如何以最有效的内存和时间方式获取大文件的行数?

def file_len(filename):
    with open(filename) as f:
        for i, _ in enumerate(f):
            pass
    return i + 1

当前回答

一句话解决方案:

import os
os.system("wc -l  filename")  

我的代码片段:

>>> os.system('wc -l *.txt')

0 bar.txt
1000 command.txt
3 test_file.txt
1003 total

其他回答

我使用的最简单和最短的方法是:

f = open("my_file.txt", "r")
len(f.readlines())

为什么不读取前100行和后100行,然后估计平均行长,然后用这些数字除以总文件大小呢?如果你不需要一个确切的值,这可以工作。

大文件的另一种选择是使用xreadlines():

count = 0
for line in open(thefilepath).xreadlines(  ): count += 1

对于Python 3,请参阅:在Python 3中什么替代xreadlines() ?

下面是一个python程序,使用多处理库将行计数分布到不同的机器/核。使用8核windows 64服务器,我的测试将一个2000万行文件的计数从26秒提高到7秒。注意:不使用内存映射会使运行速度变慢。

import multiprocessing, sys, time, os, mmap
import logging, logging.handlers

def init_logger(pid):
    console_format = 'P{0} %(levelname)s %(message)s'.format(pid)
    logger = logging.getLogger()  # New logger at root level
    logger.setLevel( logging.INFO )
    logger.handlers.append( logging.StreamHandler() )
    logger.handlers[0].setFormatter( logging.Formatter( console_format, '%d/%m/%y %H:%M:%S' ) )

def getFileLineCount( queues, pid, processes, file1 ):
    init_logger(pid)
    logging.info( 'start' )

    physical_file = open(file1, "r")
    #  mmap.mmap(fileno, length[, tagname[, access[, offset]]]

    m1 = mmap.mmap( physical_file.fileno(), 0, access=mmap.ACCESS_READ )

    #work out file size to divide up line counting

    fSize = os.stat(file1).st_size
    chunk = (fSize / processes) + 1

    lines = 0

    #get where I start and stop
    _seedStart = chunk * (pid)
    _seekEnd = chunk * (pid+1)
    seekStart = int(_seedStart)
    seekEnd = int(_seekEnd)

    if seekEnd < int(_seekEnd + 1):
        seekEnd += 1

    if _seedStart < int(seekStart + 1):
        seekStart += 1

    if seekEnd > fSize:
        seekEnd = fSize

    #find where to start
    if pid > 0:
        m1.seek( seekStart )
        #read next line
        l1 = m1.readline()  # need to use readline with memory mapped files
        seekStart = m1.tell()

    #tell previous rank my seek start to make their seek end

    if pid > 0:
        queues[pid-1].put( seekStart )
    if pid < processes-1:
        seekEnd = queues[pid].get()

    m1.seek( seekStart )
    l1 = m1.readline()

    while len(l1) > 0:
        lines += 1
        l1 = m1.readline()
        if m1.tell() > seekEnd or len(l1) == 0:
            break

    logging.info( 'done' )
    # add up the results
    if pid == 0:
        for p in range(1,processes):
            lines += queues[0].get()
        queues[0].put(lines) # the total lines counted
    else:
        queues[0].put(lines)

    m1.close()
    physical_file.close()

if __name__ == '__main__':
    init_logger( 'main' )
    if len(sys.argv) > 1:
        file_name = sys.argv[1]
    else:
        logging.fatal( 'parameters required: file-name [processes]' )
        exit()

    t = time.time()
    processes = multiprocessing.cpu_count()
    if len(sys.argv) > 2:
        processes = int(sys.argv[2])
    queues=[] # a queue for each process
    for pid in range(processes):
        queues.append( multiprocessing.Queue() )
    jobs=[]
    prev_pipe = 0
    for pid in range(processes):
        p = multiprocessing.Process( target = getFileLineCount, args=(queues, pid, processes, file_name,) )
        p.start()
        jobs.append(p)

    jobs[0].join() #wait for counting to finish
    lines = queues[0].get()

    logging.info( 'finished {} Lines:{}'.format( time.time() - t, lines ) )

凯尔的回答

num_lines = sum(1 for line in open('my_file.txt'))

最好的替代方案是什么

num_lines =  len(open('my_file.txt').read().splitlines())

这里是两者的性能比较

In [20]: timeit sum(1 for line in open('Charts.ipynb'))
100000 loops, best of 3: 9.79 µs per loop

In [21]: timeit len(open('Charts.ipynb').read().splitlines())
100000 loops, best of 3: 12 µs per loop