我必须对SODE系统进行一些模拟。由于我需要使用随机图，我认为使用python为图形生成相邻矩阵然后使用C进行模拟是个好主意。所以我转向cython。
我按照cython documentation的提示编写了我的代码，以尽可能提高速度。但是知道我真的不知道我的代码是否好。我也运行cython toast.pyx -a，但我不了解这些问题。
bruit或np.array在我的代码上定义double？请注意，我将比较矩阵的元素（0或1）以便进行总和。结果将是矩阵NxT，其中N是系统的维度，而它是我想用于模拟的时间。double[:]的文档？double？但我让我的代码为我说话：
from __future__ import division
import scipy.stats as stat
import numpy as np
import networkx as net
#C part
from libc.math cimport sin
from libc.math cimport sqrt
#cimport cython
cimport numpy as np
cimport cython
cdef double tau = 2*np.pi #http://tauday.com/
#graph
def graph(int N, double p):
"""
It generates an adjacency matrix for a Erdos-Renyi graph G{N,p} (by default not directed).
Note that this is an O(n^2) algorithm and it gives an array, not a (sparse) matrix.
Remark: fast_gnp_random_graph(n, p, seed=None, directed=False) is O(n+m), where m is the expected number of edges m=p*n*(n-1)/2.
Arguments:
N : number of edges
p : probability for edge creation
"""
G=net.gnp_random_graph(N, p, seed=None, directed=False)
G=net.adjacency_matrix(G, nodelist=None, weight='weight')
G=G.toarray()
return G
@cython.boundscheck(False)
@cython.wraparound(False)
#simulations
def simul(int N, double H, G, np.ndarray W, np.ndarray X, double d, double eps, double T, double dt, int kt_max):
"""
For details view the general description of the package.
Argumets:
N : population size
H : coupling strenght complete case
G : adjiacenty matrix
W : disorder
X : initial condition
d : diffusion term
eps : 0 for the reversibily case, 1 for the non-rev case
T : time of the simulation
dt : increment time steps
kt_max = (int) T/dt
"""
cdef int kt
#kt_max = T/dt to check
cdef np.ndarray xt = np.zeros([N,kt_max], dtype=np.float64)
cdef double S1 = 0.0
cdef double Stilde1 = 0.0
cdef double xtmp, xtilde, x_diff, xi
cdef np.ndarray bruit = d*sqrt(dt)*stat.norm.rvs(N)
cdef int i, j, k
for i in range(N): #setting initial conditions
xt[i][0]=X[i]
for kt in range(kt_max-1):
for i in range(N):
S1 = 0.0
Stilde1= 0.0
xi = xt[i][kt]
for j in range(N): #computation of the sum with the adjiacenty matrix
if G[i][j]==1:
x_diff = xt[j][kt] - xi
S2 = S2 + sin(x_diff)
xtilde = xi + (eps*(W[i]) + (H/N)*S1)*dt + bruit[i]
for j in range(N):
if G[i][j]==1:
x_diff = xt[j][kt] - xtilde
Stilde2 = Stilde2 + sin(x_diff)
#computation of xt[i]
xtmp = xi + (eps*(W[i]) + (H/N)*(S1+Stilde1)*0.5)*dt + bruit
xt[i][kt+1] = xtmp%tau
return xt
非常感谢！
我将变量定义的顺序np.array更改为double，将xt[i][j]更改为xt[i,j]，将矩阵更改为long long。现在代码非常快，而html文件中的黄色部分就在声明的周围。谢谢！
from __future__ import division
import scipy.stats as stat
import numpy as np
import networkx as net
#C part
from libc.math cimport sin
from libc.math cimport sqrt
#cimport cython
cimport numpy as np
cimport cython
cdef double tau = 2*np.pi #http://tauday.com/
#graph
def graph(int N, double p):
"""
It generates an adjacency matrix for a Erdos-Renyi graph G{N,p} (by default not directed).
Note that this is an O(n^2) algorithm and it gives an array, not a (sparse) matrix.
Remark: fast_gnp_random_graph(n, p, seed=None, directed=False) is O(n+m), where m is the expected number of edges m=p*n*(n-1)/2.
Arguments:
N : number of edges
p : probability for edge creation
"""
G=net.gnp_random_graph(N, p, seed=None, directed=False)
G=net.adjacency_matrix(G, nodelist=None, weight='weight')
G=G.toarray()
return G
@cython.boundscheck(False)
@cython.wraparound(False)
#simulations
def simul(int N, double H, long long [:, ::1] G, double[:] W, double[:] X, double d, double eps, double T, double dt, int kt_max):
"""
For details view the general description of the package.
Argumets:
N : population size
H : coupling strenght complete case
G : adjiacenty matrix
W : disorder
X : initial condition
d : diffusion term
eps : 0 for the reversibily case, 1 for the non-rev case
T : time of the simulation
dt : increment time steps
kt_max = (int) T/dt
"""
cdef int kt
#kt_max = T/dt to check
cdef double S1 = 0.0
cdef double Stilde1 = 0.0
cdef double xtmp, xtilde, x_diff
cdef double[:] bruit = d*sqrt(dt)*np.random.standard_normal(N)
cdef double[:, ::1] xt = np.zeros((N, kt_max), dtype=np.float64)
cdef double[:, ::1] yt = np.zeros((N, kt_max), dtype=np.float64)
cdef int i, j, k
for i in range(N): #setting initial conditions
xt[i,0]=X[i]
for kt in range(kt_max-1):
for i in range(N):
S1 = 0.0
Stilde1= 0.0
for j in range(N): #computation of the sum with the adjiacenty matrix
if G[i,j]==1:
x_diff = xt[j,kt] - xt[i,kt]
S1 = S1 + sin(x_diff)
xtilde = xt[i,kt] + (eps*(W[i]) + (H/N)*S1)*dt + bruit[i]
for j in range(N):
if G[i,j]==1:
x_diff = xt[j,kt] - xtilde
Stilde1 = Stilde1 + sin(x_diff)
#computation of xt[i]
xtmp = xt[i,kt] + (eps*(W[i]) + (H/N)*(S1+Stilde1)*0.5)*dt + bruit[i]
xt[i,kt+1] = xtmp%tau
return xt
答案 0 :(得分：2)
cython -a颜色编码cython源代码。如果单击某一行，则会显示相应的C源。根据经验，你不希望内循环中有任何黄色。
您的代码中会出现一些问题：
x[j][i]在每次调用时为x[j]创建一个临时数组，因此请改用x[j, i]。cdef ndarray x更好地提供维度和dtype（cdef ndarray[ndim=2, dtype=float]）或---最好---使用类型化的memoryview语法：cdef double[:, :] x。，而不是
cdef np.ndarray xt = np.zeros([N,kt_max], dtype=np.float64)
更好地使用
cdef double[:, ::1] xt = np.zeros((N, kt_max), dtype=np.float64)
double[:, ::1]并迭代数组，最后一个索引变化最快。 编辑：见http://cython.readthedocs.io/en/latest/src/userguide/memoryviews.html
对于类型的memoryview 语法double[:, ::1]等