How to fix Key error when using a custom dictionary

Question

I'm trying to do the following thing: parse all the files in my music folder and if they are among my favorites, copy them to a separate folder.

In order to achieve this, I'm parsing an xml with my loved tracks and use the info to create a custom dictionary structured like this

lovedSongs[artist][title] = 1

When I'm parsing the files from my folder, I read the song title and artist from the id3 tags and, in theory, I try to see if lovedSongs[tagArtist][tagTitle] is 1

for filename in filelist:     
        if filename[-4:] == ".mp3":
            try:
                id3r = id3reader.Reader(filename)
            except UnicodeDecodeError:
                artist = "None"
                title = "None"
            except IOError:
                artist = "None"
                title = "None"
            else:
                artist = ensureutf8andstring(id3r.getValue('performer'))
                title = ensureutf8andstring(id3r.getValue('title'))

            if artist != "None" or title != "None":
                if lovedSongs[artist][title] == 1:
                    print artist + ' - ' + title

When I try to run the .py file, I get a dictionary Key Error. I found a potential answere here: How to make a python dictionary that returns key for keys missing from the dictionary instead of raising KeyError? and applied it like below, but I still got the Key Error

class SongDict(dict):
    def __init__(self, default=None):
        self.default = default

    def __getitem__(self, key):
        if not self.has_key(key):
            self[key] = self.default()
        return dict.__getitem__(self, key)

    def __missing__(self, key):
        return 0

What would be the best solution? Use try/except perhaps instead of the simple if?

P.S. I think I should mention that I'm a newbie to python, with only a couple of hours of 'experience' and this is my way of trying to learn the language.

Answer 1

It sounds like what you actually want is a simple set

for filename in filelist:     
    if filename.endswith(".mp3"):
        try:
            id3r = id3reader.Reader(filename)
        except (UnicodeDecodeError, IOError):
            pass
        else:
            artist = ensureutf8andstring(id3r.getValue('performer'))
            title = ensureutf8andstring(id3r.getValue('title'))

            if artist != "None" or title != "None":
                if (artist, title) in lovedSongs:
                    print artist + ' - ' + title

Answer 2

One thing you could try is using a defaultdict, with the argument being a defaultdict(dict). Anytime you try to reference a value that doesn't exist, it will not throw an error and will instead return the default null value of what you specify:

In [7]: from collections import defaultdict

In [8]: lovedsongs = defaultdict(lambda: defaultdict(dict))

In [9]: lovedsongs['OAR']['Blackrock'] = 1

In [10]: lovedsongs['Timati']['Moj Put'] = 1

In [11]: lovedsongs
Out[11]: defaultdict(<function <lambda> at 0x24e5ed8>, {'Timati': defaultdict(<type 'dict'>, {'Moj Put': 1}), 'OAR': defaultdict(<type 'dict'>, {'Blackrock': 1})})

In [12]: lovedsongs['Not']['Found']
Out[12]: {}

In [13]: if lovedsongs['Not']['Found'] == 1:
   ....:     print 'Found it'
   ....:
   ....:

In [14]: if lovedsongs['OAR']['Blackrock'] == 1:
   ....:     print 'Found it'
   ....:
   ....:
Found it

The defaultdict(lambda: defaultdict(dict)) syntax is likely better understood by looking at how a single-level (i.e. 'more regular') defaultdict works:

In [1]: from collections import defaultdict

In [2]: d = defaultdict(int)

In [3]: d['one'] = 1

In [4]: d['two'] = 2

In [5]: d
Out[5]: defaultdict(<type 'int'>, {'two': 2, 'one': 1})

In [6]: d['one']
Out[6]: 1

In [7]: d['three']
Out[7]: 0

We declared the defaultdict with an argument of int - this means that if a value is not found, instead of throwing an error, we will just get 0 back (since 0 is the null value of an integer). If we used str, we would get '' back, and if we used dict, we'd get a dictionary back. However the tricky part in this situation is that you have a nested dictionary (defaultdict(dict)), which will run into the same problem if you try to reference a key in the nested dictionary that does not exist:

In [13]: d = defaultdict(dict)

In [14]: d['three']['more']
---------------------------------------------------------------------------
KeyError                                  Traceback (most recent call last)

/path/<ipython console> in <module>()

KeyError: 'more'

So in order to get our nested dictionary to become a defaultdict as well, we have to do a bit of trickery: since the argument to defaultdict is something that needs to be callable (but not yet called -> see int vs. int()), we use a lambda function to somewhat simulate this behavior. At a very simple level, think of the lambda as preventing the defaultdict(dict) from being called until it is needed. Now you will have a dictionary (with a nested dictionary) that will handle missing keys two-levels deep without throwing an error. Hope that explains a bit.