读取文件非常快。读取一个100MB的文件需要不到0.1秒(请参阅我的文章用Python读写文件)。因此，你应该完整地阅读它，然后处理单行。

大多数回答这里做的不是错，而是风格不好。打开文件应该总是用with，因为它可以确保文件再次关闭。

所以你应该这样做:

with open("path/to/file.txt") as f:
    lines = f.readlines()
print(lines[26])  # or whatever you want to do with this line
print(lines[30])  # or whatever you want to do with this line

巨大的文件

如果你有一个巨大的文件，内存消耗是一个问题，你可以逐行处理它:

with open("path/to/file.txt") as f:
    for i, line in enumerate(f):
        pass  # process line i

def getitems(iterable, items):
  items = list(items) # get a list from any iterable and make our own copy
                      # since we modify it
  if items:
    items.sort()
    for n, v in enumerate(iterable):
      if n == items[0]:
        yield v
        items.pop(0)
        if not items:
          break

print list(getitems(open("/usr/share/dict/words"), [25, 29]))
# ['Abelson\n', 'Abernathy\n']
# note that index 25 is the 26th item

f = open(filename, 'r')
totalLines = len(f.readlines())
f.close()
f = open(filename, 'r')

lineno = 1
while lineno < totalLines:
    line = f.readline()

    if lineno == 26:
        doLine26Commmand(line)

    elif lineno == 30:
        doLine30Commmand(line)

    lineno += 1
f.close()

我认为这行得通

 open_file1 = open("E:\\test.txt",'r')
 read_it1 = open_file1.read()
 myline1 = []
 for line1 in read_it1.splitlines():
 myline1.append(line1)
 print myline1[0]

文件对象有一个.readlines()方法，它将为您提供文件内容的列表，每个列表项一行。在此之后，您可以使用普通的列表切片技术。

http://docs.python.org/library/stdtypes.html#file.readlines

如果要读取的文件很大，并且你不想一次在内存中读取整个文件:

fp = open("file")
for i, line in enumerate(fp):
    if i == 25:
        # 26th line
    elif i == 29:
        # 30th line
    elif i > 29:
        break
fp.close()

注意第n行i == n-1。

在Python 2.6或更高版本中:

with open("file") as fp:
    for i, line in enumerate(fp):
        if i == 25:
            # 26th line
        elif i == 29:
            # 30th line
        elif i > 29:
            break