Question

http://docs.djangoproject.com/en/dev/ref/templates/builtins/#regroup

我可以想到一些用循环做的方法，但我特别想知道是否有一个整齐的单行。

Answer 1

将itertools.groupby与operator.itemgetter结合起来，以获得一个非常好的解决方案：

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

Answer 2

以前的答案帮助我解决了我的问题。备查，如果您有一些嵌套数据，如

{＆＃39; city_name＆＃39;：＆＃39; City1＆＃39;，＆＃39; comp_name＆＃39;：＆＃39; Company1＆＃39;，＆＃39; name＆＃39;：＆＃39;分支1＆＃39;}

你希望按城市分组，然后按公司在该城市分组：

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

我这样解决了这个问题：

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)

Answer 3

如果数据源（在这种情况下为people）已按键排序，则可以绕过sorted来电：

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

注意：如果sorted是一个普通字典，则不保证顺序;因此，您必须致电sorted。我假设sorted是collections.OrderedDict或其他类型的有序数据结构。

什么是惯用的Python相当于Django的'重新组合'模板标签？

3 个答案: