将文件分解为8192字节的块(或其他128字节的倍数)，并使用update()将它们连续地馈送给MD5。

这利用了MD5有128字节摘要块的事实(8192是128×64)。由于不是将整个文件读入内存，因此使用的内存不会超过8192个字节。

在Python 3.8+中你可以这样做

import hashlib
with open("your_filename.txt", "rb") as f:
    file_hash = hashlib.md5()
    while chunk := f.read(8192):
        file_hash.update(chunk)
print(file_hash.digest())
print(file_hash.hexdigest())  # to get a printable str instead of bytes

Bastien Semene的代码的混合，考虑了Hawkwing关于通用哈希函数的评论…

def hash_for_file(path, algorithm=hashlib.algorithms[0], block_size=256*128, human_readable=True):
    """
    Block size directly depends on the block size of your filesystem
    to avoid performances issues
    Here I have blocks of 4096 octets (Default NTFS)

    Linux Ext4 block size
    sudo tune2fs -l /dev/sda5 | grep -i 'block size'
    > Block size:               4096

    Input:
        path: a path
        algorithm: an algorithm in hashlib.algorithms
                   ATM: ('md5', 'sha1', 'sha224', 'sha256', 'sha384', 'sha512')
        block_size: a multiple of 128 corresponding to the block size of your filesystem
        human_readable: switch between digest() or hexdigest() output, default hexdigest()
    Output:
        hash
    """
    if algorithm not in hashlib.algorithms:
        raise NameError('The algorithm "{algorithm}" you specified is '
                        'not a member of "hashlib.algorithms"'.format(algorithm=algorithm))

    hash_algo = hashlib.new(algorithm)  # According to hashlib documentation using new()
                                        # will be slower then calling using named
                                        # constructors, ex.: hashlib.md5()
    with open(path, 'rb') as f:
        for chunk in iter(lambda: f.read(block_size), b''):
             hash_algo.update(chunk)
    if human_readable:
        file_hash = hash_algo.hexdigest()
    else:
        file_hash = hash_algo.digest()
    return file_hash

如果不阅读完整的内容，就无法获得它的md5。但是您可以使用update函数逐块读取文件的内容。

m.update(一个);M.update (b)等价于M.update (a+b)。

将文件分解为8192字节的块(或其他128字节的倍数)，并使用update()将它们连续地馈送给MD5。

这利用了MD5有128字节摘要块的事实(8192是128×64)。由于不是将整个文件读入内存，因此使用的内存不会超过8192个字节。

在Python 3.8+中你可以这样做

import hashlib
with open("your_filename.txt", "rb") as f:
    file_hash = hashlib.md5()
    while chunk := f.read(8192):
        file_hash.update(chunk)
print(file_hash.digest())
print(file_hash.hexdigest())  # to get a printable str instead of bytes

实现Yuval Adam对Django的回答:

import hashlib
from django.db import models

class MyModel(models.Model):
    file = models.FileField()  # Any field based on django.core.files.File

    def get_hash(self):
        hash = hashlib.md5()
        for chunk in self.file.chunks(chunk_size=8192):
            hash.update(chunk)
        return hash.hexdigest()

以下是我对Piotr Czapla方法的看法:

def md5sum(filename):
    md5 = hashlib.md5()
    with open(filename, 'rb') as f:
        for chunk in iter(lambda: f.read(128 * md5.block_size), b''):
            md5.update(chunk)
    return md5.hexdigest()