我建议这个解决方案。

with open('your_file.txt', 'w') as f:        
    list(map(lambda item : f.write("%s\n" % item),my_list))   

序列化列表到文本文件与逗号分隔值

mylist = dir()
with open('filename.txt','w') as f:
    f.write( ','.join( mylist ) )

poem = '''\
Programming is fun
When the work is done
if you wanna make your work also fun:
use Python!
'''
f = open('poem.txt', 'w') # open for 'w'riting
f.write(poem) # write text to file
f.close() # close the file

How It Works: First, open a file by using the built-in open function and specifying the name of the file and the mode in which we want to open the file. The mode can be a read mode (’r’), write mode (’w’) or append mode (’a’). We can also specify whether we are reading, writing, or appending in text mode (’t’) or binary mode (’b’). There are actually many more modes available and help(open) will give you more details about them. By default, open() considers the file to be a ’t’ext file and opens it in ’r’ead mode. In our example, we first open the file in write text mode and use the write method of the file object to write to the file and then we finally close the file.

上面的例子来自Swaroop C H编写的《A Byte of Python》一书。 swaroopch.com

我认为探索使用genexp的好处会很有趣，所以下面是我的看法。

问题中的例子使用方括号来创建一个临时列表，因此相当于:

file.writelines( list( "%s\n" % item for item in list ) )

这将不必要地构造一个包含所有将被写入的行的临时列表，这可能会消耗大量的内存，这取决于列表的大小以及str(item)的输出有多详细。

去掉方括号(相当于去掉上面的包装列表()调用)将会传递一个临时生成器给file.writelines():

file.writelines( "%s\n" % item for item in list )

该生成器将按需创建以换行符结束的项目对象表示(即当它们被写入时)。这很好，有几个原因:

内存开销很小，即使对于非常大的列表也是如此 如果str(item)很慢，则在处理每个项时，文件中都有可见的进展

这避免了内存问题，例如:

In [1]: import os

In [2]: f = file(os.devnull, "w")

In [3]: %timeit f.writelines( "%s\n" % item for item in xrange(2**20) )
1 loops, best of 3: 385 ms per loop

In [4]: %timeit f.writelines( ["%s\n" % item for item in xrange(2**20)] )
ERROR: Internal Python error in the inspect module.
Below is the traceback from this internal error.

Traceback (most recent call last):
...
MemoryError

(我通过限制Python的max触发了此错误。使用ulimit -v 102400虚拟内存到~100MB)。

把内存使用放在一边，这个方法实际上并不比原来的方法快:

In [4]: %timeit f.writelines( "%s\n" % item for item in xrange(2**20) )
1 loops, best of 3: 370 ms per loop

In [5]: %timeit f.writelines( ["%s\n" % item for item in xrange(2**20)] )
1 loops, best of 3: 360 ms per loop

(Python 2.6.2 on Linux)

你打算怎么处理这个文件?这个文件是否存在于人类，或其他具有明确互操作性要求的程序?

如果您只是试图将列表序列化到磁盘，以便稍后由同一python应用程序使用，则应该pickle该列表。

import pickle

with open('outfile', 'wb') as fp:
    pickle.dump(itemlist, fp)

再读一遍:

with open ('outfile', 'rb') as fp:
    itemlist = pickle.load(fp)

在Python 3中，你可以使用print和*进行参数解包:

with open("fout.txt", "w") as fout:
    print(*my_list, sep="\n", file=fout)