Enforce items at beginning and end of list

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How can I modify this list so that all p's appear at the beginning, the q's at the end, and the values in between are sorted alphabetically?

l = ['f','g','p','a','p','c','b','q','z','n','d','t','q']

So I would like to have:


You can use sorted with the following key:

sorted(l, key = lambda s: (s!='p', s=='q', s))
['p', 'p', 'a', 'b', 'c', 'd', 'f', 'g', 'n', 't', 'z', 'q', 'q']


To get a better idea of how this is working, the following list comprehension aims to replicate what is being returned from the lambda function defined in the key argument prior to making comparisons:

t = [(s!='p', s=='q', s) for s in pl]

[(True, False, 'f'),
 (True, False, 'g'),
 (False, False, 'p'),
 (True, False, 'a'),
 (False, False, 'p'),
 (True, False, 'c'),
 (True, False, 'b'),
 (True, True, 'q'),
 (True, False, 'z'),
 (True, False, 'n'),
 (True, False, 'd'),
 (True, False, 't'),
 (True, True, 'q')]

This will then be the key to be used to sort the items in the list, as mentioned in the documentation:

The value of the key parameter should be a function that takes a single argument and returns a key to use for sorting purposes.

So taking into account that False = 0 and True = 1, when this list of tuples is sorted the result will be the following:

[(False, False, 'p'),
 (False, False, 'p'),
 (True, False, 'a'),
 (True, False, 'b'),
 (True, False, 'c'),
 (True, False, 'd'),
 (True, False, 'f'),
 (True, False, 'g'),
 (True, False, 'n'),
 (True, False, 't'),
 (True, False, 'z'),
 (True, True, 'q'),
 (True, True, 'q')]

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One idea is to use a priority dictionary with a custom function. This is naturally extendable should you wish to include additional criteria.

L = ['f','g','p','a','p','c','b','q','z','n','d','t','q']

def sort_func(x):
    priority = {'p': 0, 'q': 2}
    return priority.get(x, 1), x

res = sorted(L, key=sort_func)


['p', 'p', 'a', 'b', 'c', 'd', 'f', 'g', 'n', 't', 'z', 'q', 'q']

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Use the key parameter in sorted:

l = ['f','g','p','a','p','c','b','q','z','n','d','t','q']

def key(c):
    if c == 'q':
        return (2, c)
    elif c == 'p':
        return (0, c)
    return (1, c)

result = sorted(l, key=key)


['p', 'p', 'a', 'b', 'c', 'd', 'f', 'g', 'n', 't', 'z', 'q', 'q']

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Just define an appropriate key function:

>>> def _key(x):
...     if x == 'p':
...         return -1
...     elif x == 'q':
...         return float('inf')
...     else:
...         return ord(x)
>>> l = ['f','g','p','a','p','c','b','q','z','n','d','t','q']
>>> sorted(l, key=_key)
['p', 'p', 'a', 'b', 'c', 'd', 'f', 'g', 'n', 't', 'z', 'q', 'q']

Note, every character is mapped to an integer >= 0, so we can just rely on ord, and since -1 will always be less than anything returned by ord, we can use that for p, and for q, we can use infinity, so it will be alway greater than something returned by ord.

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You can find all p and q elements, filter the original list, and then sort:

l = ['f','g','p','a','p','c','b','q','z','n','d','t','q']
_ps, _qs = [i for i in l if i == 'p'], [i for i in l if i == 'q']
new_l = _ps+sorted(filter(lambda x:x not in {'q', 'p'}, l))+_qs


['p', 'p', 'a', 'b', 'c', 'd', 'f', 'g', 'n', 't', 'z', 'q', 'q']

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  • Well I though of somehow sorting those items that anen't in ['p','q'], and then adding as many p and q as I found at the beginning and end. But thought that it could be done in a much easier way as it is clear now
  • You simply want a custom sort function/key where 'p' compares first, 'q' compares last, and everything else comes in-between in normal sort order.
  • Now, this is pretty cool. Can you explain a bit? sorted sorts True < characters < False?
  • @ScottBoston no, it returns a (bool, bool, str) tuple
  • @juanpa.arrivillaga Ah.... Okay. I think I see. I'm going to have to play around with this a bit to get this cemented in my head. Thanks.
  • So (False, False) comes first and (True,True) in the end. In between there will be (True,False)
  • For those of you who need this sorted([(s!='p', s=='q', s) for s in l]) is a great visual of what is going on.
  • If you're going to do it this way, you can also just count the ps and qs: ['p']*l.count('p') + sorted(filter({'q', 'p'}.isdisjoint, l)) + ['q']*l.count('q') which is the same time complexity. Edit: You can get rid of the lambda too.