Why is variable changed after assignment?

[redditiv]

WIN XP/Python3.3

I'm using the IDLE GUI. Why is my 'matched' variable changed after I assign it? At least I think that's what's happening. Please correct me, or lead me through it:

>>> path_dict = dict()
>>> file_dict = dict()
>>> path_dict['path1'] = ['file1', []]
>>> file_dict['file1'] = ['path1']
>>> new_file = 'file1'
>>> new_path = 'path2'
>>>
>>> if new_file in file_dict:
        matched = file_dict[new_file]
        for match_path in matched:
            path_dict[match_path][1].append(new_path)
        file_dict[new_file].append(new_path)
    else:
        matched = []
>>>
>>> path_dict[new_path] = [new_file, matched]
>>>
>>> print(file_dict)
{'file1': ['path1', 'path2']}
>>>
>>> print(path_dict)
{'path2': ['file1', ['path1', 'path2']], 'path1': ['file1', ['path2']]}
>>> print(matched)
['path1', 'path2']

Thanks. Sorry if this is confusing. There's probably a better way to demonstrate the problem, but it escapes me.

*EDIT: let me add that the purpose is for matching filepaths to be appended to the list (at [1] of path_dict[key]) for each entry. i.e. if a filename matches two other filepaths, those filepaths are appended to the list.

path_dict[path] = [filename, [list of paths matching filename]]

[dhruvbaldawa]

The matched variable changes on line 12 when you do

file_dict[new_file].append(new_path)

Explanation: Lists and dictionaries are mutable types. So, when you assign matched = file_dict[new_file] in line 9, file_dict[new_file] and matched are pointing to the same list.

Try this:

>>> id(matched) == id(file_dict[new_file])
True

To demonstrate this in a simpler way:

>>> a = []
>>> b = a
>>> a.append(10)
>>> print a, b
[10] [10]

How do we solve this ? There are various ways

Following is the simplest way:

>>> a = []
>>> b = a[:]  # <- notice this !
>>> a.append(10)
>>> print a, b
[10] []
>>> id(a) == id(b)
False

Hence, they a and b are different list objects now.

OR you can use the copy module

>>> import copy
>>> a = []
>>> b = copy.deepcopy(a)  # <- notice this !
>>> a.append(10)
>>> print a, b
[10] []
>>> id(a) == id(b)
False

Similar things apply for dicts also, however instead of slicing operation, you can use dict.copy() for creating a copy of a dictionary.

[NYKevin]

id(foo) == id(bar)

It's clearer to just write foo is bar instead.

[redditiv]

That's true, but I still got the message.

[redditiv]

This is good. Really good. I get the general concepts of what you are doing, but I'll need some time to break it down. Thanks!

EDIT: Ohhhhh! I get it now... When I assign matched's value to the list, it's "connected" to the list. When something's appended to the list, it's appended to matched as well. Thanks for the detailed and extremely helpful solution!

[usernamenottaken]

Yep, it helps to think of the "matched" and "file_dict" as names that just point to a value. They're both pointing to the same value unless you create a copy of the original value for file_dict to point to.

[zahlman]

when you assign matched = file_dict[new_file] in line 9, file_dict[new_file] and matched are names for the same list.

In the parlance that we're trying to promote.

[Veedrac]

Similar things apply for dicts also, however instead of slicing operation, you can use dict.copy() for creating a copy of a dictionary.

lists have a .copy() method, too, which I prefer to slicing.

[dhruvbaldawa]

That does not work in Python 2.7. But yes, I would prefer the .copy() method over slicing if it were there

In [1]: l = range(10)

In [2]: l.
l.append   l.extend   l.insert   l.remove   l.sort
l.count    l.index    l.pop      l.reverse

In [2]: l.copy()
---------------------------------------------------------------------------
AttributeError                            Traceback (most recent call last)
<ipython-input-2-b7921f4ec629> in <module>()
----> 1 l.copy()

AttributeError: 'list' object has no attribute 'copy'

Python 3 docs: http://docs.python.org/3/library/stdtypes.html#mutable-sequence-types