Extract files from zip without keeping the structure using python ZipFile?

It is possible to iterate over the ZipFile.infolist(). On the returned ZipInfo objects you can then manipulate the filename to remove the directory part and finally extract it to a specified directory.

import zipfile
import os

my_dir = "D:\\Download\\"
my_zip = "D:\\Download\\my_file.zip"

with zipfile.ZipFile(my_zip) as zip:
    for zip_info in zip.infolist():
        if zip_info.filename[-1] == '/':
            continue
        zip_info.filename = os.path.basename(zip_info.filename)
        zip.extract(zip_info, my_dir)

Just extract to bytes in memory,compute the filename, and write it there yourself, instead of letting the library do it - -mostly, just use the "read()" instead of "extract()" method:

Python 3.6+ update(2020) - the same code from the original answer, but using pathlib.Path, which ease file-path manipulation and other operations (like "write_bytes")

from pathlib import Path
import zipfile
import os

my_dir = Path("D:\\Download\\")
my_zip = my_dir / "my_file.zip"

zip_file = zipfile.ZipFile(my_zip, 'r')
for files in zip_file.namelist():
    data = zip_file.read(files, my_dir)
    myfile_path = my_dir / Path(files.filename).name
    myfile_path.write_bytes(data)
zip_file.close()

Original code in answer without pathlib:

import zipfile
import os

my_dir = "D:\\Download\\"
my_zip = "D:\\Download\\my_file.zip"

zip_file = zipfile.ZipFile(my_zip, 'r')
for files in zip_file.namelist():
    data = zip_file.read(files, my_dir)
    # I am almost shure zip represents directory separator
    # char as "/" regardless of OS, but I  don't have DOS or Windos here to test it
    myfile_path = os.path.join(my_dir, files.split("/")[-1])
    myfile = open(myfile_path, "wb")
    myfile.write(data)
    myfile.close()
zip_file.close()

This opens file handles of members of the zip archive, extracts the filename and copies it to a target file (that's how ZipFile.extract works, without taking care of subdirectories).

import os
import shutil
import zipfile

my_dir = r"D:\Download"
my_zip = r"D:\Download\my_file.zip"

with zipfile.ZipFile(my_zip) as zip_file:
    for member in zip_file.namelist():
        filename = os.path.basename(member)
        # skip directories
        if not filename:
            continue
    
        # copy file (taken from zipfile's extract)
        source = zip_file.open(member)
        target = open(os.path.join(my_dir, filename), "wb")
        with source, target:
            shutil.copyfileobj(source, target)

A similar concept to the solution of Gerhard Götz, but adapted for extracting single files instead of the entire zip:

with ZipFile(zipPath, 'r') as zipObj:
    zipInfo = zipObj.getinfo(path_in_zip))
    zipInfo.filename = os.path.basename(destination)
    zipObj.extract(zipInfo, os.path.dirname(os.path.realpath(destination)))