# Go deh!

Mainly Tech projects on Python and Electronic Design Automation.

## Saturday, February 23, 2008

### Length Sorts

Andre Roberge just informed me of a site with ninety nine Prolog problems. A quick scan and I got as far as P28 before thinking that I would like to try it. so here is my solution to P28. Note that there is much more comments than code, I decided to cut-n-paste my command line scribblings as an explanation of the second function, rather like doctests.

In doing this, I also learned that you could not take the len of a groupby object :-)

The code:

(Note: I would like to make this available under the Python License version 2.4.2)

`''' \Length sorts. From https://prof.ti.bfh.ch/hew1/informatik3/prolog/p-99/ P28A code kataAnswer in Python (C) Donald 'Paddy' McCarthy,  Feb. 23, 2008'''from itertools import groupbydef length_sort(lst):  ''' \  Sort a list-of-lists on sub-list length  Example    >>> l = [['a','b','c'],['d','e'],['f','g','h'],['d','e'],['i','j','k','l'],['m','n'],['o']]    >>> length_sort(l)    [['o'], ['d', 'e'], ['d', 'e'], ['m', 'n'], ['a', 'b', 'c'], ['f', 'g', 'h'], ['i', 'j', 'k', 'l']]         '''  return sorted(lst, key=len)def length_freq_sort(lst):  ''' \  Sort a list-of-lists on frequency of sub-lists length  Example    >>> l = [['a','b','c'],['d','e'],['f','g','h'],['d','e'],['i','j','k','l'],['m','n'],['o']]    >>> length_freq_sort(l)    [['i', 'j', 'k', 'l'], ['o'], ['a', 'b', 'c'], ['f', 'g', 'h'], ['d', 'e'], ['d', 'e'], ['m', 'n']]  '''  ## Comments give example working out  # >>> lst = [['a','b','c'],['d','e'],['f','g','h'],['d','e'],['i','j','k','l'],['m','n'],['o']]     lengths = sorted(len(x) for x in lst)  # >>> lengths  # [1, 2, 2, 2, 3, 3, 4]  # >>> [(a,len(list(b))) for a,b in groupby(lengths)]  # [(1, 1), (2, 3), (3, 2), (4, 1)]  # >>> dict([(a,len(list(b))) for a,b in groupby(lengths)])  # {1: 1, 2: 3, 3: 2, 4: 1}    len2freq = dict([(a,len(list(b))) for a,b in groupby(lengths)])  return sorted(lst, key=lambda x: len2freq[len(x)])`

1. Superfluous lambda alert: sorted(lst, key=len) works just as well.

2. Thanks Peter.

Can I adapt your solution when I get around to writing up problem 28?
(I'd like to make them available with no copyright restrictions...)

4. Will the Python license suffice.

length_sort = functools.partial(sorted, key=len)

1. Understand the question to attempt a correct answer.