小编toy*_*toy的帖子

java servlet 3.0已经完成了吗？我正在尝试实现彗星,继续使用Servlet 3.0是一个好主意.但我不确定它是否是前进的方式.

谢谢

我按照Spark的网站上的说明操作,我在亚马逊上运行了1个主服务器和1个服务器.但是,我无法使用pyspark连接到主节点

我可以使用SSH连接到主节点,没有任何问题.

这是我的命令spark-ec2 --key-pair = graph-cluster --identity-file =/Users/.ssh/pem.pem --region = us-east-1 --zone = us-east-1a launch graph -簇

我可以去http://ec2-54-152-xx-xxx.compute-1.amazonaws.com:8080/

并且看到Spark已启动并运行我也看到了Spark Master

spark://ec2-54-152-xx-xxx.compute-1.amazonaws.com:7077

但是当我运行命令时

MASTER=spark://ec2-54-152-xx-xx.compute-1.amazonaws.com:7077 pyspark

我收到这个错误

2015-09-16 15:39:31,800 ERROR actor.OneForOneStrategy (Slf4jLogger.scala:apply$mcV$sp(66)) -
java.lang.NullPointerException
    at org.apache.spark.deploy.client.AppClient$ClientActor$$anonfun$receiveWithLogging$1.applyOrElse(AppClient.scala:160)
    at scala.runtime.AbstractPartialFunction$mcVL$sp.apply$mcVL$sp(AbstractPartialFunction.scala:33)
    at scala.runtime.AbstractPartialFunction$mcVL$sp.apply(AbstractPartialFunction.scala:33)
    at scala.runtime.AbstractPartialFunction$mcVL$sp.apply(AbstractPartialFunction.scala:25)
    at org.apache.spark.util.ActorLogReceive$$anon$1.apply(ActorLogReceive.scala:59)
    at org.apache.spark.util.ActorLogReceive$$anon$1.apply(ActorLogReceive.scala:42)
    at scala.PartialFunction$class.applyOrElse(PartialFunction.scala:118)
    at org.apache.spark.util.ActorLogReceive$$anon$1.applyOrElse(ActorLogReceive.scala:42)
    at akka.actor.Actor$class.aroundReceive(Actor.scala:465)
    at org.apache.spark.deploy.client.AppClient$ClientActor.aroundReceive(AppClient.scala:61)
    at akka.actor.ActorCell.receiveMessage(ActorCell.scala:516)
    at akka.actor.ActorCell.invoke(ActorCell.scala:487)
    at akka.dispatch.Mailbox.processMailbox(Mailbox.scala:238)
    at akka.dispatch.Mailbox.run(Mailbox.scala:220)
    at akka.dispatch.ForkJoinExecutorConfigurator$AkkaForkJoinTask.exec(AbstractDispatcher.scala:393)
    at scala.concurrent.forkjoin.ForkJoinTask.doExec(ForkJoinTask.java:260)
    at scala.concurrent.forkjoin.ForkJoinPool$WorkQueue.runTask(ForkJoinPool.java:1339)
    at scala.concurrent.forkjoin.ForkJoinPool.runWorker(ForkJoinPool.java:1979)
    at scala.concurrent.forkjoin.ForkJoinWorkerThread.run(ForkJoinWorkerThread.java:107)
2015-09-16 15:39:31,804 INFO  client.AppClient$ClientActor (Logging.scala:logInfo(59)) - Connecting to master …

我是Go的新人.我想知道如何使用Reflection in Go获得映射的值.

type url_mappings struct{
    mappings map[string]string
}

func init() {
    var url url_mappings
    url.mappings = map[string]string{
        "url": "/",
        "controller": "hello"}

谢谢

我是函数式编程的新手,因为我正在读这本书.它基本上说如果你的代码包含"var",那就意味着你仍然在以命令的方式做.我不确定如何将代码更改为功能方式.请建议.

所以这个代码的基本操作就是使用一些文本来处理文本,并使用正则表达式从"taggedText"中提取特定文本,并将其添加到列表中并将列表转换为json.

    val text = params("text")
    val pattern = """(\w+)/ORGANIZATION""".r

    var list = List[String]()
    val taggedText = processText(text)
    pattern.findAllIn(taggedText).matchData foreach {
      m => list ::= m.group(1)
    }

    pretty(render(list)) // render to json

我是Ruby和RSpec的新手.我来自Java背景,这就是为什么我的测试真的看起来像junit代码.我想更多地了解RSpec的,但我不太明白subject,let,!let.所以,如果有人可以指导我清理这些代码,我将非常感激.

我有Sinatra,RSpec正在用Twitter登录.

get '/login/twitter' do
  begin
    request_token = TwitterService.new.authentication_request_token

    session[:request_token_twitter] = request_token

    redirect request_token.authorize_url
  rescue Exception => e
    logger.error(e.message)
    redirect '/'
  end  
end

get '/login/twitter/success' do
  request_token = session[:request_token_twitter]
  twitter_service = TwitterService.new
  access_token = twitter_service.authorize(request_token, params[:oauth_verifier])

  begin
    twitter_user_info = twitter_service.verify_credentials

    twitter_id = twitter_user_info["id"]
    response.set_cookie("auth_token", :value => twitter_id, :path => '/')
    response.set_cookie(@social_flag, :value => "t", :path => '/')

    expected_user = @user_manager.find_by_id(twitter_id.to_s)

    if expected_user.is_null?
      twitter_user = User.new(twitter_id, access_token.token, access_token.secret, "t")
      twitter_user.save

      logger.info("Saving ...")
      logger.info("Twitter ID #{twitter_id}") …

我是响应式设计和 CSS 的新手。这似乎是一个简单的问题，但我无法从谷歌得到直接答案。我已经尝试过http://css-tricks.com/centering-in-the-unknown/幽灵块工作完美，但它给我留下了白色背景颜色。现在我被困住了。基本上，我的徽标大小为 534x385，并且我希望该徽标在任何设备上居中。对于手机，我希望该徽标也缩小以匹配屏幕尺寸。

<div>
    <img class="logo" src="images/shapes-logo.png" />
</div>

.logo {
    position: fixed;
    top:15%;
    left: 50%;
    margin-left: -267px;
} 

html {
    background: url('../images/shapes-background.jpg') no-repeat center center fixed;
    -webkit-background-size: cover;
    -moz-background-size: cover;
    -o-background-size: cover;
    background-size: cover;
}

到目前为止，这是我的 CSS。但如果我这样做， margin-left: -267px 将在移动设备中引起问题。

我是动态编程的新手,根据我的理解,动态编程是您使用计算结果来检查函数是否正确的地方.在接受采访时我被要求实施一种方法,以检查是否n是2的力量.所以,我想出了这个.

def isPowerOfTwo(self, n):
    power_of_two = [1]
    is_power_of_two = False
    if n == 0:
        return False

    if n == 1:
        return True

    while True:
        power_of_two.append(power_of_two[-1] * 2)
        if n == power_of_two[-1]:
            is_power_of_two = True
            break

        if power_of_two[-1] > n:
            break

    return is_power_of_two

我正在尝试删除超过 7 周的数据。我将 str 时间戳转换为 time_struct 但是当我想与 timedelta 进行比较时出现此错误。

类型错误：无法排序的类型：time.struct_time() < datetime.datetime()

这是我的代码

start = new_programme.get('start')
start_time = time.strptime(start, '%Y%m%d%H%M%S -0000')
print(type(start_time))
seven_weeks = datetime.datetime.now() - datetime.timedelta(weeks=7)
print(type(seven_weeks))
is_more_than_7_weeks = start_time < seven_weeks

标准输出

<class 'time.struct_time'>
<class 'datetime.datetime'>

我正在尝试提出一个正则表达式来从字符串中提取时间来构建应用程序.

这是我到目前为止所得到的.不知道我在这里做错了什么.

https://regex101.com/r/fC0lI5/1

我可以得到一些字符串但不是所有不同的变体.

([01]?[0-9]*:?[0-9]*[AP]M?)-([01]?[0-9]*:?[0-9]*[AP]M?)


8-8:30AM MON TUES THURS FRI
8-10:30AM MON TUES THURS FRI
8:30AM-10:30AM MON TUES THURS FRI
10:30AM-11:30AM MON TUES THURS FRI
10:30AM-11:30AM MON TUES THURS FRI
8AM-8:30AM
10-11PM

我想要的是两个被捕获的团体.所以我知道第一组是FROM时间,后一组是TO时间.

我有这样的映射关键字.

categories_mapping = {
        'comics': 'Comic Books',
        'cartoons': 'Comic Books',
        'manga': 'Comic Books',
        'video and computer games': 'Video Games',
        'role playing games': 'Video Games',
        'immigration': 'Immigration',
        'police': 'Police',
        'environmental': 'Environment',
        'celebrity fan and gossip': 'Celebrity',
        'space and technology': 'NASA / Space',
        'movies and tv': 'TV and Movies',
        'elections': 'Elections',
        'referendums': 'Elections',
        'sex': 'Sex',
        'music': 'Music',
        'technology and computing': 'Technology'}

和这样的清单.

labels = ['technology and computing', 'arts and technology']

如果列表中的任何单词位于字典的键中,我想返回字典的值.

这就是我想出来的,但我认为这不是非常pythonic.

cats = []
for k,v in categories_mapping.items():
    for l in …

我已经用Regex101测试了我的正则表达式,但是当我在Python 3.5.1中尝试时它会返回None

这是我的Regex101

这是我的Python代码.不确定我是否遗漏任何东西.

Python 3.5.1 |Anaconda 2.4.0 (x86_64)| (default, Dec  7 2015, 11:24:55)
[GCC 4.2.1 (Apple Inc. build 5577)] on darwin
Type "help", "copyright", "credits" or "license" for more information.
>>> import re
>>> p = re.compile('([01]?[0-9]+:?[0-9]*(?:[AP]M)?)-([01]?[0-9]+:?[0-9]*(?:[AP]M)?)', re.MULTILINE)
>>> m = p.match('NO PARKING (SANITATION BROOM SYMBOL) 7AM-7:30AM EXCEPT SUNDAY')
>>> print(m)
None
>>>