I have built a python app with PyQt4 library. The app has a graphics scene with a picture on it. I then draw lines over the scene and I catch mouse movements to get the coordinates of the mouse location over the picture. I attach the mousehover like this: pmItem = QGraphicsPixmapItem(pm, None, self.scene) pmItem.setAcceptHoverEvents(True) pmItem.hoverMoveEvent = self.pixelSelect pmItem.hove
我已经用PyQt4库构建了一个python应用程序。 该应用程序有一个图形场景和图片。 然后我在场景上画线，然后捕捉鼠标移动以获得鼠标在画面上的位置坐标。 我附上这样的mousehover： pmItem = QGraphicsPixmapItem(pm, None, self.scene) pmItem.setAcceptHoverEvents(True) pmItem.hoverMoveEvent = self.pixelSelect pmItem.hoverLeaveEvent = self.hover_leave pixelSelect方法获取当前坐标并测量到最近的垂直和水平线的
I am using python 2.7 on Linux Mint 16. I am facing an error, if I run my IDE (tried it on Spyder and Pycharm) from a program launcher (eg. from the prompt at Alt F2 or an icon shortcut on my desktop) the modules do not load and I get the following error File "/usr/local/lib/python2.7/dist-packages/gurobipy/__init__.py", line 1, in from .gurobipy import * ImportError: libgurobi56.so: canno
我在Linux Mint 16上使用python 2.7。如果我从程序启动器运行IDE（在Spyder和Pycharm上尝试它）（例如，从Alt F2的提示或桌面上的图标快捷方式），我正面临一个错误。模块不加载，我得到以下错误 File "/usr/local/lib/python2.7/dist-packages/gurobipy/__init__.py", line 1, in from .gurobipy import * ImportError: libgurobi56.so: cannot open shared object file: No such file or directory 但是，如果我从命令
As far as I can tell, aiohttp is currently the reference for web servers built on asyncio. While it's an impressive piece of work, it doesn't quite match my particular (peculiar) taste: In addition to being a web server, it's also a framework for web applications (with routing and everything), plus the API is fairly class-heavy. I'd prefer something that is just an HTTP server,
据我所知，aiohttp目前是构建在asyncio上的Web服务器的参考。 虽然这是一项令人印象深刻的工作，但它并不完全符合我特殊的（特殊）口味：除了作为Web服务器之外，它也是Web应用程序的框架（包括路由和所有内容），此外，重。 我更喜欢东西只是一个HTTP服务器，提供一个最小的接口（基本上是一个异步的WSGI等价物）。 这样的事情是否已经存在？ 那么，你可以尝试低级的aiohttp.server而不是aiohttp.web 。 它没有路线，中
This question already has an answer here: Why not use Double or Float to represent currency? 14 answers The answer here is that you should not represent indivisible values as floats. Represent everything in cents. The reason that is the solution is that floats cannot precisely represent all decimals. Accordingly, division by different amounts will not do what you want. Instead, you need
这个问题在这里已经有了答案： 为什么不使用Double或Float来表示货币？ 14个答案 这里的答案是，你不应该将不可分割的值表示为浮点数。 以美分表示一切。 解决问题的原因是浮点数不能精确地表示所有小数点。 因此，按不同金额划分将不会做你想要的。 相反，您需要划分整数金额，并适当处理发生的剩余金额。 一般来说，浮动上的平等检查也是出于这个原因的危险：您需要决定哪些差异“足够平等”并进行测试。 把所有东
I want to install Lxml so I can then install Scrapy. When I updated my Mac today it wouldn't let me reinstall lxml, I get the following error: In file included from src/lxml/lxml.etree.c:314: /private/tmp/pip_build_root/lxml/src/lxml/includes/etree_defs.h:9:10: fatal error: 'libxml/xmlversion.h' file not found #include "libxml/xmlversion.h" ^ 1 error generated. error: command 'cc'
我想安装Lxml，以便我可以安装Scrapy。 当我今天更新我的Mac时，它不会让我重新安装lxml，我得到以下错误： In file included from src/lxml/lxml.etree.c:314: /private/tmp/pip_build_root/lxml/src/lxml/includes/etree_defs.h:9:10: fatal error: 'libxml/xmlversion.h' file not found #include "libxml/xmlversion.h" ^ 1 error generated. error: command 'cc' failed with exit status 1 我已经尝试使用bre
Here the code on Python3.3: import sys, re, math str1 = str(sys.stdin.readlines()) Data = re.findall('\b\d+\b', str1) for i in reversed (Data): print('%.4f' % math.sqrt(float(i))) as you can see, this program grabs data (multi-line random string) from input, and search for every digits this string contains. After that just returns square root of every digit it finds. Well, algorithm wor
这里是Python3.3上的代码： import sys, re, math str1 = str(sys.stdin.readlines()) Data = re.findall('\b\d+\b', str1) for i in reversed (Data): print('%.4f' % math.sqrt(float(i))) 正如你所看到的，这个程序从输入中抓取数据（多行随机字符串），并搜索这个字符串包含的每个数字。 之后，它只返回找到的每个数字的平方根。 那么，算法的作品，但速度不够快，我不知道如何优化它。 请帮助我。 我需要做什么
For some reason, numpy's QR decomposition is always faster than scipy's which seems strange because scipy is supposed to include all of numpy plus more advanced features. Any idea why? import numpy.linalg as nla import scipy.linalg as la A = np.random.randn(4000,4000) %timeit -n 3 -r 3 Q,R = nla.qr(A) %timeit -n 3 -r 3 Q,R = la.qr(A) %timeit -n 3 -r 3 Q,R = nla.qr(A) %timeit -n 3 -r 3
出于某种原因，numpy的QR分解总是比scipy更快，这似乎很奇怪，因为scipy应该包含所有numpy和更高级的功能。 任何想法为什么？ import numpy.linalg as nla import scipy.linalg as la A = np.random.randn(4000,4000) %timeit -n 3 -r 3 Q,R = nla.qr(A) %timeit -n 3 -r 3 Q,R = la.qr(A) %timeit -n 3 -r 3 Q,R = nla.qr(A) %timeit -n 3 -r 3 Q,R = la.qr(A) 3个循环，最好为3：每循环1.17秒 3个回路，最好3个：每回路1
I'm experimenting with NumPy to see how and where it is faster than using generic list comprehensions in Python. Here's a standard coding question I'm using for this experiment. Find the sum of all the multiples of 3 or 5 below 1000000. I have written three functions to compute this number. def fA(M): sum = 0 for x in range(M): if x % 3 == 0 or x % 5 == 0:
我正在尝试使用NumPy来查看在Python中如何以及在哪里使用通用列表解析。 这是我用于此实验的标准编码问题。 Find the sum of all the multiples of 3 or 5 below 1000000. 我写了三个函数来计算这个数字。 def fA(M): sum = 0 for x in range(M): if x % 3 == 0 or x % 5 == 0: sum += x return sum def fB(M): multiples_3 = range(0, M, 3) multiples_5 = range(0, M, 5) mult
I am comparing performance of numpy vs matlab, in several cases I observed that numpy is significantly slower (indexing, simple operations on arrays such as absolute value, multiplication, sum, etc.). Let's look at the following example, which is somehow striking, involving the function digitize (which I plan to use for synchronizing timestamps): import numpy as np import time scale=np.aran
我在比较numpy与matlab的性能，在几种情况下，我观察到numpy显着较慢（索引，对数组的简单操作，如绝对值，乘法，和等）。 让我们看看下面的例子，这是一个惊人的，涉及数字化函数（我打算用于同步时间戳）： import numpy as np import time scale=np.arange(1,1e+6+1) y=np.arange(1,1e+6+1,10) t1=time.time() ind=np.digitize(scale,y) t2=time.time() print 'Time passed is %2.2f seconds' %(t2-t1) 结果是： 时间
Lets start with three arrays of dtype=np.double . Timings are performed on a intel CPU using numpy 1.7.1 compiled with icc and linked to intel's mkl . A AMD cpu with numpy 1.6.1 compiled with gcc without mkl was also used to verify the timings. Please note the timings scale nearly linearly with system size and are not due to the small overhead incurred in the numpy functions if statements
让我们从三个dtype=np.double数组开始。 计时是在英特尔CPU上使用nccu 1.7.1与icc编译并链接到intel的mkl 。 使用gcc而不使用mkl编译的numpy 1.6.1的AMD cpu也用于验证时序。 请注意，时序规模几乎与系统规模成线性关系，并不是因为numpy函数的小开销，因为if声明这些差异将以毫秒而非毫秒显示： arr_1D=np.arange(500,dtype=np.double) large_arr_1D=np.arange(100000,dtype=np.double) arr_2D=np.arange(500**2,dtype=np.do