It is possible to 'fill' an array in Python like so: > [0] * 10 [0, 0, 0, 0, 0, 0, 0, 0, 0, 0] I wanted to use this sample principle to quickly create a list of similar objects: > a = [{'key': 'value'}] * 3 > a [{'key': 'value'}, {'key': 'value'}, {'key': 'value'}] But it appears these objects are linked with one another: > a[0]['key'] = 'another value' > a [{'key': 'a
用Python填充数组是可能的，如下所示： > [0] * 10 [0, 0, 0, 0, 0, 0, 0, 0, 0, 0] 我想使用这个示例原则来快速创建一个类似对象的列表： > a = [{'key': 'value'}] * 3 > a [{'key': 'value'}, {'key': 'value'}, {'key': 'value'}] 但看起来这些对象是相互关联的： > a[0]['key'] = 'another value' > a [{'key': 'another value'}, {'key': 'another value'}, {'key': 'another value'}] 鉴于Python没
I am a newbie in Python. I installed Python 2.7.10 and it comes already with pip . I tried to run this command to download the NumPy library. D:workspacepython>pip install numpy Collecting numpy Could not find a version that satisfies the requirement numpy (from versions: ) No matching distribution found for numpy But I got this error. How can I solve this? I also want to know the dif
我是Python的新手。 我安装了Python 2.7.10，它已经与pip 。 我试图运行这个命令来下载NumPy库。 D:workspacepython>pip install numpy Collecting numpy Could not find a version that satisfies the requirement numpy (from versions: ) No matching distribution found for numpy 但是我得到了这个错误。 我该如何解决这个问题？ 我也想知道easy_install和pip之间的区别。 我建议下载NumPy安装程序，而不是从
I'm trying to mock a singluar method from the boto3 s3 client object to throw and exception. But I need all other methods for this class to work as normal. This is so I can test a singular Exception test when and error occurs performing a upload_part_copy 1st Attempt import boto3 from mock import patch with patch('botocore.client.S3.upload_part_copy', side_effect=Exception('Error Uploa
我试图从boto3 s3客户端对象中嘲讽singluar方法来抛出异常。 但我需要所有其他方法来正常工作。 这样，我可以在执行upload_part_copy时发生错误时测试单个异常测试 第一次尝试 import boto3 from mock import patch with patch('botocore.client.S3.upload_part_copy', side_effect=Exception('Error Uploading')) as mock: client = boto3.client('s3') # Should return actual result o = client.get_object
I have a tornado server which I am ok with running synchronously. One of the callback handlers calls a method in another module which internally uses generators to iterate over results, using the yield keyword. Calling the function directly works however triggering it from the tornado server seems to stop execution as soon as it reaches the yield keyword. Is it possible to use both? This is
我有一个龙卷风服务器，我可以同步运行。 其中一个回调处理程序在另一个模块中调用一个方法，该方法使用yield关键字在内部使用生成器遍历结果。 直接调用该函数可以起作用，但从龙卷风服务器触发它似乎会在达到yield关键字时立即停止执行。 可以同时使用吗？ 这是一个与使用简单的Python生成器作为Tornado异步处理程序中的协同例程相反的问题。 从龙卷风末端调用代码例如： class SearchHandler(tornado.web.RequestHan
Reading the Tornado documentation, it's very clear how to call an async function to return a response: class GenAsyncHandler(RequestHandler): @gen.coroutine def get(self): http_client = AsyncHTTPClient() response = yield http_client.fetch("http://example.com") do_something_with_response(response) self.render("template.html") What's lacking is how
阅读Tornado文档，非常清楚如何调用异步函数来返回响应： class GenAsyncHandler(RequestHandler): @gen.coroutine def get(self): http_client = AsyncHTTPClient() response = yield http_client.fetch("http://example.com") do_something_with_response(response) self.render("template.html") 缺乏的是如何调用一个与当前请求无关的后台任务： class GenAsyncHandler(RequestHa
I'm trying to write a Tornado request handler which makes asynchronous HTTP requests, and returns data to the client as it receives it from it's async requests. Unfortunately, I'm unable to get Tornado to return any data to the client until all of it's Async HTTPRequests have completed. A demo of my request handler is below. class StreamingHandler(web.RequestHandler): al
我正在尝试编写一个Tornado请求处理程序，它使得异步HTTP请求成为可能，并且在它从异步请求中接收数据时将数据返回给客户端。 不幸的是，我无法让Tornado将任何数据返回给客户端，直到它的所有异步HTTP请求都完成为止。 下面是我的请求处理程序的演示。 class StreamingHandler(web.RequestHandler): all_requested = False requests = [] @web.asynchronous def get(self): http_client = httpc
I have a tornado app and I want to to use a blocking library to accomplish something. In cases where it's not possible to rewrite the library in an async manner, what's the way to execute it in tornado? For example, I'd like to be able to put an @asynchronous decorator on a request handler, in it start some long running function that will just return a response once it's done.
我有一个龙卷风应用程序，我想用一个阻塞库来完成一些事情。 如果无法以异步方式重写库，那么在龙卷风中执行它的方式是什么？ 例如，我希望能够在请求处理程序中放置一个@asynchronous装饰器，并在其中启动一个长时间运行的函数，该函数一旦完成就会返回响应。 我不能只是回拨。 最简单的例子当然是什么才是正确的方式睡10秒，而不会阻止龙卷风的唯一线程？ 看起来我想要的仅仅是创建一个新的线程/进程，并且回调龙卷风的
OpenSSL provides a popular (but insecure – see below!) command line interface for AES encryption: openssl aes-256-cbc -salt -in filename -out filename.enc Python has support for AES in the shape of the PyCrypto package, but it only provides the tools. How to use Python/PyCrypto to decrypt files that have been encrypted using OpenSSL? Notice This question used to also concern encryption in
OpenSSL为AES加密提供了一个流行的（但不安全 - 见下文！）命令行界面： openssl aes-256-cbc -salt -in filename -out filename.enc Python支持PyCrypto软件包形式的AES，但它只提供工具。 如何使用Python / PyCrypto解密使用OpenSSL加密的文件？ 注意 这个问题也涉及到使用相同方案的Python中的加密。 此后我除去了那部分，以阻止任何人使用它。 不要以这种方式对更多数据进行加密，因为它不符合当今的标准。 除了BA
Background: I'm attempting to warp one face to another of a different shape. In order to warp one image to another, I'm using a delaunay triangulation of facial landmarks and warping the triangles of one portrait to the corresponding triangles of the second portrait. I'm using a barycentric coordinate system to map a point within a triangle to its corresponding warped location on t
背景：我试图将一张脸歪曲到另一个不同的形状。 为了将一幅图像转换为另一幅图像，我使用了面部标志的delaunay三角剖分，并将一幅肖像的三角形翘曲为第二幅肖像的相应三角形。 我使用重心坐标系将三角形内的一个点映射到另一个三角形上相应的变形位置。 我的第一种方法是用逆乘法方法求解系统Ax = b，其中A由三角形的三个角组成，b代表当前点，x代表此点的重心坐标（alpha，beta和gamma） ）。 我发现矩阵A的逆矩阵每个三
I have experienced that in other languages. Now I have the same problem in Python. I have a dictionary that has a lot of CRUD actions. One would assume that deleting elements from a dictionary should decrease the memory footprint of it. It's not the case. Once a dictionary grows in size (doubling usually), it never(?) releases allocated memory back. I have run this experiment: import r
我已经在其他语言中体验过。 现在我在Python中遇到了同样的问题。 我有一个有很多CRUD操作的字典。 人们会假设从字典中删除元素应该减少它的内存占用量。 情况并非如此。 一旦字典大小增加（通常加倍），它永远不会（？）释放分配的内存。 我已经运行了这个实验： import random import sys import uuid a= {} for i in range(0, 100000): a[uuid.uuid4()] = uuid.uuid4() if i % 1000 == 0: print sys.g