Python: How to custom order a list?

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]])

More generally, there could be elements of the mylist that are not in the specified fixed order. This will order according to the rule, but leave alone the relative order of everything outside of the rule:

def orderListByRule(alist,orderRule,listKeys=None,dropIfKey=None):
    ###
    #######################################################################################
    """ Reorder alist according to the order specified in orderRule. The orderRule lists the order to be imposed on a set of keys. The keys are alist, if listkeys==None, or listkeys otherwise.  That is, the length of listkeys must be the same as of alist. That is, listkeys are the tags on alist which determine the ordering.  orderRule is a list of those same keys and maybe more which specifies the desired ordering.
    There is an optional dropIfKey which lists keys of items that should be dropped outright.
    """
    maxOR = len(orderRule)
    orDict = dict(zip(orderRule, range(maxOR)))
    alDict = dict(zip(range(maxOR, maxOR+len(alist)),
                      zip(alist if listKeys is None else listKeys, alist)))
    outpairs = sorted(  [[orDict.get(b[0],a),(b)] for a,b in alDict.items()]  )
    if dropIfKey is None: dropIfKey=[]
    outL = [b[1] for a,b in outpairs if b[0] not in dropIfKey]
    return outL

def test_orderListByRule():
    L1 = [1,2,3,3,5]
    L2 = [3,4,5,10]
    assert orderListByRule(L1, L2) == [3, 3, 5, 1, 2]
    assert orderListByRule(L1, L2, dropIfKey=[2,3]) == [5, 1,]
    Lv = [c for c in 'abcce']
    assert orderListByRule(Lv, L2, listKeys=L1) == ['c', 'c', 'e', 'a', 'b']
    assert orderListByRule(Lv, L2, listKeys=L1, dropIfKey=[2,3]) == ['e','a']

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.

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])