Python:将名称列表划分为大小相等的子列表 [英] Python: partition list of names into equally sized sublists

问题描述

我有一个名字列表，例如['Agrajag', 'Colin', 'Deep Thought', ... , 'Zaphod Beeblebrox', 'Zarquon'].现在，我想将此列表划分为大小大致相等的子列表，以使子组的边界位于名称的首字母，例如AF，GL，MP，QZ，而不是A-Fe，Fi-Mo，Mu-Pra ，Z之前.

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.

我只能提出一个静态大小的分区，而不考虑子组的大小:

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)

还有更好的主意吗?

P.S .:这听起来有点像作业，但实际上是针对网页的，其中成员应显示在5-10个相同大小的组的标签中.

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.

推荐答案

这可能有效.我确信有一种更简单的方法……可能涉及到itertools.请注意，num_pages仅大致确定您实际将获得的页面数.

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.

糟糕！有一个错误-它切断了最后一组！下面的内容应该是固定的，但是请注意，最后一页的长度会有些不可预测.另外，我添加了.upper()来考虑可能的小写字母名称.

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.

以前定义letter_groups的方法效率很低；以下基于dict的代码更具可扩展性:

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)

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