pythonic way to filter list for elements with unique length

Question

I want to filter a list, leaving only first elements with unique length. I wrote a function for it, but I believe there should be a simpler way of doing it:

def uniq_len(_list):
    from itertools import groupby
    uniq_lens = list(set([x for x, g in groupby(_list, len)]))
    all_goods = []
    for elem in _list:
        elem_len = len(elem)
        try:
            good = uniq_lens.pop([i for i, x in enumerate(uniq_lens) if x==elem_len][0])
            if good:
                all_goods.append(elem)
        except IndexError as _e:
            #print all_goods
            pass
    return all_goods

In [97]: jones
Out[97]: ['bob', 'james', 'jim', 'jon', 'bill', 'susie', 'jamie']

In [98]: uniq_len(jones)
Out[98]: ['bob', 'james', 'bill']

Answer 1

If you just want any arbitrary string for each length, in arbitrary order, the easy way to do this is to first convert to a dict mapping lengths to strings, then just read off the values:

>>> {len(s): s for s in jones}.values()
dict_values(['jon', 'bill', 'jamie'])

If you want the first for each length, and you need to preserve the order, then that's just unique_everseen from the itertools recipes, with len as the key:

>>> from more_itertools import unique_everseen
>>> list(unique_everseen(lst, key=len))
['bob', 'james', 'bill']

(If you pip install more-itertools, it includes all of the recipes from the itertools docs, plus a bunch of other helpful things.)

Answer 2

Getting the first item of the list with unique length (not necessarily in the same order as they appear in the list).

>>> lst = ['bob', 'james', 'jim', 'jon', 'bill', 'susie', 'jamie']
>>> list({len(x): x for x in reversed(lst)}.values())
['bob', 'bill', 'james']

Respecting the order of the original list, you can use an auxiliary set:

>>> seen = set()
>>> [x for x in lst if len(x) not in seen and seen.add(len(x)) is None]
['bob', 'james', 'bill']

For the above expression to work properly in succession, you have to make sure you reset seen to an empty set each time.