【问题标题】:Django How do I annotate the total number of instances of a keyword over multiple models AFTER filtering?Django 如何在过滤后注释多个模型上的关键字实例总数?
【发布时间】:2017-02-03 01:26:05
【问题描述】:

好的,所以我有一个“关键字”模型来存储我的其他 4 个模型的关键字,每个模型都有一个“key_list”字段,它是一个 ManyToManyField 指向我的“关键字”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:

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)

这将返回一个包含所有关键字的字典,这些关键字分配给我的 4 个模型中包含我选择的术语的所有条目。这里的想法是创建一个过滤器,在与我的搜索查询匹配的对象的key_list 中直观地显示所有关键字的词频。我希望它将它附加到要输出到上下文的字典中,以便我可以调用该值并将其用于字体大小,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,这应该创建我的关键字的关键字“云”,它会过滤并直观地显示给定关键字在我的模型结果中的频率,但我完全不知道如何让它仅在我的过滤器的结果。我不知道如何跨多个模型实现这一目标。

我的模型供参考:

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

【问题讨论】:

    标签: python css django python-3.x


    【解决方案1】:

    找到了一个可以使用 groupby 的答案 here by Lauritz V. Thaulow

    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 %}
    

    【讨论】:

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