我将发送一个c++数组到python函数,numpy array然后再返回另一个numpy array.在查阅了numpy文档和其他一些线程并调整代码后,最后代码正在运行,但我想知道这段代码是否以最佳方式编写,考虑到:
c++和之间不必要地复制数组numpy (python).C++代码:
// python_embed.cpp : Defines the entry point for the console application.
//
#include "stdafx.h"
#define NPY_NO_DEPRECATED_API NPY_1_7_API_VERSION
#include "Python.h"
#include "numpy/arrayobject.h"
#include<iostream>
using namespace std;
int _tmain(int argc, _TCHAR* argv[])
{
Py_SetProgramName(argv[0]);
Py_Initialize();
import_array()
// Build the 2D array
PyObject *pArgs, *pReturn, *pModule, *pFunc;
PyArrayObject *np_ret, *np_arg;
const int SIZE{ 10 };
npy_intp dims[2]{SIZE, SIZE};
const int ND{ 2 }; …Run Code Online (Sandbox Code Playgroud) 我正在用matplotlib创建几个pdf图,它由400个子图组成.每个只有5个数据点.在一台好的电脑上需要420秒才能保存5张pdf图片.有没有办法优化代码,或者matplotlib是正常的?
绘图的部分代码:
plot_cnt = 1
for k in np.arange(K_min, K_max + 1):
for l in np.arange(L_min, L_max + 1):
ax = plt.subplot(grid[0], grid[1], plot_cnt)
plot_cnt += 1
plt.setp(ax, 'frame_on', False)
ax.set_ylim([-0.1, 1.1])
ax.set_xlabel('K={},L={}'.format(k, l), size=3)
ax.set_xlim([-0.1, 4.1])
ax.set_xticks([])
ax.set_yticks([])
ax.grid('off')
ax.plot(np.arange(5), (data['S1']['Azimuth'][:, k - 1, l + offset_l] + \
data['S1']['Delta Speed'][:, k - 1, l + offset_l] + \
data['S1']['Speed'][:, k - 1, l + offset_l]) / 3,
'r-o', ms=1, mew=0, mfc='r')
ax.plot(np.arange(5), data['S2'][case][:, k - 1, l + …Run Code Online (Sandbox Code Playgroud) matlab-shell当我打印到stdout时,我在缓冲区中收到此警告:
Warning (undo): Buffer `*MATLAB*' undo info was 12268000 bytes long.
The undo info was discarded because it exceeded `undo-outer-limit'.
This is normal if you executed a command that made a huge change
to the buffer. In that case, to prevent similar problems in the
future, set `undo-outer-limit' to a value that is large enough to
cover the maximum size of normal changes you expect a single
command to make, but not so large that it might exceed the …Run Code Online (Sandbox Code Playgroud) 我有一个二阶微分方程,我想在 python 中解决它。问题是,对于其中一个变量,我没有初始条件,0而只有无穷大的值。谁能告诉我应该提供哪些参数scipy.integrate.odeint?能解决吗?
方程:

需要在时间方面找到 Theta。它的一阶导数在 处为零t=0。theta 未知,t=0但它在足够长的时间内变为零。其余的都是已知的。由于近似值I可以设置为零,因此消除了二阶导数,这将使问题更容易。
下面的代码用于空窗口,但在我的Intel i3上显示相对较高的CPU使用率25%.我也试过setFramerateLimit没有改变.有没有办法减少CPU使用率?
#include<SFML/Window.hpp>
void processEvents(sf::Window& window);
int main()
{
sf::Window window(sf::VideoMode(800, 600), "My Window", sf::Style::Close);
window.setVerticalSyncEnabled(true);
while (window.isOpen())
{
processEvents(window);
}
return 0;
}
void processEvents(sf::Window& window)
{
sf::Event event;
window.pollEvent(event);
switch (event.type)
{
case sf::Event::Closed:
window.close();
break;
}
}
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