能会做ziph.write(os.path.join(路径,文件),arcname =文件)更感兴趣,以便存档里面的文件名是不是相对于硬盘驱动器
克里斯托夫布林

1
啊,我错过了.close()电话!
information_interchange

1059

最简单的方法是使用shutil.make_archive。它支持zip和tar格式。

import shutil
shutil.make_archive(output_filename, 'zip', dir_name)

如果您需要做的事情比压缩整个目录还要复杂(例如跳过某些文件),那么您将需要zipfile按照其他人的建议深入研究该模块。


113
shutil是标准python库的一部分。这应该是最佳答案
AlexG

4
这是这里最简洁的答案,并且还具有直接将所有子目录和文件添加到存档中的优点,而不是将所有内容都包含在顶级文件夹中(解压缩后在文件夹结构中导致冗余级别)。
aitch-hat

3
@cmcginty您能否更具体一点说明它的哪些方面不是线程安全的?一个调用时运行多个线程会导致解释器崩溃吗?
std''OrgnlDave

13
请注意,在Python 3.4之前,shutil.make_archive不支持ZIP64,并且在创建大于2GB的ZIP文件时将失败。
azdev

2
@Teekin否。如果您查看错误报告(bugs.python.org/issue30511),就会看到它的shutil.make_archive使用os.chdir()。据我了解os.chdir(),它在全球范围内运作。
山姆·马来耶克

65

要将内容添加mydirectory到新的zip文件中,包括所有文件和子目录:

import os
import zipfile

zf = zipfile.ZipFile("myzipfile.zip", "w")
for dirname, subdirs, files in os.walk("mydirectory"):
    zf.write(dirname)
    for filename in files:
        zf.write(os.path.join(dirname, filename))
zf.close()

对我来说,此代码引发以下错误TypeError:无效文件:<zipfile.ZipFile [关闭]>
Nishad Up

10
您能不能使用a with而不是close()最后必须给自己打电话?
ArtOfWarfare

50

如何在Python中创建目录结构的zip存档?

在Python脚本中

在Python 2.7+中,shutil具有make_archive功能。

from shutil import make_archive
make_archive(
  'zipfile_name', 
  'zip',           # the archive format - or tar, bztar, gztar 
  root_dir=None,   # root for archive - current working dir if None
  base_dir=None)   # start archiving from here - cwd if None too

此处的压缩存档将命名为zipfile_name.zip。如果base_dir距离较远root_dir,它将排除不在中的文件base_dir,但仍将文件归档在父目录中,直到root_dir

我在使用2.7的Cygwin上测试时确实遇到了问题-它需要一个root_dir参数,用于cwd:

make_archive('zipfile_name', 'zip', root_dir='.')

从外壳使用Python

您还可以使用以下zipfile模块从外壳使用Python :

$ python -m zipfile -c zipname sourcedir

zipname您想要的目标文件的名称在哪里(.zip如果需要,可以添加,它将不会自动添加),而sourcedir是目录的路径。

压缩Python(或者只是不希望父目录):

如果你想拉上一个Python包用__init__.py__main__.py,和你不想要的父目录,它是

$ python -m zipfile -c zipname sourcedir/*

$ python zipname

将运行该软件包。(请注意,您不能将子包作为压缩存档的入口点运行。)

压缩Python应用程式:

如果您拥有python3.5 +,并且特别想压缩一个Python包,请使用zipapp

$ python -m zipapp myapp
$ python myapp.pyz

31

此功能将递归压缩目录树,压缩文件,并在存档中记录正确的相对文件名。存档条目与生成的条目相同zip -r output.zip source_dir

import os
import zipfile
def make_zipfile(output_filename, source_dir):
    relroot = os.path.abspath(os.path.join(source_dir, os.pardir))
    with zipfile.ZipFile(output_filename, "w", zipfile.ZIP_DEFLATED) as zip:
        for root, dirs, files in os.walk(source_dir):
            # add directory (needed for empty dirs)
            zip.write(root, os.path.relpath(root, relroot))
            for file in files:
                filename = os.path.join(root, file)
                if os.path.isfile(filename): # regular files only
                    arcname = os.path.join(os.path.relpath(root, relroot), file)
                    zip.write(filename, arcname)

17

使用shutil,它是python标准库集的一部分。使用shutil非常简单(请参见下面的代码):

  • 第一个参数:生成的zip / tar文件的文件名,
  • 第二个参数:zip / tar,
  • 第三个参数:dir_name

码:

import shutil
shutil.make_archive('/home/user/Desktop/Filename','zip','/home/username/Desktop/Directory')

12

要将压缩添加到生成的zip文件中,请查看此链接

您需要更改:

zip = zipfile.ZipFile('Python.zip', 'w')

zip = zipfile.ZipFile('Python.zip', 'w', zipfile.ZIP_DEFLATED)

5

我对Mark Byers给出的代码进行了一些更改。如果有空目录,下面的函数还会添加空目录。通过示例可以更清楚地了解添加到zip的路径是什么。

#!/usr/bin/env python
import os
import zipfile

def addDirToZip(zipHandle, path, basePath=""):
    """
    Adding directory given by \a path to opened zip file \a zipHandle

    @param basePath path that will be removed from \a path when adding to archive

    Examples:
        # add whole "dir" to "test.zip" (when you open "test.zip" you will see only "dir")
        zipHandle = zipfile.ZipFile('test.zip', 'w')
        addDirToZip(zipHandle, 'dir')
        zipHandle.close()

        # add contents of "dir" to "test.zip" (when you open "test.zip" you will see only it's contents)
        zipHandle = zipfile.ZipFile('test.zip', 'w')
        addDirToZip(zipHandle, 'dir', 'dir')
        zipHandle.close()

        # add contents of "dir/subdir" to "test.zip" (when you open "test.zip" you will see only contents of "subdir")
        zipHandle = zipfile.ZipFile('test.zip', 'w')
        addDirToZip(zipHandle, 'dir/subdir', 'dir/subdir')
        zipHandle.close()

        # add whole "dir/subdir" to "test.zip" (when you open "test.zip" you will see only "subdir")
        zipHandle = zipfile.ZipFile('test.zip', 'w')
        addDirToZip(zipHandle, 'dir/subdir', 'dir')
        zipHandle.close()

        # add whole "dir/subdir" with full path to "test.zip" (when you open "test.zip" you will see only "dir" and inside it only "subdir")
        zipHandle = zipfile.ZipFile('test.zip', 'w')
        addDirToZip(zipHandle, 'dir/subdir')
        zipHandle.close()

        # add whole "dir" and "otherDir" (with full path) to "test.zip" (when you open "test.zip" you will see only "dir" and "otherDir")
        zipHandle = zipfile.ZipFile('test.zip', 'w')
        addDirToZip(zipHandle, 'dir')
        addDirToZip(zipHandle, 'otherDir')
        zipHandle.close()
    """
    basePath = basePath.rstrip("\\/") + ""
    basePath = basePath.rstrip("\\/")
    for root, dirs, files in os.walk(path):
        # add dir itself (needed for empty dirs
        zipHandle.write(os.path.join(root, "."))
        # add files
        for file in files:
            filePath = os.path.join(root, file)
            inZipPath = filePath.replace(basePath, "", 1).lstrip("\\/")
            #print filePath + " , " + inZipPath
            zipHandle.write(filePath, inZipPath)

上面是一个简单函数,适用于简单情况。您可以在我的Gist中找到更优雅的课程:https : //gist.github.com/Eccenux/17526123107ca0ac28e6


1
使用os.path可以大大简化路径处理。看我的答案。
George V. Reilly

错误:zipHandle.write(os.path.join(root,“。”))没有考虑basePath。
佩特2014年

是的,您可能是对的。我稍后对此进行了一些补充
Nux 2014年

4

现代Python(3.6+)使用该pathlib模块进行类似于OOP的简洁路径处理和pathlib.Path.rglob()递归glob。据我所知,这相当于George V. Reilly的答案:压缩压缩,最上面的元素是目录,保留空目录,使用相对路径。

from pathlib import Path
from zipfile import ZIP_DEFLATED, ZipFile

from os import PathLike
from typing import Union


def zip_dir(zip_name: str, source_dir: Union[str, PathLike]):
    src_path = Path(source_dir).expanduser().resolve(strict=True)
    with ZipFile(zip_name, 'w', ZIP_DEFLATED) as zf:
        for file in src_path.rglob('*'):
            zf.write(file, file.relative_to(src_path.parent))

注意:如可选类型提示所指示,zip_name不能是Path对象(将在3.6.2+中修复)。


1
太棒了!简洁!现代!
ingyhere

3

我有另一个使用python3,pathlib和zipfile可能会有所帮助的代码示例。它应该可以在任何操作系统上运行。

from pathlib import Path
import zipfile
from datetime import datetime

DATE_FORMAT = '%y%m%d'


def date_str():
    """returns the today string year, month, day"""
    return '{}'.format(datetime.now().strftime(DATE_FORMAT))


def zip_name(path):
    """returns the zip filename as string"""
    cur_dir = Path(path).resolve()
    parent_dir = cur_dir.parents[0]
    zip_filename = '{}/{}_{}.zip'.format(parent_dir, cur_dir.name, date_str())
    p_zip = Path(zip_filename)
    n = 1
    while p_zip.exists():
        zip_filename = ('{}/{}_{}_{}.zip'.format(parent_dir, cur_dir.name,
                                             date_str(), n))
        p_zip = Path(zip_filename)
        n += 1
    return zip_filename


def all_files(path):
    """iterator returns all files and folders from path as absolute path string
    """
    for child in Path(path).iterdir():
        yield str(child)
        if child.is_dir():
            for grand_child in all_files(str(child)):
                yield str(Path(grand_child))


def zip_dir(path):
    """generate a zip"""
    zip_filename = zip_name(path)
    zip_file = zipfile.ZipFile(zip_filename, 'w')
    print('create:', zip_filename)
    for file in all_files(path):
        print('adding... ', file)
        zip_file.write(file)
    zip_file.close()


if __name__ == '__main__':
    zip_dir('.')
    print('end!')


1

这是Nux给出的答案的变体,它对我有用:

def WriteDirectoryToZipFile( zipHandle, srcPath, zipLocalPath = "", zipOperation = zipfile.ZIP_DEFLATED ):
    basePath = os.path.split( srcPath )[ 0 ]
    for root, dirs, files in os.walk( srcPath ):
        p = os.path.join( zipLocalPath, root [ ( len( basePath ) + 1 ) : ] )
        # add dir
        zipHandle.write( root, p, zipOperation )
        # add files
        for f in files:
            filePath = os.path.join( root, f )
            fileInZipPath = os.path.join( p, f )
            zipHandle.write( filePath, fileInZipPath, zipOperation )

1

试试下面的一个对我有用

import zipfile, os
zipf = "compress.zip"  
def main():
    directory = r"Filepath"
    toZip(directory)
def toZip(directory):
    zippedHelp = zipfile.ZipFile(zipf, "w", compression=zipfile.ZIP_DEFLATED )

    list = os.listdir(directory)
    for file_list in list:
        file_name = os.path.join(directory,file_list)

        if os.path.isfile(file_name):
            print file_name
            zippedHelp.write(file_name)
        else:
            addFolderToZip(zippedHelp,file_list,directory)
            print "---------------Directory Found-----------------------"
    zippedHelp.close()

def addFolderToZip(zippedHelp,folder,directory):
    path=os.path.join(directory,folder)
    print path
    file_list=os.listdir(path)
    for file_name in file_list:
        file_path=os.path.join(path,file_name)
        if os.path.isfile(file_path):
            zippedHelp.write(file_path)
        elif os.path.isdir(file_name):
            print "------------------sub directory found--------------------"
            addFolderToZip(zippedHelp,file_name,path)


if __name__=="__main__":
    main()

1

如果要使用任何通用图形文件管理器的compress文件夹之类的功能,则可以使用以下代码,它使用zipfile模块。使用此代码,您将获得带有路径的zip文件作为其根文件夹。

import os
import zipfile

def zipdir(path, ziph):
    # Iterate all the directories and files
    for root, dirs, files in os.walk(path):
        # Create a prefix variable with the folder structure inside the path folder. 
        # So if a file is at the path directory will be at the root directory of the zip file
        # so the prefix will be empty. If the file belongs to a containing folder of path folder 
        # then the prefix will be that folder.
        if root.replace(path,'') == '':
                prefix = ''
        else:
                # Keep the folder structure after the path folder, append a '/' at the end 
                # and remome the first character, if it is a '/' in order to have a path like 
                # folder1/folder2/file.txt
                prefix = root.replace(path, '') + '/'
                if (prefix[0] == '/'):
                        prefix = prefix[1:]
        for filename in files:
                actual_file_path = root + '/' + filename
                zipped_file_path = prefix + filename
                zipf.write( actual_file_path, zipped_file_path)


zipf = zipfile.ZipFile('Python.zip', 'w', zipfile.ZIP_DEFLATED)
zipdir('/tmp/justtest/', zipf)
zipf.close()

1

为了提供更大的灵活性,例如,按名称选择目录/文件,请使用:

import os
import zipfile

def zipall(ob, path, rel=""):
    basename = os.path.basename(path)
    if os.path.isdir(path):
        if rel == "":
            rel = basename
        ob.write(path, os.path.join(rel))
        for root, dirs, files in os.walk(path):
            for d in dirs:
                zipall(ob, os.path.join(root, d), os.path.join(rel, d))
            for f in files:
                ob.write(os.path.join(root, f), os.path.join(rel, f))
            break
    elif os.path.isfile(path):
        ob.write(path, os.path.join(rel, basename))
    else:
        pass

对于文件树:

.
├── dir
   ├── dir2
      └── file2.txt
   ├── dir3
      └── file3.txt
   └── file.txt
├── dir4
   ├── dir5
   └── file4.txt
├── listdir.zip
├── main.py
├── root.txt
└── selective.zip

您可以例如仅选择dir4root.txt

cwd = os.getcwd()
files = [os.path.join(cwd, f) for f in ['dir4', 'root.txt']]

with zipfile.ZipFile("selective.zip", "w" ) as myzip:
    for f in files:
        zipall(myzip, f)

或者只是listdir在脚本调用目录中,然后从此处添加所有内容:

with zipfile.ZipFile("listdir.zip", "w" ) as myzip:
    for f in os.listdir():
        if f == "listdir.zip":
            # Creating a listdir.zip in the same directory
            # will include listdir.zip inside itself, beware of this
            continue
        zipall(myzip, f)

这会压缩,但不会压缩。
亚历克斯

1

假设您要压缩当前目录中的所有文件夹(子目录)。

for root, dirs, files in os.walk("."):
    for sub_dir in dirs:
        zip_you_want = sub_dir+".zip"
        zip_process = zipfile.ZipFile(zip_you_want, "w", zipfile.ZIP_DEFLATED)
        zip_process.write(file_you_want_to_include)
        zip_process.close()

        print("Successfully zipped directory: {sub_dir}".format(sub_dir=sub_dir))

1

为了将文件夹层次结构保留在要归档的父目录下的简洁方法:

import glob
import zipfile

with zipfile.ZipFile(fp_zip, "w", zipfile.ZIP_DEFLATED) as zipf:
    for fp in glob(os.path.join(parent, "**/*")):
        base = os.path.commonpath([parent, fp])
        zipf.write(fp, arcname=fp.replace(base, ""))

如果需要,可以将其更改为pathlib 用于文件globbing


1

这里有这么多答案,我希望我可以为自己的版本做出贡献,该版本基于原始答案(顺便说一句),但具有更多图形化的视角,还为每个zipfile设置和排序使用了上下文os.walk(),以便获得有序输出。

具有这些文件夹及其文件(以及其他文件夹),我想.zip为每个cap_文件夹创建一个:

$ tree -d
.
├── cap_01
|    ├── 0101000001.json
|    ├── 0101000002.json
|    ├── 0101000003.json
|
├── cap_02
|    ├── 0201000001.json
|    ├── 0201000002.json
|    ├── 0201001003.json
|
├── cap_03
|    ├── 0301000001.json
|    ├── 0301000002.json
|    ├── 0301000003.json
| 
├── docs
|    ├── map.txt
|    ├── main_data.xml
|
├── core_files
     ├── core_master
     ├── core_slave

这是我应用的内容,并带有注释,以使您更好地理解该过程。

$ cat zip_cap_dirs.py 
""" Zip 'cap_*' directories. """           
import os                                                                       
import zipfile as zf                                                            


for root, dirs, files in sorted(os.walk('.')):                                                                                               
    if 'cap_' in root:                                                          
        print(f"Compressing: {root}")                                           
        # Defining .zip name, according to Capítulo.                            
        cap_dir_zip = '{}.zip'.format(root)                                     
        # Opening zipfile context for current root dir.                         
        with zf.ZipFile(cap_dir_zip, 'w', zf.ZIP_DEFLATED) as new_zip:          
            # Iterating over os.walk list of files for the current root dir.    
            for f in files:                                                     
                # Defining relative path to files from current root dir.        
                f_path = os.path.join(root, f)                                  
                # Writing the file on the .zip file of the context              
                new_zip.write(f_path) 

基本上,每次迭代过os.walk(path),我打开了情境zipfile设置,之后,迭代循环访问files,这是一个list从文件root目录,形成了基于当前的每个文件的相对路径root的目录,附加到zipfile其运行的背景下。

输出显示如下:

$ python3 zip_cap_dirs.py
Compressing: ./cap_01
Compressing: ./cap_02
Compressing: ./cap_03

要查看每个.zip目录的内容,可以使用以下less命令:

$ less cap_01.zip

Archive:  cap_01.zip
 Length   Method    Size  Cmpr    Date    Time   CRC-32   Name
--------  ------  ------- ---- ---------- ----- --------  ----
  22017  Defl:N     2471  89% 2019-09-05 08:05 7a3b5ec6  cap_01/0101000001.json
  21998  Defl:N     2471  89% 2019-09-05 08:05 155bece7  cap_01/0101000002.json
  23236  Defl:N     2573  89% 2019-09-05 08:05 55fced20  cap_01/0101000003.json
--------          ------- ---                           -------
  67251             7515  89%                            3 files

0

这是使用pathlib和上下文管理器的一种现代方法。将文件直接放在zip中,而不放在子文件夹中。

def zip_dir(filename: str, dir_to_zip: pathlib.Path):
    with zipfile.ZipFile(filename, 'w', zipfile.ZIP_DEFLATED) as zipf:
        # Use glob instead of iterdir(), to cover all subdirectories.
        for directory in dir_to_zip.glob('**'):
            for file in directory.iterdir():
                if not file.is_file():
                    continue
                # Strip the first component, so we don't create an uneeded subdirectory
                # containing everything.
                zip_path = pathlib.Path(*file.parts[1:])
                # Use a string, since zipfile doesn't support pathlib  directly.
                zipf.write(str(file), str(zip_path))

0

我通过将Mark Byers的解决方案与Reimund和Morten Zilmer的注释(相对路径,包括空目录)合并在一起来准备函数。最佳实践with是在ZipFile的文件构造中使用。

该函数还准备一个默认的zip文件名,带有压缩的目录名和'.zip'扩展名。因此,它仅适用于一个参数:要压缩的源目录。

import os
import zipfile

def zip_dir(path_dir, path_file_zip=''):
if not path_file_zip:
    path_file_zip = os.path.join(
        os.path.dirname(path_dir), os.path.basename(path_dir)+'.zip')
with zipfile.ZipFile(path_file_zip, 'wb', zipfile.ZIP_DEFLATED) as zip_file:
    for root, dirs, files in os.walk(path_dir):
        for file_or_dir in files + dirs:
            zip_file.write(
                os.path.join(root, file_or_dir),
                os.path.relpath(os.path.join(root, file_or_dir),
                                os.path.join(path_dir, os.path.pardir)))

0
# import required python modules
# You have to install zipfile package using pip install

import os,zipfile

# Change the directory where you want your new zip file to be

os.chdir('Type your destination')

# Create a new zipfile ( I called it myfile )

zf = zipfile.ZipFile('myfile.zip','w')

# os.walk gives a directory tree. Access the files using a for loop

for dirnames,folders,files in os.walk('Type your directory'):
    zf.write('Type your Directory')
    for file in files:
        zf.write(os.path.join('Type your directory',file))

0

好了,在阅读建议之后,我想到了一种与2.7.x相似的方式,而不创建“有趣的”目录名称(类似绝对的名称),并且只会在zip中创建指定的文件夹。

或者,以防万一您需要您的zip包含一个包含所选目录内容的文件夹。

def zipDir( path, ziph ) :
 """
 Inserts directory (path) into zipfile instance (ziph)
 """
 for root, dirs, files in os.walk( path ) :
  for file in files :
   ziph.write( os.path.join( root, file ) , os.path.basename( os.path.normpath( path ) ) + "\\" + file )

def makeZip( pathToFolder ) :
 """
 Creates a zip file with the specified folder
 """
 zipf = zipfile.ZipFile( pathToFolder + 'file.zip', 'w', zipfile.ZIP_DEFLATED )
 zipDir( pathToFolder, zipf )
 zipf.close()
 print( "Zip file saved to: " + pathToFolder)

makeZip( "c:\\path\\to\\folder\\to\\insert\\into\\zipfile" )

0

创建zip文件的功能。

def CREATEZIPFILE(zipname, path):
    #function to create a zip file
    #Parameters: zipname - name of the zip file; path - name of folder/file to be put in zip file

    zipf = zipfile.ZipFile(zipname, 'w', zipfile.ZIP_DEFLATED)
    zipf.setpassword(b"password") #if you want to set password to zipfile

    #checks if the path is file or directory
    if os.path.isdir(path):
        for files in os.listdir(path):
            zipf.write(os.path.join(path, files), files)

    elif os.path.isfile(path):
        zipf.write(os.path.join(path), path)
    zipf.close()

请举例说明,以便我纠正我的答案
sushh

然而,zip文件(从“目前还不能创建一个加密文件” docs.python.org/3.9/library/zipfile.html
乔治·

0

使用zipfly

import zipfly

paths = [
    {
        'fs': '/path/to/large/file'
    },
]

zfly = zipfly.ZipFly( paths = paths )

with open("large.zip", "wb") as f:
    for i in zfly.generator():
        f.write(i)
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