我正在使用 python,当然你不能很快地遍历大图像的每个像素,所以我遵循 C DLL。
我想做这样的事情:
img = QImage("myimage.png").constBits()
imgPtr = c_void_p(img)
found = ctypesDLL.myImageSearchMethod(imgPtr, width, height)
Run Code Online (Sandbox Code Playgroud)
但是这一行 imgPtr = c_void_p(img) ylds
builtins.TypeError:无法转换为指针
我不需要修改位。请教我你在这方面的绝地方法。
如果您查看https://en.wikipedia.org/wiki/Clique_problem,您会注意到派系和最大派系之间存在区别。最大集团只包含在其自身之外,不包含其他集团。所以我想要那些派系,但networkx似乎只提供:
networkx.algorithms.clique.enumerate_all_cliques(G)
所以我尝试了一个简单的for循环过滤机制(见下文)。
def filter_cliques(self, cliques):
# TODO: why do we need this? Post in forum...
res = []
for C in cliques:
C = set(C)
for D in res:
if C.issuperset(D) and len(C) != len(D):
res.remove(D)
res.append(C)
break
elif D.issuperset(C):
break
else:
res.append(C)
res1 = []
for C in res:
for D in res1:
if C.issuperset(D) and len(C) != len(D):
res1.remove(D)
res1.append(C)
elif D.issuperset(C):
break
else:
res1.append(C)
return res1
Run Code Online (Sandbox Code Playgroud)
我想过滤掉所有合适的小派系。但正如你所看到的,它很糟糕,因为我必须过滤两次。这不是很优雅。所以,问题是,给定一个对象列表(整数、字符串),它们是图中的节点标签;enumerate_all_cliques(G)
准确返回此标签列表列表。现在,给定这个列表列表,过滤掉所有适当的子团。例如:
[[a, b, c], [a, b], …
我需要将args和kwargs存储在一个元组中以便稍后调用,所以在这种情况下,元组中的适当值是*args还是args?换句话说,这会工作:
def __init__(self, *args, **kwargs):
self._buildCalls = [
(self.__init__, args, kwargs)
]
self._building = False
def __getattr__(self, attr):
if self.building():
if hasmethod(self, attr):
return lambda *args, **kwargs: self.argSaver(attr, *args, **kwargs)
else:
return super().__getattr__(attr)
def argSaver(self, method, *args, **kwargs):
self._buildCalls.append((method, args, kwargs))
return method(*args, **kwargs)
Run Code Online (Sandbox Code Playgroud) main = do line <- getLine
let line' = reverse line
putStrLn $ "You said " ++ line' ++ " backwards!"
putStrLn $ "Yes, you really said " ++ line ++ " backwards!"
Run Code Online (Sandbox Code Playgroud)
错误:
$ stack runhaskell "c:\Users\FruitfulApproach\Desktop\Haskell\test.hs"
C:\Users\FruitfulApproach\Desktop\Haskell\test.hs:4:5: error:
parse error on input `putStrLn'
|
4 | putStrLn $ "You said " ++ line' ++ " backwards!"
| ^^^^^^^^
Run Code Online (Sandbox Code Playgroud)
我也尝试过在缩进线上使用单个选项卡。
这是我的标签->空格设置:
我也尝试过重新启动VSCode。
提前致谢!
MWE 演示这个问题是这样的:
#include <QMainWindow>
#include <QApplication>
#include <QWidget>
#include <QWebEngineView>
#include <QGridLayout>
int main(int argc, char *argv[])
{
QApplication a(argc, argv);
QMainWindow w;
auto w1 = new QWidget();
w1->setLayout(new GridLayout());
auto view = new QWebEngineView();
view->load(QUrl("file://C:\\Users\\FruitfulApproach\\Desktop\\AbstractSpacecraft\\MathEnglishApp\\KaTeX_template.html"));
view->show();
w1->layout()->addWidget(view);
w.setCentralWidget(w1);
w.show();
return a.exec();
}
Run Code Online (Sandbox Code Playgroud)
KaTeX_template.html:
<!DOCTYPE html>
<html>
<head>
<title>MathJax TeX Test Page</title>
<script type="text/x-mathjax-config">
MathJax.Hub.Config({tex2jax: {inlineMath: [['$','$'], ['\\(','\\)']]}});
</script>
<script type="text/javascript" async
src="https://example.com/mathjax/MathJax.js?config=TeX-AMS_CHTML">
</script>
</head>
<body>
When $a \ne 0$, there are two solutions to \(ax^2 + bx + c …
Run Code Online (Sandbox Code Playgroud) python ×3
python-3.x ×3
qt ×2
algorithm ×1
args ×1
c++ ×1
ctypes ×1
dereference ×1
graph-theory ×1
haskell ×1
kwargs ×1
networkx ×1
pyqt ×1