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python - Django 如何在过滤后注释多个模型上关键字的实例总数?

转载 作者:行者123 更新时间:2023-11-28 04:24:58 25 4
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好的,所以我有一个“关键字”模型,它存储我的其他 4 个模型的关键字,每个模型都有一个“key_list”字段,它是一个指向我的“关键字”的 ManyToManyField ' 模型。我的每个模型都有多个关键字,我正在搜索它们并成功找到它们,如下所示:

keys_selected='term1$term2$term3$'
keys_selected = keys_selected.rstrip('$')
keys_selected = keys_selected.split('$')
goal = len(keys_selected)
remaining = list()
remaining += Keyword.objects.filter(employee__key_list__key__in=keys_selected).annotate(num_keys=Count('key')).filter(num_keys=goal).distinct()
remaining += Keyword.objects.filter(vendor__key_list__key__in=keys_selected).annotate(num_keys=Count('key')).filter(num_keys=goal).distinct()
remaining += Keyword.objects.filter(application__key_list__key__in=keys_selected).annotate(num_keys=Count('key')).filter(num_keys=goal).distinct()
remaining += Keyword.objects.filter(machine__key_list__key__in=keys_selected).annotate(num_keys=Count('key')).filter(num_keys=goal).distinct()
key_list = list()
for x in remaining:
if x not in key_list:
key_list.append(x)

这将返回一个字典,其中包含分配给我的 4 个模型中包含我选择的术语的所有条目的所有关键字。这里的想法是创建一个过滤器,直观地显示与我的搜索查询匹配的对象的 key_list 中所有关键字的词频。我希望它将它附加到正在输出到上下文的字典中,这样我就可以调用该值并将它用于字体大小,如下所示:

{% for key in key_list %}
<a href="{% url 'keysearch:index' %}?keys_selected={{ key }}${{ keys_selected }}" style="font-size: {{ key.num_keys }}px;">({{ key }})</a>
{% endfor %}

换句话说,这应该为我的关键字创建一个关键字“云”,它可以过滤并直观地显示给定关键字在我的模型结果中的出现频率,但我完全不知道如何让它只在我的过滤器的结果。我不知道如何在多个模型中实现这一目标。

我的引用模型:

class Keyword(models.Model):
key = models.CharField(max_length=2000, unique=True)

def __str__(self):
return self.key

class Meta:
ordering = ('key',)


class Entries(models.Model):
name = models.CharField("Name", max_length=200)
updated = models.DateTimeField("Last Updated", auto_now=True)
key_list = models.ManyToManyField(Keyword, blank=True, verbose_name="Keywords")
description = models.TextField("Description", blank=True)

class Meta:
abstract = True
ordering = ('name',)


class Employee(Entries):
uid = models.SlugField("Employee User ID", max_length=6, unique=True, blank=True)
manager = models.SlugField("Manager's User ID", max_length=6)

def __str__(self):
return self.name


class Vendor(Entries):
company = models.CharField("Vendor Company", max_length=200)
email = models.EmailField("Vendor Company Email Address", max_length=254, unique=True)
vend_man_name = models.CharField("Manager's Name", max_length=200)
vend_man_email = models.EmailField("Manager's Email Address", max_length=254)

def __str__(self):
return self.name


class Application(Entries):
app_url = models.URLField("Application URL", max_length=800, unique=True)

def __str__(self):
return self.name


class Machine(Entries):
address = models.CharField("Machine Address", max_length=800, unique=True)
phys_loc = models.TextField("Physical Location", blank=True)

def __str__(self):
return self.name

最佳答案

找到答案 here by Lauritz V. Thaulow可以使用 groupby。

我的解决方案(包括他的代码):

from django.db.models import Count
from keysearch.models import Employee, Vendor, Application, Machine, Keyword
from itertools import groupby


def unique_keys(input):
output = []
for x in input:
if x not in output:
output.append(x)
return output


def canonicalize_dict(x):
return sorted(x.items(), key=lambda x: hash(x[0]))


def unique_and_count(lst):
grouper = groupby(sorted(map(canonicalize_dict, lst)))
return [dict(k + [("count", int(len(list(g)) * 2.3 + 16))]) for k, g in grouper]


def keycount(key_list='', keys_selected=''):
goal = len(keys_selected)
key_ref = list()
db_list = [Employee, Vendor, Application, Machine]
for db in db_list:
if keys_selected == '':
source = db.objects.all()
else:
source = db.objects.filter(key_list__key__in=keys_selected).annotate(num_keys=Count('key_list')).filter(num_keys=goal).distinct()
for entry in source:
key_ref += entry.key_list.values()
key_list = unique_and_count(key_ref)
return key_list

@register.inclusion_tag('keysearch/key_cloud.html')
def key_cloud(keys_selected=''):
if keys_selected == '':
key_list = Keyword.objects.all()
key_list = keycount(key_list, keys_selected)
else:
keys_selected = keys_selected.rstrip('$')
keys_selected = keys_selected.split('$')
keys_selected = unique_keys(keys_selected)
goal = len(keys_selected)
remaining = list()
remaining += Keyword.objects.filter(employee__key_list__key__in=keys_selected).annotate(num_keys=Count('key')).filter(num_keys=goal)
remaining += Keyword.objects.filter(vendor__key_list__key__in=keys_selected).annotate(num_keys=Count('key')).filter(num_keys=goal)
remaining += Keyword.objects.filter(application__key_list__key__in=keys_selected).annotate(num_keys=Count('key')).filter(num_keys=goal)
remaining += Keyword.objects.filter(machine__key_list__key__in=keys_selected).annotate(num_keys=Count('key')).filter(num_keys=goal)
key_list = unique_keys(remaining)
key_list = keycount(key_list, keys_selected)
keys_selected = "$".join(keys_selected)
keys_selected += '$'
context = {'key_list': key_list, 'keys_selected': keys_selected}
return context

然后在模板中:

{% for key in key_list %}
<a href="{% url 'keysearch:index' %}?keys_selected={{ key.key }}${{ keys_selected }}" style="font-size: {{ key.count }}px;">({{ key.key }})</a>
{% empty %}
<div class="w3-card-2 w3-black w3-center w3-rest"><h4>There are no key queries with this combination.</h4></div>
{% endfor %}

关于python - Django 如何在过滤后注释多个模型上关键字的实例总数?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/42015291/

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