我正在模拟二维随机游走,方向为0
a=np.zeros((1000,2), dtype=np.float)
print a # Prints array with zeros as entries
# Single random walk
def randwalk(x,y): # Defines the randwalk function
theta=2*math.pi*rd.rand()
x+=math.cos(theta);
y+=math.sin(theta);
return (x,y) # Function returns new (x,y) coordinates
x, y = 0., 0. # Starting point is the origin
for i in range(1000): # Walk contains 1000 steps
x, y = randwalk(x,y)
a[i,:] = x, y # Replaces entries of a with (x,y) coordinates
# Repeating random walk 12 times
fn_base = "random_walk_%i.txt" # Saves each run to sequentially named .txt
for j in range(12):
rd.seed() # Uses different random seed for every run
x, y = 0., 0.
for i in range(1000):
x, y = randwalk(x,y)
a[i,:] = x, y
fn = fn_base % j # Allocates fn to the numbered file
np.savetxt(fn, a) # Saves run data to appropriate text file
现在,我要计算所有12个步行路程的均方位移。为此,我最初的想法是将每个文本文件中的数据导入到一个numpy数组中,例如:
infile="random_walk_0.txt"
rw0dat=np.genfromtxt(infile)
print rw0dat
然后以某种方式操纵阵列以找到均方位移。
是否有更有效的方法来找到我拥有的MSD?
最佳答案
这是计算均方差(MSD)的快速摘要。
路径似乎是由时间上等距的点组成的
为了你的兰德沃克。您可以将其放置在12步for循环中,并为每个a [i ,:]计算它
#input path =[ [x1,y1], ... ,[xn,yn] ].
def compute_MSD(path):
totalsize=len(path)
msd=[]
for i in range(totalsize-1):
j=i+1
msd.append(np.sum((path[0:-j]-path[j::])**2)/float(totalsize-j))
msd=np.array(msd)
return msd
关于python - 用Python计算2d随机游动的均方位移,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/26472653/