python lines that start with @

Possible Duplicate: Understanding Python decorators I was reading a django app source code where I find this @login_required def activities(request = None, project_id = 0, task_id = 0, ... What does the line that start with @ mean? Please check out Python Decorators Explained. It has an amazing answer that will explain everything. It's a decorato

以@开头的python行

可能重复: 了解Python装饰器 我正在阅读一个Django应用程序的源代码,我发现这一点 @login_required def activities(request = None, project_id = 0, task_id = 0, ... 以@开头的行是什么意思? 请检查Python装饰器解释。 它有一个惊人的答案,将解释一切。 这是一个装饰者。 它所做的基本上是包装这个功能。 它与此代码等效: def activities(request = None,

Python decorators just syntactic sugar?

Possible Duplicate: Understanding Python decorators I am quite new on using Python decorators and from what I understand on my first impression that they are just syntactic sugar. Is there a more profound use of them for more complex uses ? Yes it is syntactic sugar. Everything can be achieved without them, but with a few more lines of code. But it helps you write more concise code. E

Python装饰只是语法糖?

可能重复: 了解Python装饰器 我在使用Python装饰器方面颇为新颖,从我对第一印象的理解来看,它们只是语法糖。 他们是否有更深入的用途来进行更复杂的用途? 是的,它是句法糖。 一切都可以在没有它们的情况下实现,但需要更多的代码。 但它可以帮助您编写更简洁的代码。 例子: from functools import wraps def requires_foo(func): @wraps(func) def wrapped(self, *args, **kwargs): if not h

How can I make my Python script faster?

I'm pretty new to Python, and I have written a (probably very ugly) script that is supposed to randomly select a subset of sequences from a fastq-file. A fastq-file stores information in blocks of four rows each. The first row in each block starts with the character "@". The fastq file I use as my input file is 36 GB, containing about 14,000,000 lines. I tried to rewrite an alr

我怎样才能让我的Python脚本更快?

我对Python很陌生,并且写了一个(可能非常难看的)脚本,它应该从fastq文件中随机选择一部分序列。 一个fastq文件将信息存储在每行四个块中。 每个块的第一行以字符“@”开头。 我用作输入文件的fastq文件是36 GB,包含大约1400万行。 我试图重写一个已经存在的使用太多内存的脚本,并且我设法减少了很多内存使用。 但脚本需要永远运行,我不明白为什么。 parser = argparse.ArgumentParser() parser.add_argument("infile"

Cython not recognizing c++11 commands

I'm wrapping a C++ class with Python, and I cannot compile any C++11 features with the Cython module. Everything compiles okay when compiling the C++ alone. But when I run this setup.py below: setup( ext_modules = cythonize( "marketdata.pyx", # our Cython source sources=["cpp/OBwrapper.cpp, cpp/OrderBook/orderbook.h, cpp/OrderBook/orderbook.cpp"], # additiona

Cython不识别c ++ 11命令

我使用Python封装了一个C ++类,并且我无法使用Cython模块编译任何C ++ 11功能。 编译C ++时,一切都编译好。 但是当我运行下面的setup.py时: setup( ext_modules = cythonize( "marketdata.pyx", # our Cython source sources=["cpp/OBwrapper.cpp, cpp/OrderBook/orderbook.h, cpp/OrderBook/orderbook.cpp"], # additional source file(s) language="c++", # generate

Separate objects from the background. ( Using contour detection )

I want to separate objects from the background. The edge image was taken using the sobel edge detection with some filters. Then using the function "find contours" in opencv I get all the contours in the image. Then remove the contours that are not in the specific threshold size. (area of the contours should be greater than the 1/1000 th of the image area and smaller then 1/25 th of

从背景中分离物体。 (使用轮廓检测​​)

我想分离背景中的物体。 边缘图像是使用某些滤波器的sobel边缘检测拍摄的。 然后使用opencv中的“查找轮廓”功能,我可以获得图像中的所有轮廓。 然后删除不在特定阈值大小的轮廓。 (等高线的面积应大于图像面积的1/1000,小于图像面积的1/25) 主要问题是边缘检测的输出不是很精确。 有几个物体在图像中非常接近和模糊。 在计算轮廓时,两个物体都将被视为一个轮廓,那么整个轮廓就会被拒绝,因为它不在正确的范围内。

Improving Rock Extraction Analysis in OpenCV

I've been experimenting with OpenCV and trying my hand at some image analysis with NASA images. My goal is to extract rocks from scenes from the Mars rovers. I've managed to extract a few, but the hit rate is pretty poor as seen in these pictures: Original: After Processing: I have been using a Canny edge detector, finding the enclosed contours, and then filtering those by size.

改进OpenCV中的岩石提取分析

我一直在试用OpenCV,并试图用NASA图像进行一些图像分析。 我的目标是从火星漫游车的场景中提取岩石。 我设法提取了一些,但是在这些图片中看到的命中率很差: 原版的: 处理后: 我一直在使用Canny边缘检测器,找到封闭的轮廓,然后按大小对其进行过滤。 我尝试过使用一些像扩张和侵蚀这样的形态学操作来提高我的命中率,但它似乎没有帮助。 这是我的提取尝试: original_img = cv2.imread('mars_sample_image.jpg',

How can I change the Python that my Django project is using?

I have 2 versions installed in my server. I used virtualenv to install Python 2.7. I am using WSGI to deploy my project. WSGIPythonPath /home/ENV/lib/python2.7/site-packages WSGIScriptAlias / /var/www/html/my_project/wsgi.py My http.conf is like that. python -V gives Python 2.7.3 But in my projects Debug window, it says Django is using 2.6.8. Where am I wrong? UPDATE: Here is my wsg

我该如何更改我的Django项目正在使用的Python?

我有2个版本安装在我的服务器上。 我使用virtualenv来安装Python 2.7。 我正在使用WSGI来部署我的项目。 WSGIPythonPath /home/ENV/lib/python2.7/site-packages WSGIScriptAlias / /var/www/html/my_project/wsgi.py 我的http.conf就是这样。 蟒蛇-V给 Python 2.7.3 但在我的项目调试窗口中,它说Django使用2.6.8。 我错在哪里? 更新:这是我的wsgi文件 import os import sys sys.path.append('/var/www/html')

Any idea why i am getting this error?

I am trying to install django-cms in my ubuntu pc. I am getting this error: Traceback (most recent call last): File "manage.py", line 14, in <module> execute_manager(settings) File "/usr/local/lib/python2.6/dist-packages/django/core/management/__init__.py", line 438, in execute_manager utility.execute() File "/usr/local/lib/python2.6/dist-packages/django/core/management/

任何想法,为什么我得到这个错误?

我正试图在我的ubuntu pc上安装django-cms。 我得到这个错误: Traceback (most recent call last): File "manage.py", line 14, in <module> execute_manager(settings) File "/usr/local/lib/python2.6/dist-packages/django/core/management/__init__.py", line 438, in execute_manager utility.execute() File "/usr/local/lib/python2.6/dist-packages/django/core/management/__init__.py", lin

Implicit Module Search in Python?

I'm working on a more complete understanding of Python Virtual Environment for portablity issues across cheap hosting solutions and development platforms and there are pieces I don't understand. I've been building virtual environments, but I'm not seeing the COMPLETE ISOLATION I need for copying virtual environments to hosts assuming the hosts don't have python installed T

Python中的隐式模块搜索?

我正在更全面地了解Python虚拟环境,以了解廉价托管解决方案和开发平台中的可移植性问题,并且有些内容我不明白。 我一直在构建虚拟环境,但我没有看到完整的隔离,我需要将虚拟环境复制到主机,假设主机没有安装python 问题是,我将建立,这不是完全从装置P隔离的平台P -斌/ python2.6的- LIB / python2.6的/站点包 PATH更改为仅包含P / bin / PYTHONPATH是空白的 完美的作品。 但是,它仍然访问系统目录来解决不在

virtualenv confusion

So I open a terminal, cd to my desktop, and run: virtualenv test_env I then create the following file in my normal environment: /home/jesse/.local/lib/python2.6/site-packages/foo_package/__init__.py This file contains one line: print "importing from normal env" In the test_env I create: /home/jesse/Desktop/test_env/lib/python2.6/site-packages/foo_package/__init__.py Cont

virtualenv困惑

所以我打开一个终端,cd到我的桌面,然后运行: virtualenv test_env 然后我在我的正常环境中创建以下文件: /home/jesse/.local/lib/python2.6/site-packages/foo_package/__init__.py 该文件包含一行: print "importing from normal env" 在test_env中创建: /home/jesse/Desktop/test_env/lib/python2.6/site-packages/foo_package/__init__.py 含: print "importing from test env"