从某一行读出:

n = 4   # for reading from 5th line
with open("write.txt",'r') as t:
     for i,line in enumerate(t):
         if i >= n:             # i == n-1 for nth line
            print(line)

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

你可以用已经有人提到过的语法很简单地做到这一点，但这是迄今为止最简单的方法:

inputFile = open("lineNumbers.txt", "r")
lines = inputFile.readlines()
print (lines[0])
print (lines[2])

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()

其中一些很可爱，但可以做得更简单:

start = 0 # some starting index
end = 5000 # some ending index
filename = 'test.txt' # some file we want to use

with open(filename) as fh:
    data = fin.readlines()[start:end]

print(data)

这将使用简单的列表切片，它加载整个文件，但大多数系统将适当地最小化内存使用，它比上面给出的大多数方法都快，并且适用于我的10G+数据文件。好运！

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]