在python 2.7中打开文件的正确方法是什么？

Question

I'm trying to open a txt file in IDLE but it gives me an error. I can't figure out what happens to the f in my file name or why the single '\\' becomes double in the error message.

>>>f=open('D:\programs\python 2.7.10\programs\foo.txt','r')

Traceback (most recent call last):
  File "<pyshell#94>", line 1, in <module>
    f=open('D:\programs\python 2.7.10\programs\foo.txt','r')
IOError: [Errno 22] invalid mode ('r') or filename: 'D:\\programs\\python 2.7.10\\programs\x0coo.txt'

Answer 1

Backslashes are used for escape sequences - in your case the culprit is \\f which is the form-feed character. You can also use forward slashes on modern Windows systems as well as an alternative.

Use a raw string:

f=open(r'D:\programs\python 2.7.10\programs\foo.txt','r')

Ideally though, you should use the with statement so that it automatically closes the file in case of exceptions or when the with block exits, eg:

with open(r'D:\programs\python 2.7.10\programs\foo.txt','r') as f:
    # do stuff with `f`

Answer 2

You have a funny character "\\x0c" in your path. Its "f" in hex. Python doesn't understand. That's why ASCII gives an error. Rename your file into something nicer and you will be fine.