监控 ZIP 文件提取 Python

发布于 2024-09-29 12:53:36 字数 153 浏览 1 评论 0原文

我需要解压缩 .ZIP 存档。我已经知道如何解压缩它，但它是一个巨大的文件，需要一些时间才能解压。如何打印提取完成的百分比？我想要这样的东西：

Extracting File
1% Complete
2% Complete
etc, etc

原文

I need to unzip a .ZIP archive. I already know how to unzip it, but it is a huge file and takes some time to extract. How would I print the percentage complete for the extraction? I would like something like this:

Extracting File
1% Complete
2% Complete
etc, etc

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雨落星ぅ辰 2024-10-06 12:53:37

这是一个您可以开始的示例，它没有经过优化：

import zipfile

zf = zipfile.ZipFile('test.zip')

uncompress_size = sum((file.file_size for file in zf.infolist()))

extracted_size = 0

for file in zf.infolist():
    extracted_size += file.file_size
    print "%s %%" % (extracted_size * 100/uncompress_size)
    zf.extract(file)

为了使其更美观，在打印时执行以下操作：

 print "%s %%\r" % (extracted_size * 100/uncompress_size),

here an example that you can start with, it's not optimized:

import zipfile

zf = zipfile.ZipFile('test.zip')

uncompress_size = sum((file.file_size for file in zf.infolist()))

extracted_size = 0

for file in zf.infolist():
    extracted_size += file.file_size
    print "%s %%" % (extracted_size * 100/uncompress_size)
    zf.extract(file)

to make it more beautiful do this when printing:

 print "%s %%\r" % (extracted_size * 100/uncompress_size),

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梦巷 2024-10-06 12:53:37

您可以使用 tqdm() 监控每个文件的提取进度：

from zipfile import ZipFile
from tqdm import tqdm

# Open your .zip file
with ZipFile(file=path) as zip_file:

    # Loop over each file
    for file in tqdm(iterable=zip_file.namelist(), total=len(zip_file.namelist())):

        # Extract each file to another directory
        # If you want to extract to current working directory, don't specify path
        zip_file.extract(member=file, path=directory)

You can just monitor the progress of each file being extracted with tqdm():

from zipfile import ZipFile
from tqdm import tqdm

# Open your .zip file
with ZipFile(file=path) as zip_file:

    # Loop over each file
    for file in tqdm(iterable=zip_file.namelist(), total=len(zip_file.namelist())):

        # Extract each file to another directory
        # If you want to extract to current working directory, don't specify path
        zip_file.extract(member=file, path=directory)

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御弟哥哥 2024-10-06 12:53:37

在 python 2.6 ZipFile 对象中有一个 open 方法，可以打开一个zip 中的命名文件作为文件对象，您可以起诉它以分块读取数据

import zipfile
import os

def read_in_chunks(zf, name):
    chunk_size= 4096
    f = zf.open(name)
    data_list = []
    total_read = 0
    while 1:
        data = f.read(chunk_size)
        total_read += len(data)
        print "read",total_read
        if not data:
            break
        data_list.append(data)

    return "".join(data_list)

zip_file_path = r"C:\Users\anurag\Projects\untitled-3.zip"
zf = zipfile.ZipFile(zip_file_path, "r")
for name in zf.namelist():
    data = read_in_chunks(zf, name)

编辑：要获取总大小，您可以执行类似的操作

total_size = sum((file.file_size for file in zf.infolist()))

现在您可以打印总进度和每个文件的进度，例如假设您只有 1 zip 中的大文件，其他方法（例如仅计算文件大小并提取）根本不会提供任何进展。

In python 2.6 ZipFile object has a open method which can open a named file in zip as a file object, you can sue that to read data in chunks

import zipfile
import os

def read_in_chunks(zf, name):
    chunk_size= 4096
    f = zf.open(name)
    data_list = []
    total_read = 0
    while 1:
        data = f.read(chunk_size)
        total_read += len(data)
        print "read",total_read
        if not data:
            break
        data_list.append(data)

    return "".join(data_list)

zip_file_path = r"C:\Users\anurag\Projects\untitled-3.zip"
zf = zipfile.ZipFile(zip_file_path, "r")
for name in zf.namelist():
    data = read_in_chunks(zf, name)

Edit: To get the total size you can do something like this

total_size = sum((file.file_size for file in zf.infolist()))

So now you can print the total progress and progress per file, e.g. suppose you have only 1 big file in zip, other methods(e.g. just counting file sizes and extract) will not give any progress at all.

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