使用 Python 2.7.5 将文件夹中的所有压缩文件解压缩到同一文件夹 [英] Unzip all zipped files in a folder to that same folder using Python 2.7.5

问题描述

我想编写一个简单的脚本来遍历文件夹中的所有文件，并将那些已压缩 (.zip) 的文件解压缩到同一文件夹中.对于这个项目，我有一个包含近 100 个压缩 .las 文件的文件夹，我希望有一种简单的方法来批量解压缩它们.我尝试使用以下脚本

import os, zipfile文件夹 = 'D:/GISData/LiDAR/SomeFolder'扩展名 = ".zip"对于 os.listdir(folder) 中的项目:如果 item.endswith(extension):zipfile.ZipFile.extract(item)

但是，当我运行脚本时，出现以下错误:

回溯(最近一次调用最后一次):文件D:/GISData/Tools/MO_Tools/BatchUnzip.py"，第 10 行，在 <module> 中提取 = zipfile.ZipFile.extract(item)类型错误:必须使用 ZipFile 实例作为第一个参数调用未绑定的方法 extract()(改为使用 str 实例)

我使用的是 python 2.7.5 解释器.我查看了 zipfile 模块的文档(https://docs.python.org/2/library/zipfile.html#module-zipfile)，我想了解我做错了什么.

在我看来，这个过程应该是这样的:

1. 获取文件夹名称
2. 遍历文件夹并找到 zip 文件
3. 将 zip 文件解压到文件夹

谢谢 Marcus，但是，在实施建议时，我又遇到了另一个错误:

回溯(最近一次调用最后一次):文件D:/GISData/Tools/MO_Tools/BatchUnzip.py"，第 12 行，在 <module> 中zipfile.ZipFile(item).extract()文件C:\Python27\ArcGIS10.2\lib\zipfile.py"，第 752 行，在 __init__ 中self.fp = 打开(文件，modeDict[mode])IOError: [Errno 2] 没有这样的文件或目录:'JeffCity_0752.las.zip'

当我使用打印语句时，我可以看到文件在那里.例如:

 用于 os.listdir(文件夹)中的项目:如果 item.endswith(extension):打印 os.path.abspath(item)文件名 = os.path.basename(item)打印文件名

产量:

D:\GISData\Tools\MO_Tools\JeffCity_0752.las.zipJeffCity_0752.las.zipD:\GISData\Tools\MO_Tools\JeffCity_0753.las.zipJeffCity_0753.las.zip

据我了解文档，

zipfile.ZipFile(file[, mode[, compression[, allowZip64]]])

<块引用>

打开一个 ZIP 文件，其中 file 可以是文件的路径(字符串)或类似文件的对象

在我看来，一切都已存在并已考虑在内.我只是不明白我做错了什么.

有什么建议吗?

谢谢

解决方案

以下是对我有用的代码:

import os, zipfiledir_name = 'C:\\SomeDirectory'扩展名 = ".zip"os.chdir(dir_name) # 将目录从工作目录更改为带有文件的目录for item in os.listdir(dir_name): # 遍历目录中的项目if item.endswith(extension): # 检查.zip"扩展名file_name = os.path.abspath(item) # 获取文件的完整路径zip_ref = zipfile.ZipFile(file_name) # 创建 zipfile 对象zip_ref.extractall(dir_name) # 提取文件到目录zip_ref.close() # 关闭文件os.remove(file_name) # 删除压缩文件

回顾我修改过的代码，目录和脚本的目录混淆了.

以下也有效，同时不会破坏工作目录.先去掉一行

os.chdir(dir_name) # 将目录从工作目录更改为带有文件的目录

然后将 file_name 赋值为

file_name = dir_name + "/" + item

I would like to write a simple script to iterate through all the files in a folder and unzip those that are zipped (.zip) to that same folder. For this project, I have a folder with nearly 100 zipped .las files and I'm hoping for an easy way to batch unzip them. I tried with following script

import os, zipfile

folder = 'D:/GISData/LiDAR/SomeFolder'
extension = ".zip"

for item in os.listdir(folder):
    if item.endswith(extension):
        zipfile.ZipFile.extract(item)

However, when I run the script, I get the following error:

Traceback (most recent call last):
  File "D:/GISData/Tools/MO_Tools/BatchUnzip.py", line 10, in <module>
    extract = zipfile.ZipFile.extract(item)
TypeError: unbound method extract() must be called with ZipFile instance as first argument (got str instance instead)

I am using the python 2.7.5 interpreter. I looked at the documentation for the zipfile module (https://docs.python.org/2/library/zipfile.html#module-zipfile) and I would like to understand what I'm doing incorrectly.

I guess in my mind, the process would go something like this:

1. Get folder name
2. Loop through folder and find zip files
3. Extract zip files to folder

Thanks Marcus, however, when implementing the suggestion, I get another error:

Traceback (most recent call last):
  File "D:/GISData/Tools/MO_Tools/BatchUnzip.py", line 12, in <module>
    zipfile.ZipFile(item).extract()
  File "C:\Python27\ArcGIS10.2\lib\zipfile.py", line 752, in __init__
    self.fp = open(file, modeDict[mode])
IOError: [Errno 2] No such file or directory: 'JeffCity_0752.las.zip'

When I use print statements, I can see that the files are in there. For example:

for item in os.listdir(folder):
    if item.endswith(extension):
        print os.path.abspath(item)
        filename = os.path.basename(item)
        print filename

yields:

D:\GISData\Tools\MO_Tools\JeffCity_0752.las.zip
JeffCity_0752.las.zip
D:\GISData\Tools\MO_Tools\JeffCity_0753.las.zip
JeffCity_0753.las.zip

As I understand the documentation,

zipfile.ZipFile(file[, mode[, compression[, allowZip64]]])

Open a ZIP file, where file can be either a path to a file (a string) or a file-like object

It appears to me like everything is present and accounted for. I just don't understand what I'm doing wrong.

Any suggestions?

Thank You

解决方案

Below is the code that worked for me:

import os, zipfile

dir_name = 'C:\\SomeDirectory'
extension = ".zip"

os.chdir(dir_name) # change directory from working dir to dir with files

for item in os.listdir(dir_name): # loop through items in dir
    if item.endswith(extension): # check for ".zip" extension
        file_name = os.path.abspath(item) # get full path of files
        zip_ref = zipfile.ZipFile(file_name) # create zipfile object
        zip_ref.extractall(dir_name) # extract file to dir
        zip_ref.close() # close file
        os.remove(file_name) # delete zipped file

Looking back at the code I had amended, the directory was getting confused with the directory of the script.

The following also works while not ruining the working directory. First remove the line

os.chdir(dir_name) # change directory from working dir to dir with files

Then assign file_name as

file_name = dir_name + "/" + item

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