使用python3.9、pathlib&zipfile模块，您可以从系统中的任何位置创建zip文件。

def zip_dir(dir: Union[Path, str], filename: Union[Path, str]):
    """Zip the provided directory without navigating to that directory using `pathlib` module"""

    # Convert to Path object
    dir = Path(dir)

    with zipfile.ZipFile(filename, "w", zipfile.ZIP_DEFLATED) as zip_file:
        for entry in dir.rglob("*"):
            zip_file.write(entry, entry.relative_to(dir))

它整洁、打字，代码更少。

使用pathlib.Path的解决方案，它独立于所使用的操作系统：

import zipfile
from pathlib import Path

def zip_dir(path: Path, zip_file_path: Path):
    """Zip all contents of path to zip_file"""
    files_to_zip = [
        file for file in path.glob('*') if file.is_file()]
    with zipfile.ZipFile(
        zip_file_path, 'w', zipfile.ZIP_DEFLATED) as zip_f:
        for file in files_to_zip:
            print(file.name)
            zip_f.write(file, file.name)

current_dir = Path.cwd()  
zip_dir = current_dir / "test"
tools.zip_dir(
    zip_dir, current_dir / 'Zipped_dir.zip')

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)

要向生成的zip文件添加压缩，请查看此链接。

您需要更改：

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

to

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

要将mydirectory的内容添加到新的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()

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。