Question

我需要重新组合一个列表，而不是按每个结果列表的项目数显示排序结果。

类似的东西：

{%  regroup car_models by make as car_models_by_make %}

{% for make in car_models_by_make|dictsort:"???" --> the problem %}
    {{ make.grouper }}  ({{ make.list|length }}) <br>
{% endfor %}

所以如果数据是：

[
('panda','fiat'),
('500','fiat'),
('focus','ford')
]

结果应该是：

fiat 2
ford 1

我试过：dictsort：“make.list.count”，“make.list.length”，“make.list | length”..没有做过工作..

Answer 1

我认为dictsort不适合这里。

>>> l = [('panda', 'fiat'), ('500', 'fiat'), ('focus', 'ford')]
>>> values = [i[1] for i in l]
>>> from collections import Counter
>>> c = Counter(values)
>>> c.most_common()
[('fiat', 2), ('ford', 1)]

您可以定义自己的template tags，

from django import template
from collections import Counter

register = template.Library()

@register.filter
def my_sort(l):
    values = [i[1] for i in l]
    c = Counter(values)
    return c.most_common()

Answer 2

您应该将数据作为dicts的排序列表传递给模板：

>>> data = [ ('panda','fiat'),
... ('500','fiat'),
... ('focus','ford') ]
>>> 
>>> from itertools import groupby

# note that we need to sort data before using groupby
>>> groups = groupby(sorted(data, key=lambda x: x[1]), key=lambda x:x[1])

# make the generators into lists
>>> group_list = [(make, list(record)) for make, record in groups]

# sort the group by the number of car records it has
>>> sorted_group_list = sorted(group_list, key=lambda x: len(x[1]), reverse=True)

# build the final dict to send to the template
>>> sorted_car_model = [{'make': make, "model": r[0]} for (make, records) in sorted_group_list for r in records]
>>> sorted_car_model
[{'make': 'fiat', 'model': 'panda'}, {'make': 'fiat', 'model': '500'}, {'make': 'ford', 'model': 'focus'}]

这里的目标是对列表进行排序，以便按照他们生产的模型数量对汽车制造商进行排名。

然后，使用：

{% regroup car_models by make as make_list %}

{% for item in make_list %}
    {{ item.grouper }} ({{ item.list|length }}) <br>
{% endfor %}

得到：

fiat 2
ford 1

参考：regroup

如果您的数据与评论中的数据类似：

>>> car_models = [ 
...     {'make': 'ford', 'model': 'focus'},
...     {'make': 'fiat', 'model': 'panda'},
...     {'make': 'ford', 'model': '500'},
...     {'make': 'fiat', 'model': '500'},
...     {'make': 'opel', 'model': '500'},
...     {'make': 'opel', 'model': '500'},
...     {'make': 'opel', 'model': '500'},
... ] 
>>> 
>>> groups = groupby(sorted(car_models, key=lambda x:x['make']), key=lambda x:x['make'])
>>> groups = [(make, list(x)) for make, x in groups]
>>> sorted_group_list = sorted(groups, key=lambda x:len(x[1]), reverse=True)
>>> sorted_car_model = [{'make': make, 'model': r['model']} for (make, records) in sorted_group_list for r in records]

>>> sorted_car_model
[{'make': 'opel', 'model': '500'}, {'make': 'opel', 'model': '500'}, {'make': 'opel', 'model': '500'}, {'make': 'fiat', 'model': 'panda'}, {'make': 'fiat', 'model': '500'}, {'make': 'ford', 'model': 'fo
cus'}, {'make': 'ford', 'model': '500'}]

或者，您可以将结果传递给模板，如下所示：

>>> from collections import Counter
>>> l = [r['make'] for r in car_models]
>>> c = Counter(l)
>>> result = [{k: v} for k,v in dict(c).iteritems()]
>>> result = sorted(result, key=lambda x: x.values(), reverse=True)
[{'opel': 3}, {'fiat': 2}, {'ford': 2}]

然后，将此结果传递给模板，只需渲染结果：

{% for r in result %}
    {% for k,v in r.items %}
        {{ k }} {{ v }}
    {% endfor %}j
{% endfor %}

然后你会得到：

opel 3
fiat 2
ford 2

Answer 3

使用此模板标签我做了：

@register.filter()
def groups_sort(groups):

    groups.sort(key=lambda group: len(group['list']))
    return groups

@register.filter()
def groups_sort_reversed(groups):
    groups.sort(key=lambda group: len(group['list']), reverse=True)
    return groups

可以这样做：

{% load custom_templatetags %}
{%  regroup car_models by make as car_models_by_make %}

{% for make in car_models_by_make|groups_sort_reversed %}
    {{ make.grouper }}  ({{ make.list|length }}) <br>
{% endfor %}

会给你

fiat 2
ford 1

使用groups_sort将为您提供：

ford 1
fiat 2

这当然适用于Django模型

享受

django按照列表长度重新组合而不是dictsort

3 个答案: