我们有一个很大的原始数据文件,我们想把它修剪成指定的大小。

如何在python中获取文本文件的前N行?所使用的操作系统对实现有任何影响吗?


当前回答

这适用于Python 2和3:

from itertools import islice

with open('/tmp/filename.txt') as inf:
    for line in islice(inf, N, N+M):
        print(line)

其他回答

Python 3:

with open("datafile") as myfile:
    head = [next(myfile) for x in range(N)]
print(head)

Python 2:

with open("datafile") as myfile:
    head = [next(myfile) for x in xrange(N)]
print head

下面是另一种方法(Python 2和3都是):

from itertools import islice

with open("datafile") as myfile:
    head = list(islice(myfile, N))
print(head)
N = 10
with open("file.txt", "a") as file:  # the a opens it in append mode
    for i in range(N):
        line = next(file).strip()
        print(line)

这里有另一个不错的解决方案与列表理解:

file = open('file.txt', 'r')

lines = [next(file) for x in range(3)]  # first 3 lines will be in this list

file.close()

我想通过读取整个文件来处理小于n行的文件

def head(filename: str, n: int):
    try:
        with open(filename) as f:
            head_lines = [next(f).rstrip() for x in range(n)]
    except StopIteration:
        with open(filename) as f:
            head_lines = f.read().splitlines()
    return head_lines

这要归功于约翰·拉·鲁伊和伊莲·伊利耶夫。使用异常句柄函数以获得最佳性能

修改1:感谢FrankM的反馈,处理文件存在和读取权限我们可以进一步增加

import errno
import os

def head(filename: str, n: int):
    if not os.path.isfile(filename):
        raise FileNotFoundError(errno.ENOENT, os.strerror(errno.ENOENT), filename)  
    if not os.access(filename, os.R_OK):
        raise PermissionError(errno.EACCES, os.strerror(errno.EACCES), filename)     
   
    try:
        with open(filename) as f:
            head_lines = [next(f).rstrip() for x in range(n)]
    except StopIteration:
        with open(filename) as f:
            head_lines = f.read().splitlines()
    return head_lines

您可以使用第二个版本,也可以使用第一个版本,稍后再处理文件异常。从性能的角度来看,检查是快速的,而且大部分是免费的

这适用于Python 2和3:

from itertools import islice

with open('/tmp/filename.txt') as inf:
    for line in islice(inf, N, N+M):
        print(line)