print open('file.txt', 'r').read().count("\n") + 1

这个呢

def file_len(fname):
  counts = itertools.count()
  with open(fname) as f: 
    for _ in f: counts.next()
  return counts.next()

这是我用的，看起来很干净:

import subprocess

def count_file_lines(file_path):
    """
    Counts the number of lines in a file using wc utility.
    :param file_path: path to file
    :return: int, no of lines
    """
    num = subprocess.check_output(['wc', '-l', file_path])
    num = num.split(' ')
    return int(num[0])

更新:这比使用纯python略快，但以内存使用为代价。子进程在执行您的命令时将派生一个与父进程具有相同内存占用的新进程。

这是对其他一些答案的元评论。

The line-reading and buffered \n-counting techniques won't return the same answer for every file, because some text files have no newline at the end of the last line. You can work around this by checking the last byte of the last nonempty buffer and adding 1 if it's not b'\n'. In Python 3, opening the file in text mode and in binary mode can yield different results, because text mode by default recognizes CR, LF, and CRLF as line endings (converting them all to '\n'), while in binary mode only LF and CRLF will be counted if you count b'\n'. This applies whether you read by lines or into a fixed-size buffer. The classic Mac OS used CR as a line ending; I don't know how common those files are these days. The buffer-reading approach uses a bounded amount of RAM independent of file size, while the line-reading approach could read the entire file into RAM at once in the worst case (especially if the file uses CR line endings). In the worst case it may use substantially more RAM than the file size, because of overhead from dynamic resizing of the line buffer and (if you opened in text mode) Unicode decoding and storage. You can improve the memory usage, and probably the speed, of the buffered approach by pre-allocating a bytearray and using readinto instead of read. One of the existing answers (with few votes) does this, but it's buggy (it double-counts some bytes). The top buffer-reading answer uses a large buffer (1 MiB). Using a smaller buffer can actually be faster because of OS readahead. If you read 32K or 64K at a time, the OS will probably start reading the next 32K/64K into the cache before you ask for it, and each trip to the kernel will return almost immediately. If you read 1 MiB at a time, the OS is unlikely to speculatively read a whole megabyte. It may preread a smaller amount but you will still spend a significant amount of time sitting in the kernel waiting for the disk to return the rest of the data.

def count_text_file_lines(path):
    with open(path, 'rt') as file:
        line_count = sum(1 for _line in file)
    return line_count

已经有很多答案了，但不幸的是，它们中的大多数只是一个几乎不可优化的问题上的微型经济……

在我参与的几个项目中，行数是软件的核心功能，以最快的速度处理大量文件是至关重要的。

行数的主要瓶颈是I/O访问，因为您需要读取每一行以检测行返回字符，因此没有其他方法。第二个潜在的瓶颈是内存管理:一次加载的内存越多，处理的速度就越快，但与第一个瓶颈相比，这个瓶颈可以忽略不计。

因此，除了禁用gc收集和其他微管理技巧等微小优化外，还有3种主要方法可以减少行计数函数的处理时间:

Hardware solution: the major and most obvious way is non-programmatic: buy a very fast SSD/flash hard drive. By far, this is how you can get the biggest speed boosts. Data preparation solution: if you generate or can modify how the files you process are generated, or if it's acceptable that you can pre-process them, first convert the line return to unix style (\n) as this will save 1 character compared to Windows or MacOS styles (not a big save but it's an easy gain), and secondly and most importantly, you can potentially write lines of fixed length. If you need variable length, you can always pad smaller lines. This way, you can calculate instantly the number of lines from the total filesize, which is much faster to access. Often, the best solution to a problem is to pre-process it so that it better fits your end purpose. Parallelization + hardware solution: if you can buy multiple hard disks (and if possible SSD flash disks), then you can even go beyond the speed of one disk by leveraging parallelization, by storing your files in a balanced way (easiest is to balance by total size) among disks, and then read in parallel from all those disks. Then, you can expect to get a multiplier boost in proportion with the number of disks you have. If buying multiple disks is not an option for you, then parallelization likely won't help (except if your disk has multiple reading headers like some professional-grade disks, but even then the disk's internal cache memory and PCB circuitry will likely be a bottleneck and prevent you from fully using all heads in parallel, plus you have to devise a specific code for this hard drive you'll use because you need to know the exact cluster mapping so that you store your files on clusters under different heads, and so that you can read them with different heads after). Indeed, it's commonly known that sequential reading is almost always faster than random reading, and parallelization on a single disk will have a performance more similar to random reading than sequential reading (you can test your hard drive speed in both aspects using CrystalDiskMark for example).

如果这些都不是选择，那么你只能依靠微观管理技巧来提高行数函数的速度，但不要指望有什么真正重要的东西。相反，您可以预期，与您将看到的速度改进回报相比，您花费在调整上的时间将是不均衡的。