显而易见的方法是使用shutil，就像第二个顶级答案所说的那样，但如果出于某种原因，您仍然希望使用ZipFile，并且如果您在执行此操作时遇到一些问题（如Windows等中的ERR 13），您可以使用此修复程序：

import os
import zipfile

def retrieve_file_paths(dirName):
  filePaths = []
  for root, directories, files in os.walk(dirName):
    for filename in files:
        filePath = os.path.join(root, filename)
        filePaths.append(filePath)
  return filePaths
 
def main(dir_name, output_filename):
  filePaths = retrieve_file_paths(dir_name)
   
  zip_file = zipfile.ZipFile(output_filename+'.zip', 'w')
  with zip_file:
    for file in filePaths:
      zip_file.write(file)

main("my_dir", "my_dir_archived")

该方法递归地遍历给定文件夹中的每个子文件夹/文件，并将它们写入zip文件，而不是尝试直接压缩文件夹。

显而易见的方法是使用shutil，就像第二个顶级答案所说的那样，但如果出于某种原因，您仍然希望使用ZipFile，并且如果您在执行此操作时遇到一些问题（如Windows等中的ERR 13），您可以使用此修复程序：

import os
import zipfile

def retrieve_file_paths(dirName):
  filePaths = []
  for root, directories, files in os.walk(dirName):
    for filename in files:
        filePath = os.path.join(root, filename)
        filePaths.append(filePath)
  return filePaths
 
def main(dir_name, output_filename):
  filePaths = retrieve_file_paths(dir_name)
   
  zip_file = zipfile.ZipFile(output_filename+'.zip', 'w')
  with zip_file:
    for file in filePaths:
      zip_file.write(file)

main("my_dir", "my_dir_archived")

该方法递归地遍历给定文件夹中的每个子文件夹/文件，并将它们写入zip文件，而不是尝试直接压缩文件夹。

要提供更大的灵活性，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

例如，您可以只选择dir4和root.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)

或者只需在脚本调用目录中列出目录，然后从中添加所有内容：

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)

您可能想看看zipfile模块；文档位于http://docs.python.org/library/zipfile.html.

您可能还需要os.walk（）来索引目录结构。

这是一种现代方法，使用pathlib和上下文管理器。将文件直接放在zip文件中，而不是放在子文件夹中。

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))

试试下面的一个。它对我有用。

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()