好的，我是新手。我终于使用swig和numpy.i成功地包装了我的python程序中最昂贵的部分。该程序是2D波PDE的有限差分方案。我的问题是我现在如何使用它？我在IPython中导入后可以看到它。
In [1]: import wave2
In [2]: wave2.wave_prop
Out[2]: <function _wave2.wave_prop>
但是，当我去使用它时，我得到一个错误说：
TypeError: in method 'wave_prop', argument 1 of type 'float **'
如何将2D numpy数组转换为某种形式，使我可以使用它。还有另一个非常相似的stackoverflow没有帮助我，虽然我在这个过程中找到了很多帮助。
以下是标题：
void wave_prop(float** u_prev ,int Lx, int Ly,float** u ,int Lx2, int Ly2,float** u_next,int Lx3,int Ly3 );
这是c代码：
#include<string.h>
#include<stdlib.h>
#include<stdio.h>
#include<math.h>
#define n 100
void wave_prop(float** u_prev ,int Lx,int Ly,float** u ,int Lx2,int Ly2,float** u_next,int Lx3,int Ly3 ){
int dx=1;
int dy=1;
float c=1;
float dt =1;
int t_old=0;int t=0;int t_end=150;
int x[Lx];
int y[Ly];
for(int i=0;i<=99;i++){
x[i]=i;
y[i]=i;
}
while(t<t_end){
t_old=t; t +=dt;
//the wave steps through time
for (int i=1;i<99;i++){
for (int j=1;j<99;j++){
u_next[i][j] = - u_prev[i][j] + 2*u[i][j] + \
(c*dt/dx)*(c*dt/dx)*u[i-1][j] - 2*u[i][j] + u[i+1][j] + \
(c*dt/dx)*(c*dt/dx)*u[i][j-1] - 2*u[i][j] + u[i][j+1];
}
}
//set boundary conditions to 0
for (int j=0;j<=99;j++){ u_next[0][j] = 0;}
for (int i=0;i<=99;i++){ u_next[i][0] = 0;}
for (int j=0;j<=99;j++){ u_next[Lx-1][j] = 0;}
for (int i=0;i<=99;i++){ u_next[i][Ly-1] = 0;}
//memcpy(dest, src, sizeof (mytype) * rows * coloumns);
memcpy(u_prev, u, sizeof (float) * Lx * Ly);
memcpy(u, u_next, sizeof (float) * Lx * Ly);
}
}
这是我的界面：
%module wave2
%{
#define SWIG_FILE_WITH_INIT
#include "wave2.h"
%}
%include "numpy.i"
%init %{
import_array();
%}
%include "wave2.h"
%apply (float** INPLACE_ARRAY2, int DIM1, int DIM2) { (float** u_prev,int Lx,int Ly ),(float** u,int Lx2,int Ly2),(float* u_next,int Lx3,int Ly3)}
这些是我用来编译和链接的命令：
$ swig -python wave2.i
$ gcc -c -fpic wave2.c wave2_wrap.c -I/usr/include/python2.7 -std=c99
$ gcc -shared wave2.o wave2_wrap.o -o _wave2.so
没有任何错误或警告。在互联网上缺乏像这样的中间例子，相信我，我已经搜索过了！所以如果WE能够使这个工作，它可以作为一个很好的教程给某人。请不要将我的问题标记下来，然后走到深夜。如果您认为我的部分编码需要改进，请告诉我我现在基本上都在教自己一切......非常感谢您的帮助
哦，这也是我试图使用它的脚本。我也尝试在IPython中以其他方式使用该函数...
'''George Lees Jr.
2D Wave pde '''
from numpy import *
import numpy as np
import matplotlib.pyplot as plt
from wave2 import *
import wave2
#declare variables
#need 3 arrays u_prev is for previous time step due to d/dt
Lx=Ly = (100)
n=100
dx=dy = 1
x=y = np.array(xrange(Lx))
u_prev = np.array(zeros((Lx,Ly),float))
u = np.array(zeros((Lx,Ly),float))
u_next = np.array(zeros((Lx,Ly),float))
c = 1 #constant velocity
dt = (1/float(c))*(1/sqrt(1/dx**2 + 1/dy**2))
t_old=0;t=0;t_end=150
#set Initial Conditions and Boundary Points
#I(x) is initial shape of the wave
#f(x,t) is outside force that creates waves set =0
def I(x,y): return exp(-(x-Lx/2.0)**2/2.0 -(y-Ly/2.0)**2/2.0)
def f(x,t,y): return 0
#set up initial wave shape
for i in xrange(100):
for j in xrange(100):
u[i,j] = I(x[i],y[j])
#copy initial wave shape for printing later
u1=u.copy()
#set up previous time step array
for i in xrange(1,99):
for j in xrange(1,99):
u_prev[i,j] = u[i,j] + 0.5*((c*dt/dx)**2)*(u[i-1,j] - 2*u[i,j] + u[i+1,j]) + \
0.5*((c*dt/dy)**2)*(u[i,j-1] - 2*u[i,j] + u[i,j+1]) + \
dt*dt*f(x[i], y[j], t)
#set boundary conditions to 0
for j in xrange(100): u_prev[0,j] = 0
for i in xrange(100): u_prev[i,0] = 0
for j in xrange(100): u_prev[Lx-1,j] = 0
for i in xrange(100): u_prev[i,Ly-1] = 0
wave2.wave_prop( u_prev ,Lx ,Ly , u , Lx, Ly, u_next,Lx,Ly )
#while t<t_end:
# t_old=t; t +=dt
#the wave steps through time
# for i in xrange(1,99):
# for j in xrange(1,99):
# u_next[i,j] = - u_prev[i,j] + 2*u[i,j] + \
# ((c*dt/dx)**2)*(u[i-1,j] - 2*u[i,j] + u[i+1,j]) + \
# ((c*dt/dx)**2)*(u[i,j-1] - 2*u[i,j] + u[i,j+1]) + \
# dt*dt*f(x[i], y[j], t_old)
#
# #set boundary conditions to 0
#
# for j in xrange(100): u_next[0,j] = 0
# for i in xrange(100): u_next[i,0] = 0
# for j in xrange(100): u_next[Lx-1,j] = 0
# for i in xrange(100): u_next[i,Ly-1] = 0
#set prev time step equal to current one
# u_prev = u.copy(); u = u_next.copy();
fig = plt.figure()
plt.imshow(u,cmap=plt.cm.ocean)
plt.colorbar()
plt.show()
print u_next
同样是的，我检查确保数组都是numpy nd数组类型
答案 0 :(得分：0)
好的，所以我能够在Cython中完成我想要的，感谢上帝。在这个过程中，我发现Cython更强大的工具！
因此，结果是使用有限差分求解的2D波PDE。主计算已导出为Cythonized函数。该函数接收三个2D np.ndarrays并返回一个。使用numpy和其他数据类型也更容易。
这是Cythonized函数：
from numpy import *
cimport numpy as np
def cwave_prop( np.ndarray[double,ndim=2] u_prev, np.ndarray[double,ndim=2] u, np.ndarray[double,ndim=2] u_next):
cdef double t = 0
cdef double t_old = 0
cdef double t_end = 100
cdef int i,j
cdef double c = 1
cdef double Lx = 100
cdef double Ly = 100
cdef double dx = 1
cdef double dy = 1
cdef double dt = (1/(c))*(1/(sqrt(1/dx**2 + 1/dy**2)))
while t<t_end:
t_old=t; t +=dt
#wave steps through time and space
for i in xrange(1,99):
for j in xrange(1,99):
u_next[i,j] = - u_prev[i,j] + 2*u[i,j] + \
((c*dt/dx)**2)*(u[i-1,j] - 2*u[i,j] + u[i+1,j]) + \
((c*dt/dx)**2)*(u[i,j-1] - 2*u[i,j] + u[i,j+1])
#set boundary conditions of grid to 0
for j in xrange(100): u_next[0,j] = 0
for i in xrange(100): u_next[i,0] = 0
for j in xrange(100): u_next[Lx-1,j] = 0
for i in xrange(100): u_next[i,Ly-1] = 0
#set prev time step equal to current one
for i in xrange(100):
for j in xrange(100):
u_prev[i,j] = u[i,j];
u[i,j] = u_next[i,j];
print u_next
这是我的python脚本调用它并返回它然后绘制结果。 欢迎任何有关编写更好代码的建议......
'''George Lees Jr.
2D Wave pde '''
from numpy import *
import numpy as np
import matplotlib.pyplot as plt
import cwave2
np.set_printoptions(threshold=np.nan)
#declare variables
#need 3 arrays u_prev is for previous time step due to time derivative
Lx=Ly = (100) #Length of x and y dims of grid
dx=dy = 1 #derivative of x and y respectively
x=y = np.array(xrange(Lx)) #linspace to set the initial condition of wave
u_prev=np.ndarray(shape=(Lx,Ly), dtype=np.double) #u_prev 2D grid for previous time step needed bc of time derivative
u=np.ndarray(shape=(Lx,Ly), dtype=np.double) #u 2D grid
u_next=np.ndarray(shape=(Lx,Ly), dtype=np.double) #u_next for advancing the time step #also these are all numpy ndarrays
c = 1 #setting constant velocity of the wave
dt = (1/float(c))*(1/sqrt(1/dx**2 + 1/dy**2)) #we have to set dt specifically to this or numerical approx will fail!
print dt
#set Initial Conditions and Boundary Points
#I(x) is initial shape of the wave
def I(x,y): return exp(-(x-Lx/2.0)**2/2.0 -(y-Ly/2.0)**2/2.0)
#set up initial wave shape
for i in xrange(100):
for j in xrange(100):
u[i,j] = I(x[i],y[j])
#set up previous time step array
for i in xrange(1,99):
for j in xrange(1,99):
u_prev[i,j] = u[i,j] + 0.5*((c*dt/dx)**2)*(u[i-1,j] - 2*u[i,j] + u[i+1,j]) + \
0.5*((c*dt/dy)**2)*(u[i,j-1] - 2*u[i,j] + u[i,j+1])
#set boundary conditions to 0
for j in xrange(100): u_prev[0,j] = 0
for i in xrange(100): u_prev[i,0] = 0
for j in xrange(100): u_prev[Lx-1,j] = 0
for i in xrange(100): u_prev[i,Ly-1] = 0
#call C function from Python
cwave2.cwave_prop( u_prev , u , u_next )
#returned u (2D np.ndarray)
from tempfile import TemporaryFile
outfile = TemporaryFile()
np.save(outfile,u)
fig = plt.figure()
plt.imshow(u,cmap=plt.cm.ocean)
plt.colorbar()
plt.show()