小编Dar*_*one的帖子

为什么这个字符串不会在每个"\n"上拆分？(红宝石)

"ADVERTISING [7310]\n\t\tIRS NUMBER:\t\t\t\t061340408\n\t\tSTATE OF INCORPORATION:\t\t\tDE\n\t\tFISCAL YEAR END:\t\t\t0331\n\n\tFILING VALUES:\n\t\tFORM TYPE:\t\t10-Q\n\t\tSEC ACT:\t\t1934 Act\n\t".split('\n')
>> ["ADVERTISING [7310]\n\t\tIRS NUMBER:\t\t\t\t061340408\n\t\tSTATE OF INCORPORATION:\t\t\tDE\n\t\tFISCAL YEAR END:\t\t\t0331\n\n\tFILING VALUES:\n\t\tFORM TYPE:\t\t10-Q\n\t\tSEC ACT:\t\t1934 Act\n\t"]

现在已经摆弄了Python的multicore功能超过一个小时,试图使用Process和并行化一个相当复杂的图遍历函数Manager:

import networkx as nx
import csv
import time 
from operator import itemgetter
import os
import multiprocessing as mp

cutoff = 1

exclusionlist = ["cpd:C00024"]

DG = nx.read_gml("KeggComplete.gml", relabel = True)

for exclusion in exclusionlist:
    DG.remove_node(exclusion)

#checks if 'memorizedPaths exists, and if not, creates it
fn = os.path.join(os.path.dirname(__file__), 'memorizedPaths' + str(cutoff+1))
if not os.path.exists(fn):
    os.makedirs(fn)

manager = mp.Manager()
memorizedPaths = manager.dict()
filepaths = manager.dict()
degreelist = sorted(DG.degree_iter(),key=itemgetter(1),reverse=True)

def _all_simple_paths_graph(item, DG, cutoff, memorizedPaths, filepaths):
    source …

我正在尝试在newick格式化的树上翻译节点,而我在更换权限方面遇到了麻烦.说我有HashMap:

"(1:" : "(30:"
",1:" : ",30:" 
"(30:" : "(6:"
",30:" : ",6:"

而树:

(30:0.07,(1:0.06,2:0.76))

传统智慧会暗示多重replaceAll,但这会带来一个问题:

replaceAll("(1:", "(30:") >> (30:0.07,(30:0.06,2:0.76))
replaceAll("(30:", "(6:") >> (6:0.07,(6:0.06,2:0.76))

这里的问题是我们已经替换了之前被替换的节点.正确的树应如下所示:

(6:0.07,(30:0.06,2:0.76))

现在我已经在Python中完成了这个:

def multiple_replace(taxa, text): 
    regex = re.compile("|".join(map(re.escape, taxa.keys())))
    return regex.sub(lambda mo: taxa[mo.group(0)], text) 

但是我的Java实现遇到了问题:

private String convertTree (String treeOld, HashMap<String, String> conv) {
        Pattern pattern = Pattern.compile("\\(\\d+:|,\\d+:");
        Matcher matcher = pattern.matcher(treeOld);
        StringBuilder sbt = new StringBuilder(treeOld);
        while (matcher.find()) {
            String replace = conv.get(matcher.group());
            System.out.println(matcher.group() + "||" +replace + " || …

目前我的分页看起来像这样:

我正在尝试找出如何删除按钮之间的空间,但一直没有成功.在css文件中,我可以找到分页的唯一引用是:

jquery.dataTables.css(319 - 394):

}
.dataTables_wrapper .dataTables_paginate {
  float: right;
  text-align: right;
  padding-top: 0.25em;
}
.dataTables_wrapper .dataTables_paginate .paginate_button {
  box-sizing: border-box;
  display: inline-block;
  min-width: 1.5em;
  padding: 0.5em 1em;
  margin-left: 2px;
  text-align: center;
  text-decoration: none !important;
  cursor: pointer;
  *cursor: hand;
  color: #333333 !important;
  border: 1px solid transparent;
}
.dataTables_wrapper .dataTables_paginate .paginate_button.current, .dataTables_wrapper .dataTables_paginate .paginate_button.current:hover {
  color: #333333 !important;
  background-color: white;
  background: -webkit-gradient(linear, left top, left bottom, color-stop(0%, white), color-stop(100%, gainsboro));
  /* Chrome,Safari4+ */
  background: -webkit-linear-gradient(top, white 0%, gainsboro 100%);
  /* …

所以我从CRUD生成器复制了基本删除操作:

   /**
    * @Route("category/delete/{id}", name="category_delete")
    * @Method("DELETE")
    */
    public function deleteAction(Request $request, $id)
    {
        $repository = $this->getDoctrine()->getRepository('AppBundle:Category');
        $category = $repository->find($id);

        $form = $this->createDeleteForm($category);
        $form->handleRequest($request);

        if ($form->isSubmitted() && $form->isValid()) {
            $em = $this->getDoctrine()->getManager();
            $em->remove($category);
            $em->flush();
        }

        return $this->redirectToRoute('category_index');
    }

    /**
     *
     * @param Category $category
     *
     * @return \Symfony\Component\Form\Form The form
     */
    private function createDeleteForm(Category $category)
    {
        return $this->createFormBuilder()
            ->setAction($this->generateUrl('category_delete', array('id' => $category->getId())))
            ->setMethod('DELETE')
            ->getForm()
            ;
    } 

但是,我不确定如何实际使用动作本身.我想在两个地方删除,我不确定正确的方法:

1. 在编辑操作中删除 - 如何向表单构建器添加删除按钮？我是否必须在树枝模板中进行此操作？
2. 在索引中删除 - 我知道如何调用操作(例如<a href="{{ path('category_edit', {'id': cat.id}) …

如果我QuerySet从以下命令创建:

data = ModelA.objects.values('date').annotate(total=Sum('amount'), average=Avg('amount'))
# <QuerySet [{'date': datetime.datetime(2016, 7, 15, 0, 0, tzinfo=<UTC>), 'total': 19982.0, 'average': 333.03333333333336}, {'date': datetime.datetime(2016, 7, 15, 0, 30, tzinfo=<UTC>), 'total': 18389.0, 'average': 306.48333333333335}]>

是否可以使用从内部的每个哈希data创建ModelB对象QuerySet？

我意识到我可以迭代并解压缩QuerySet并像这样做:

   for q in data:
        ModelB.objects.create(date=q['date'], total=q['total'], average=q['average'])

但是有更优雅的方式吗？当它们几乎处于bulk_create格式时,迭代和创建似乎是多余的.

所以我有一组按钮，我希望它们位于页面的右侧，但是justify-end/justify='end'不起作用v-row。

<!DOCTYPE html>
<html>

<head>
  <link href="https://fonts.googleapis.com/css?family=Roboto:100,300,400,500,700,900" rel="stylesheet">
  <link href="https://cdn.jsdelivr.net/npm/@mdi/font@3.x/css/materialdesignicons.min.css" rel="stylesheet">
  <link href="https://cdn.jsdelivr.net/npm/vuetify@2.x/dist/vuetify.min.css" rel="stylesheet">
  <meta name="viewport" content="width=device-width, initial-scale=1, maximum-scale=1, user-scalable=no, minimal-ui">
</head>

<body>
  <div id="app">
    <v-app>
      <v-container>
        <v-form>
          <v-row justify='end'>
            <v-col>
              <v-btn>Button 1</v-btn>
              <v-btn>Button 1</v-btn>
              <v-btn>Button 1</v-btn>
            </v-col>
          </v-row>
        </v-form>
      </v-container>
    </v-app>
  </div>

  <script src="https://cdn.jsdelivr.net/npm/vue@2.x/dist/vue.js"></script>
  <script src="https://cdn.jsdelivr.net/npm/vuetify@2.x/dist/vuetify.js"></script>
  <script>
    new Vue({
      el: '#app',
      vuetify: new Vuetify(),
    })
  </script>
</body>

我看过这个问题，但是使用 text align 似乎很麻烦，我想知道是否有更好的解决方案？

我想做的是构建一个自定义版本，cache_page让我可以更好地控制缓存键，但我什至无法完成响应的基本缓存：

from django.core.cache import cache
from rest_framework import viewsets
from rest_framework.response import Response

from app import models

class BaseViewSet(viewsets.GenericViewSet):
    queryset = models.Items.objects.all()

    def get_queryset(self):
        return models.Items.objects.all()

    def list(self, request, **kwargs):
        response = Response({})
        cache.set('test', response, 10)
        return response

我的相关部分settings.py设置为：

REST_FRAMEWORK = {
    'DEFAULT_RENDERER_CLASSES': [
        'rest_framework.renderers.JSONRenderer',
    ],
    'DEFAULT_VERSIONING_CLASS': 'rest_framework.versioning.NamespaceVersioning'
}
CACHES = {
    "default": {
        "BACKEND": "django_redis.cache.RedisCache",
        "LOCATION": f"redis://127.0.0.1:6729/1",
    },
}

当我尝试调用端点时，我得到：

django.template.response.ContentNotRenderedError: The response content must be rendered before it can be pickled.

然后，如果我将该行更改为：

cache.set('test', response.render(), …

我一直在学习有关嵌入表单集合的 Symfony 3 教程，并且我想将这个想法扩展到额外的嵌套级别。我环顾四周，有 Symfony 2 的部分答案，但没有任何全面的答案（也没有 3 的答案）。

如果我们的教程Task有很多Tag例子，我将如何编码它以便它扩展到：Task有很多Tag有很多SubTag？

到目前为止，我想我理解了 Form 类：

任务：

class TaskType extends AbstractType
{
    public function buildForm(FormBuilderInterface $builder, array $options)
    {
        $builder->add('description');

        $builder->add('tags', CollectionType::class, array(
            'entry_type' => TagType::class,
            'allow_add' => true,
            'by_reference' => false,
            'allow_delete' => true
        ));
    }

    public function configureOptions(OptionsResolver $resolver)
    {
        $resolver->setDefaults(array(
            'data_class' => 'AppBundle\Entity\Task',
        ));
    }
}

标签：

class TagType extends AbstractType
    {
        public function buildForm(FormBuilderInterface $builder, array $options) …

所以我已经知道我可以在preUpdate生命周期事件中对特定实体进行更改:

/**
 * Captures pre-update events.
 * @param PreUpdateEventArgs $args
 */
 public function preUpdate(PreUpdateEventArgs $args)
 {
     $entity = $args->getEntity();


     if ($entity instanceof ParentEntity) {
            $changes = $args->getEntityChangeSet();
     }
 }

但是,有没有办法同时获得任何相关实体的更改？例如,假设ParentEntity有这样的关系设置:

/**
 * @ORM\OneToMany(targetEntity="ChildEntity", mappedBy="parentEntity", cascade={"persist", "remove"})
 */
 private $childEntities;

并且ChildEntity还有:

/**
 * @ORM\OneToMany(targetEntity="GrandChildEntity", mappedBy="childEntity", cascade={"persist", "remove"})
 */
 private $grandChildEntities;

有没有办法让在所有相关的变化preUpdate的ParentEntity？