Python: partition list of names into equally sized

Question

I have a list of names, e.g. ['Agrajag', 'Colin', 'Deep Thought', ... , 'Zaphod Beeblebrox', 'Zarquon']. Now I want to partition this list into approximately equally sized sublists, so that the boundaries of the subgroups are at the first letter of the names, e.g A-F, G-L, M-P, Q-Z, not A-Fe, Fi-Mo, Mu-Pra, Pre-Z.

I could only come up with a statically sized parition that doesn't take size of the subgroups into account:

import string, itertools

def _group_by_alphabet_key(elem):
    char = elem[0].upper()
    i = string.ascii_uppercase.index(char)
    if i > 19:
        to_c = string.ascii_uppercase[-1];
        from_c = string.ascii_uppercase[20]
    else:
        from_c = string.ascii_uppercase[i/5*5]
        to_c = string.ascii_uppercase[i/5*5 + 4]
    return "%s - %s" % (from_c, to_c)

subgroups = itertools.groupby(name_list, _group_by_alphabet_key)

Any better ideas?

P.S.: this may sound somewhat like homework, but it actually is for a webpage where members should be displayed in 5-10 tabs of equally sized groups.

Answer 1

Here's something that might work. I feel certain there's a simpler way though... probably involving itertools. Note that num_pages only roughly determines how many pages you'll actually get.

EDIT: Whoops! There was a bug -- it was cutting off the last group! The below should be fixed, but note that the length of the last page will be slightly unpredictable. Also, I added .upper() to account for possible lowercase names.

EDIT2: The previous method of defining letter_groups was inefficient; the below dict-based code is more scalable:

names = ['Agrajag', 'Colin', 'Deep Thought', 'Ford Prefect' , 'Zaphod Beeblebrox', 'Zarquon']
num_pages = 3

def group_names(names, num_pages):
    letter_groups = defaultdict(list)
    for name in names: letter_groups[name[0].upper()].append(name)
    letter_groups = [letter_groups[key] for key in sorted(letter_groups.keys())]
    current_group = []
    page_groups = []
    group_size = len(names) / num_pages
    for group in letter_groups:
        current_group.extend(group)
        if len(current_group) > group_size:
            page_groups.append(current_group)
            current_group = []
    if current_group: page_groups.append(current_group)

    return page_groups

print group_names(names, num_pages)

Answer 2

Since your name_list has to be sorted for groupby to work, can't you just check every Nth value and build your divisions that way?

right_endpoints = name_list[N-1::N]

And using "A" as your leftmost endpoint and "Z" as your rightmost endpoint, you can construct the N divisions accordingly and they should all have the same size.

1. So, the first left endpoint would be "A", the first right endpoint would be right_endpoints[0].
2. The next left endpoint would be the character after right_endpoints[0], the next right endpoint would be right_endpoints[1].
3. Etc., until you hit the Nth range and that has a set endpoint of "Z".

The issue you may run into is what if two of these right_endpoints are the same...

edit: example

>>> names = ['Aaron', 'Abel', 'Cain', 'Daniel', 'Darius', 'David', 'Ellen', 'Gary', 'James', 'Jared', 'John', 'Joseph', 'Lawrence', 'Michael', 'Nicholas', 'Terry', 'Victor', 'Zulu']
>>> right_ends, left_ends = names[2::3], names[3::3]
>>> left_ends = ['A'] + left_ends
>>> left_ends, right_ends
>>> ["%s - %s" % (left, right) for left, right in zip(left_ends, right_ends)]
['A - Cain', 'Daniel - David', 'Ellen - James', 'Jared - Joseph', 'Lawrence - Nicholas', 'Terry - Zulu']