从 zip 打开文件而不用 Python 解压？

Question

我正在编写一个脚本，该脚本使用 tje 请求库从 URL 获取 zip 文件。该 zip 文件包含一个 csv 文件。我正在尝试读取该 csv 文件而不保存它。但是在解析时它给了我这个错误：_csv.Error: iterator should return strings, not bytes (did you open the file in text mode?)

import csv
import requests
from io import BytesIO, StringIO
from zipfile import ZipFile

response = requests.get(url)
zip_file = ZipFile(BytesIO(response.content))
files = zip_file.namelist()
with zip_file.open(files[0]) as csvfile:
    csvreader = csv.reader(csvfile)

    # _csv.Error: iterator should return strings, not bytes (did you open the file in text mode?)

    for row in csvreader:
        print(row)

Answer 1

试试这个：

import pandas as pd
import requests
from io import BytesIO, StringIO
from zipfile import ZipFile

response = requests.get(url)
zip_file = ZipFile(BytesIO(response.content))
files = zip_file.namelist()
with zip_file.open(files[0]) as csvfile:   
    print(pd.read_csv(csvfile, encoding='utf8', sep=","))

Answer 2

正如@Aran-Fey 所暗示的那样：

import zipfile
import csv
import io

with open('/path/to/archive.zip', 'r') as f:
    with zipfile.ZipFile(f) as zf:
        csv_filename = zf.namelist()[0]  # see namelist() for the list of files in the archive
        with zf.open(csv_filename) as csv_f:
            csv_f_as_text = io.TextIOWrapper(csv_f)
            reader = csv.reader(csv_f_as_text)

csv.reader（和csv.DictReader）需要以文本模式打开的类似文件的对象。通常，当在'r' 模式下简单地使用open(...)ing 文件时，这不是问题，正如Python 3 文档所说，文本模式是默认模式：“默认模式是'r'（打开以读取文本，'rt 的同义词')”。但是如果你在ZipFile 上尝试rt 和open，你会看到一个错误：ZipFile.open() requires mode "r" or "w":

        with zf.open(csv_filename, 'rt') as csv_f:
---------------------------------------------------------------------------
ValueError                                Traceback (most recent call last)
...    
ValueError: open() requires mode "r" or "w"

这就是io.TextIOWrapper 的用途——将字节流包装成文本可读，动态解码它们。