Python中与Django的“重组”模板标签相对应的惯用术语是什么?

结合 itertools.groupby 具有 operator.itemgetter 要获得一个相当不错的解决方案:

from operator import itemgetter
from itertools import groupby

key = itemgetter('gender')
iter = groupby(sorted(people, key=key), key=key)

for gender, people in iter:
    print '===', gender, '==='
    for person in people:
        print person

如果数据来源( people 在这种情况下)已经按关键字排序,您可以绕过 sorted 呼叫:

iter = groupby(people, key=lambda x:x['gender'])
for gender, people in iter:
    print '===', gender, '==='
    for person in people:
        print person

注:如果 把…分类 是一本普通的词典,没有秩序的保证;因此,你必须打电话 把…分类 在这里,我假设 把…分类 是a collections.OrderedDict 或某种其他类型的有序数据结构。

以前的答案帮助我解决了问题。为了将来参考, 如果你有一些嵌套数据,比如

{“城市名”:“城市1”,“公司名”:《公司1》,“名称”:《分支机构1》}

你想按城市分组,然后按该城市的公司分组,比如:

City1
 Company 1
   Branch 1
   Branch 2
 Company 2
   Branch 1
 Company 3
   Branch 1
City2
 Company 2
   Branch 1
 Company 3
   Branch 1
   Branch 2
City3
 Company 1
   Branch 1
 Company 2
   Branch 1
   Branch 2

我通过这样做解决了这个问题:

key = itemgetter('city_name')    
iter = groupby(queryset, key=key) # assuming queryset is already sorted by city_name

for key, group in iter:
    print(key)
    key2 = itemgetter('company_name')
    iter2 = groupby(sorted(group, key=key2), key=key2) # now we must sort by company_name
    for comp, branch in iter2:
        print(comp)
        for b in branch:
            print(b)