小编Anu*_*rma的帖子

我有一个像这样的滑块类,我想改变样式属性style ="left:336px"

<div id="range-cont">
<div class="slider">
<div class="progress" style="width: 350px;"></div>
    <a class="handle" href="#" **style="left: 336px;"**></a>
</div>
<input id="percentage" class="range" type="range" value="14" name="percentage" min="0" max="100">
<p id="percent-label">%</p>
</div>
</div>

我试过 $('.handle').css({'style':'left: 300px'})但它不起作用.不知道我在这里做错了什么

我有一个mongo'_id'列表,我想删除它.目前我正在这样做

# inactive_users -->  list of inactive users 
for item in inactive_users:
    db.users.remove({'_id' : item})

但我的问题是列表太大了......(它可能会超过100,000+).因此查询列表中的每个项目只会增加服务器上的负载.他们是一种在mongo查询中传递整个列表的方法,这样我就不必一次又一次地触发查询.

谢谢

fminunc在python中是否有替代函数(来自octave/matlab)？我有一个二元分类器的成本函数.现在我想运行梯度下降来获得theta的最小值.八度/ matlab实现将如下所示.

%  Set options for fminunc
options = optimset('GradObj', 'on', 'MaxIter', 400);

%  Run fminunc to obtain the optimal theta
%  This function will return theta and the cost 
[theta, cost] = ...
    fminunc(@(t)(costFunction(t, X, y)), initial_theta, options);

我已经使用numpy库在python中转换了我的costFunction,并在numpy中查找fminunc或任何其他梯度下降算法实现.

我试图在python中使用list comprehension来压缩列表.我的名单有点像

[[1, 2, 3], [4, 5, 6], 7, 8]

只是为了打印这个列表中的单个项目我写了这段代码

   def flat(listoflist):
     for item in listoflist:
             if type(item) != list:
                     print item
             else:
                     for num in item:
                             print num  
>>> flat(list1)
1
2
3
4
5
6
7
8

然后我使用相同的逻辑通过列表理解展平我的列表我得到以下错误

    list2 = [item if type(item) != list else num for num in item for item in list1]
    Traceback (most recent call last):
    File "<stdin>", line 1, in <module>
    TypeError: 'int' object is not iterable

如何使用列表理解来展平这种类型的列表？

我有一个类将字典转换为这样的对象

class Dict2obj(dict):
    __getattr__= dict.__getitem__

    def __init__(self, d):
        self.update(**dict((k, self.parse(v))
                           for k, v in d.iteritems()))

   @classmethod
   def parse(cls, v):
    if isinstance(v, dict):
        return cls(v)
    elif isinstance(v, list):
        return [cls.parse(i) for i in v]
    else:
        return v

当我尝试制作对象的深层副本时,我收到了此错误

import copy
my_object  = Dict2obj(json_data)
copy_object = copy.deepcopy(my_object)

File "/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/copy.py", line 172, in deepcopy
copier = getattr(x, "__deepcopy__", None)
KeyError: '__deepcopy__'

但是我在类I中重写getattr函数Dict2obj能够进行深层复制操作.见下面的例子

class Dict2obj(dict):

    __getattr__= dict.__getitem__

    def __init__(self, d):
        self.update(**dict((k, self.parse(v))
                           for k, v in d.iteritems()))

    def __getattr__(self, key): …

我想阅读服务总线的订阅消息.我正在使用qpid-protonpython库.我正在关注此链接以接收消息Proton-Python-Example-Simple-Receive.我正在通过此网址接收来自服务总线的消息 -

url = 'amqps://mynamespace.servicebus.windows.net/SharedAccessKeyName=xxxx/SharedAccessKey=xxxxxxxxx/python-test/Subscriptions/AllMessages'
# python-test is the name of the topic
# AllMessages is the name of the subscription

我收到以下错误 - ERROR:root:The messaging entity 'sb://mynamespace.servicebus.windows.net/sharedaccesskeyname=xxxxx/sharedaccesskey=xxxxxxxxxxxxx/python-test/subscriptions/allmessages' could not be found. TrackingId:c1e4a39edbd44040b2fd48a552d6ae2b_G2, SystemTracker:gateway6, Timestamp:7/19/2017 7:58:51 AM

这是因为上述URL未正确形成.我在网上搜索过,在这方面没有提供适当的文件.通过qpid读取订阅消息的正确URL格式是什么.

我有这样的HTML

<div id="all-stories" class="book"> 
<ul>

<li title="Book1"  ><a href="book1_url">Book1</a></li>

<li title="Book2"  ><a href="book2_url">Book2</a></li>
</ul>

</div>

我想使用xpath获取书籍及其各自的URL,但似乎我的方法不起作用.为简单起见,我试图提取"li"标签下的所有元素,如下所示

lis = tree.xpath('//div[@id="all-stories"]/div/text()')

我想在django中更改request.GET querydict对象.我尝试了这个,但我所做的改变没有反映出来.我试过这个

tempdict = self.request.GET.copy() # Empty initially
tempdict['state'] = ['XYZ',]
tempdict['ajaxtype'] = ['facet',]
print self.request.GET

我明白了

<QueryDict: {}> as my output

是否可以在django中更改request.GET querydict对象

我有一个表测试,列名具有整数和非整数值

+------------+
| names      |
+------------+
| 123        |
| 123abc     |
| 89         |
| dkkjdk     |
| dkdn       |
+------------+

我想像这样在一行中显示整数和非整数值的计数

integer_count non_integer_count
2               3

我试图使用此查询打印整数值

select cast(names as signed) as int_val from test;

但我得到了这个结果

+---------+
| int_val |
+---------+
|     123 |
|     123 |
|      89 |
|       0 |
|       0 |
+---------+

name字段是varchar(20)字段.

我有一些文件,我必须从mongodb获取并将其设置为memcache.这是代码

import memcache
from pymongo import MongoClient

db = mongo_client.job_db.JobParsedData
jobs = db.find().sort("JobId", 1)

def set_to_memcache_raw(jobs):
    print("Setting raw message to memcache")
    count = 0
    for item in jobs:
        job_id = item.get('JobId')
        job_details = item.get('JobDetails')
        if job_id.strip():
            count += 1
            memcache_obj.set(job_id, job_details, time=72000)
            if count % 1000 == 0:
                print("Inserted {} keys in memcache".format(count))
            if count >= 1000000:
                break

但是,经过一些奇数次的迭代,代码抛出了这个错误 -

 Traceback (most recent call last):
  File "/home/dimension/.virtualenvs/docparser/lib/python3.5/site-packages/pymongo/pool.py", line 450, in receive_message
    self.sock, operation, request_id, self.max_message_size)
  File "/home/dimension/.virtualenvs/docparser/lib/python3.5/site-packages/pymongo/network.py", line 137, in …