I'm having trouble opening a Python file :(

I saved a file as DictionaryE.txt in a Modules folder I created within Python. Then I type:

fh = open("Dic开发者_如何学运维tionaryE.txt")

I get this error message:

Traceback (most recent call last):
  File "<pyshell#17>", line 1, in <module>
    fh = open("DictionaryE.txt")
IOError: [Errno 2] No such file or directory: 'DictionaryE.txt'

What am I doing wrong? Could someone please describe a specific, detailed step-by-step instruction on what to do? Thanks.

As other answers suggested, you need to specify the file's path, not just the name.

For example, if you know the file is in C:\Blah\Modules, use

fh = open('c:/Blah/Modules/DictionaryE.txt')

Note that I've turned the slashes "the right way up" (Unix-style;-) rather than "the Windows way". That's optional, but Python (and actually the underlying C runtime library) are perfectly happy with it, and it saves you trouble on many occasions (since \, in Python string literals just as in C ones, is an "escape marker", once in a while, if you use it, the string value you've actually entered is not the one you think -- with '/' instead, zero problems).

Use the full path to the file? You are trying to open the file in the current working directory.

probably something like:

import os
dict_file = open(os.path.join(os.path.dirname(__file__), 'Modules', 'DictionaryE.txt'))

It's hard to know without knowing your project structure and the context of your code. Fwiw, when you just "open" a file, it will be looking in whatever directory you're running the python program in, and __file__is the full path to ... the python file.

To complement Alex's answer, you can be more specific and explicit with what you want to do with DictionaryE.txt. The basics:

READ (this is default):

fh = open("C:/path/to/DictionaryE.txt", "r")

WRITE:

fh = open("C:/path/to/DictionaryE.txt", "w")

APPEND:

fh = open("C:/path/to/DictionaryE.txt", "a")

More info can be found here: Built-in Functions - open()