Python: How to custom order a list?

Question

Obs: I know lists in python are not order-fixed, but think that this one will be. And I'm using Python 2.4

I have a list, like (for example) this one:

mylist = [ ( u'Article', {"...some_data..."}    ) ,
           ( u'Report' , {"...some_data..."}    ) ,
           ( u'Book'   , {"...another_data..."} ) ,
...#continue
]

This variable mylist is obtained from a function, and the 'order' of the list returned will vary. So, sometimes it will be like on the example. Sometimes, the 'Report' will come before 'Article', etc.

I have a fixed order that I want on this list (and isn't the alphabetical).

Let's say that my fixed order is: 'Report', 'Article', 'Book', ...

So, what I want is that: whatever order 'mylist' is instantiated, I want to reorder it making 'Report' stay on front, 'Article' on second, etc...

What's the best approach to reorder my list (taking the first element of the tuple of each item on list) using my 'custom' order?

Answer:

I ended up with this:

mylist became a list of dicts, like this:

mylist = [{'id':'Article', "...some_data..."} ,
        ...etc
]

each dict having a 'id' that had to be sorted.

Saving the correct order on a listAssigning the correct_order on a list:

correct_order = ['Report', 'Article', 'Book', ...]

and doing:

results = sorted([item for item in results], cmp=lambda x,y:cmp(correct_order.index(x['id']), correct_order.index(y['id'])))

Answer 1

You could use a dictionary, that would allow you to access "Book", "Article", etc. without having to care about the order. I would put the data from that list into a dict that look like this:

mydict = { u'Article': "somedata",
           u'Report': "someotherdata", ...}

If you really want to sort your list in the way you described, you can use the list.sort with a key function that represents your particular sort order (Documentation). You need the key function as you need to access only the first element and your sorting order also is not alphabetical.

Answer 2

You could use a dictionary that would map every first element to its "weight" and then check this dictionary inside a sorting function.

Something like:

d = { "Report": 1,
      "Article": 2,
       "Book": 3 }
result = sorted(mylist, key=lambda x:d[x[0]])

Answer 3

This way creates a dict and pulls the items from it in order

mylist = [ ( u'Article', {"...some_data..."}    ) ,
           ( u'Report' , {"...some_data..."}    ) ,
           ( u'Book'   , {"...another_data..."} ) ,
]

mydict = dict(mylist)
ordering = [u'Report', u'Article', u'Book']

print [(k,mydict[k]) for k in ordering]

This way uses sort with O(1) lookups for the ordering

mylist = [ ( u'Article', {"...some_data..."}    ) ,
           ( u'Report' , {"...some_data..."}    ) ,
           ( u'Book'   , {"...another_data..."} ) ,
]

mydict = dict(mylist)
ordering = dict((k,v) for v,k in enumerate([u'Report', u'Article', u'Book']))

print sorted(mydict.items(), key=lambda (k,v): ordering[k])