344

我有一个非常大的 4GB 文件,当我尝试读取它时,我的计算机挂起。所以我想一块一块地读取它,在处理完每一块后将处理后的块存储到另一个文件中并读取下一块。

yield这些碎片有什么方法吗?

我很想有一个懒惰的方法

4

12 回答 12

507

要编写惰性函数,只需使用yield

def read_in_chunks(file_object, chunk_size=1024):
    """Lazy function (generator) to read a file piece by piece.
    Default chunk size: 1k."""
    while True:
        data = file_object.read(chunk_size)
        if not data:
            break
        yield data


with open('really_big_file.dat') as f:
    for piece in read_in_chunks(f):
        process_data(piece)

另一种选择是使用iter一个辅助函数:

f = open('really_big_file.dat')
def read1k():
    return f.read(1024)

for piece in iter(read1k, ''):
    process_data(piece)

如果文件是基于行的,则文件对象已经是行的惰性生成器:

for line in open('really_big_file.dat'):
    process_data(line)
于 2009-02-06T09:20:08.293 回答
42

file.readlines()接受一个可选的大小参数,该参数近似于返回的行中读取的行数。

bigfile = open('bigfilename','r')
tmp_lines = bigfile.readlines(BUF_SIZE)
while tmp_lines:
    process([line for line in tmp_lines])
    tmp_lines = bigfile.readlines(BUF_SIZE)
于 2010-01-21T18:27:59.283 回答
41

如果您的计算机、操作系统和 python 是 64 位的,那么您可以使用mmap 模块将文件的内容映射到内存中,并使用索引和切片访问它。这是文档中的一个示例:

import mmap
with open("hello.txt", "r+") as f:
    # memory-map the file, size 0 means whole file
    map = mmap.mmap(f.fileno(), 0)
    # read content via standard file methods
    print map.readline()  # prints "Hello Python!"
    # read content via slice notation
    print map[:5]  # prints "Hello"
    # update content using slice notation;
    # note that new content must have same size
    map[6:] = " world!\n"
    # ... and read again using standard file methods
    map.seek(0)
    print map.readline()  # prints "Hello  world!"
    # close the map
    map.close()

如果您的计算机、操作系统或 python 是 32 位的,那么 mmap-ing 大文件可能会保留大部分地址空间并使的程序内存不足。

于 2009-02-06T09:41:29.467 回答
40

已经有很多很好的答案,但是如果您的整个文件在一行上并且您仍然想处理“行”(而不是固定大小的块),那么这些答案对您没有帮助。

99% 的时间,可以逐行处理文件。然后,如this answer中所建议的,您可以将文件对象本身用作惰性生成器:

with open('big.csv') as f:
    for line in f:
        process(line)

但是,可能会遇到没有行分隔符的非常大的文件'\n'(常见情况是'|')。

  • 转换'|''\n'处理前可能不是一个选项,因为它可能会弄乱可能合法包含的字段'\n'(例如自由文本用户输入)。
  • 使用 csv 库也被排除在外,因为至少在 lib 的早期版本中,它是硬编码来逐行读取输入的

对于这种情况,我创建了以下代码段 [2021 年 5 月更新,适用于 Python 3.8+]:

def rows(f, chunksize=1024, sep='|'):
    """
    Read a file where the row separator is '|' lazily.

    Usage:

    >>> with open('big.csv') as f:
    >>>     for r in rows(f):
    >>>         process(r)
    """
    row = ''
    while (chunk := f.read(chunksize)) != '':   # End of file
        while (i := chunk.find(sep)) != -1:     # No separator found
            yield row + chunk[:i]
            chunk = chunk[i+1:]
            row = ''
        row += chunk
    yield row

[对于旧版本的python]:

def rows(f, chunksize=1024, sep='|'):
    """
    Read a file where the row separator is '|' lazily.

    Usage:

    >>> with open('big.csv') as f:
    >>>     for r in rows(f):
    >>>         process(r)
    """
    curr_row = ''
    while True:
        chunk = f.read(chunksize)
        if chunk == '': # End of file
            yield curr_row
            break
        while True:
            i = chunk.find(sep)
            if i == -1:
                break
            yield curr_row + chunk[:i]
            curr_row = ''
            chunk = chunk[i+1:]
        curr_row += chunk

我能够成功地使用它来解决各种问题。它已经过广泛的测试,具有各种块大小。这是我正在使用的测试套件,供那些需要说服自己的人使用:

test_file = 'test_file'

def cleanup(func):
    def wrapper(*args, **kwargs):
        func(*args, **kwargs)
        os.unlink(test_file)
    return wrapper

@cleanup
def test_empty(chunksize=1024):
    with open(test_file, 'w') as f:
        f.write('')
    with open(test_file) as f:
        assert len(list(rows(f, chunksize=chunksize))) == 1

@cleanup
def test_1_char_2_rows(chunksize=1024):
    with open(test_file, 'w') as f:
        f.write('|')
    with open(test_file) as f:
        assert len(list(rows(f, chunksize=chunksize))) == 2

@cleanup
def test_1_char(chunksize=1024):
    with open(test_file, 'w') as f:
        f.write('a')
    with open(test_file) as f:
        assert len(list(rows(f, chunksize=chunksize))) == 1

@cleanup
def test_1025_chars_1_row(chunksize=1024):
    with open(test_file, 'w') as f:
        for i in range(1025):
            f.write('a')
    with open(test_file) as f:
        assert len(list(rows(f, chunksize=chunksize))) == 1

@cleanup
def test_1024_chars_2_rows(chunksize=1024):
    with open(test_file, 'w') as f:
        for i in range(1023):
            f.write('a')
        f.write('|')
    with open(test_file) as f:
        assert len(list(rows(f, chunksize=chunksize))) == 2

@cleanup
def test_1025_chars_1026_rows(chunksize=1024):
    with open(test_file, 'w') as f:
        for i in range(1025):
            f.write('|')
    with open(test_file) as f:
        assert len(list(rows(f, chunksize=chunksize))) == 1026

@cleanup
def test_2048_chars_2_rows(chunksize=1024):
    with open(test_file, 'w') as f:
        for i in range(1022):
            f.write('a')
        f.write('|')
        f.write('a')
        # -- end of 1st chunk --
        for i in range(1024):
            f.write('a')
        # -- end of 2nd chunk
    with open(test_file) as f:
        assert len(list(rows(f, chunksize=chunksize))) == 2

@cleanup
def test_2049_chars_2_rows(chunksize=1024):
    with open(test_file, 'w') as f:
        for i in range(1022):
            f.write('a')
        f.write('|')
        f.write('a')
        # -- end of 1st chunk --
        for i in range(1024):
            f.write('a')
        # -- end of 2nd chunk
        f.write('a')
    with open(test_file) as f:
        assert len(list(rows(f, chunksize=chunksize))) == 2

if __name__ == '__main__':
    for chunksize in [1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024]:
        test_empty(chunksize)
        test_1_char_2_rows(chunksize)
        test_1_char(chunksize)
        test_1025_chars_1_row(chunksize)
        test_1024_chars_2_rows(chunksize)
        test_1025_chars_1026_rows(chunksize)
        test_2048_chars_2_rows(chunksize)
        test_2049_chars_2_rows(chunksize)
于 2015-06-11T08:23:16.987 回答
13
f = ... # file-like object, i.e. supporting read(size) function and 
        # returning empty string '' when there is nothing to read

def chunked(file, chunk_size):
    return iter(lambda: file.read(chunk_size), '')

for data in chunked(f, 65536):
    # process the data

更新:该方法在https://stackoverflow.com/a/4566523/38592中得到了最好的解释

于 2012-03-31T01:50:18.347 回答
7

参考python的官方文档 https://docs.python.org/3/library/functions.html#iter

也许这种方法更pythonic:

from functools import partial

"""A file object returned by open() is a iterator with
read method which could specify current read's block size"""
with open('mydata.db', 'r') as f_in:

    part_read = partial(f_in.read, 1024*1024)
    iterator = iter(part_read, b'')

    for index, block in enumerate(iterator, start=1):
        block = process_block(block)    # process your block data
        
        with open(f'{index}.txt', 'w') as f_out:
            f_out.write(block)
于 2019-06-23T07:49:25.093 回答
6

在 Python 3.8+中,您可以.read()while循环中使用:

with open("somefile.txt") as f:
    while chunk := f.read(8192):
        do_something(chunk)

当然,你可以使用任何你想要的块大小,你不必使用8192( 2**13) 字节。除非您的文件大小恰好是您的块大小的倍数,否则最后一个块将小于您的块大小。

于 2020-07-20T19:17:31.673 回答
4

我想我们可以这样写:

def read_file(path, block_size=1024): 
    with open(path, 'rb') as f: 
        while True: 
            piece = f.read(block_size) 
            if piece: 
                yield piece 
            else: 
                return

for piece in read_file(path):
    process_piece(piece)
于 2013-11-06T02:15:10.477 回答
2

由于我的声誉低,我不允许发表评论,但是使用 file.readlines([sizehint]) SilentGhosts 解决方案应该更容易

python文件方法

编辑: SilentGhost 是对的,但这应该比:

s = "" 
for i in xrange(100): 
   s += file.next()
于 2009-02-06T10:37:22.397 回答
1

我的情况有点类似。目前尚不清楚您是否知道块大小(以字节为单位);我通常不知道,但所需的记录(行)数是已知的:

def get_line():
     with open('4gb_file') as file:
         for i in file:
             yield i

lines_required = 100
gen = get_line()
chunk = [i for i, j in zip(gen, range(lines_required))]

更新:谢谢 nosklo。这就是我的意思。它几乎可以工作,只是它在“块之间”丢失了一条线。

chunk = [next(gen) for i in range(lines_required)]

这个技巧不会丢失任何线条,但看起来不太好。

于 2009-02-06T10:12:47.327 回答
0

要逐行处理,这是一个优雅的解决方案:

  def stream_lines(file_name):
    file = open(file_name)
    while True:
      line = file.readline()
      if not line:
        file.close()
        break
      yield line

只要没有空行。

于 2012-05-01T23:12:15.620 回答
-2

您可以使用以下代码。

file_obj = open('big_file') 

open() 返回一个文件对象

然后使用 os.stat 获取大小

file_size = os.stat('big_file').st_size

for i in range( file_size/1024):
    print file_obj.read(1024)
于 2015-06-18T13:20:52.373 回答