蟒蛇，scipy-功課 - 如何保持粒子的位置的歷史網格

Question

我有以下問題：

創建一個程序，其中一個粒子將執行這兩個爲N = 1000步隨機遊走情況：i）在一維繫統中ii）在二維繫統中。 程序必須計算平均值（S），其中S是粒子訪問至少一次的網格位置數。您將進行10000次運行並找到10個點（每100步一次，從0到1000），其中將是10000次運行的手段。做平均值（S）相對於時間t的圖。

我這樣做代碼：

import scipy as sc 
import matplotlib.pyplot as plt 
import random 

plegma=1000 
grid=sc.ones(plegma) # grid full of available positions(ones)  

for p in range(10000): 
    #-------------------Initialize problem------------------------------------------------- 
    his_pos=[]     # list which holds the position of the particle in the grid 
    in_pos = int(sc.random.randint(0,len(grid),1))   #initial position of particle 
    means=[]             #list which holds the means 
    #-------------------------------------------------------------------------------------- 
    for i in range(0,1000,100): 
     step=2*sc.random.random_integers(0,1)-1  #the step of the particle can be -1 or 1 
     # Check position for edges and fix if required 
     # Move by step 
     in_pos += step 
     #Correct according to periodic boundaries 
     in_pos = in_pos % len(grid) 

     #Keep track of random walk 
     his_pos.append(in_pos) 
     history=sc.array(his_pos) 
     mean_his=sc.mean(history) 
     means.append(mean_his) 


plt.plot(means,'bo') 
plt.show() 

修訂--------------------------------- ----

import scipy as sc 
import matplotlib.pyplot as plt 
import random 

plegma=1000 
his_pos=[] # list which holds the number of visited cells in the grid 
means=[] #list which holds the means 

for p in range(10000): 
    #-------------------Initialize problem------------------------------------------------- 
    grid=sc.ones(plegma) # grid full of available positions(ones)  
    in_pos = int(sc.random.randint(0,len(grid),1))   #initial position of particle 
    num_cells=[]  # list which holds number of visited cells during run       
    #-------------------------------------------------------------------------------------- 
    for i in range(1000): 
     step=2*sc.random.random_integers(0,1)-1 #the step of the particle can be -1 or 1 

     # Check position for edges and fix if required 
     # Move by step 
     in_pos += step 
     #Correct according to periodic boundaries 
     in_pos = in_pos % len(grid) 

     grid[in_pos]=0 # mark particle position on grid as "visited" 

     if (i+1) % 100 == 0: 

      number=1000-sc.sum(grid) # count the number of "visited positions" in grid 
      num_cells.append(number) # append it to num_cells 

     his_pos.append(num_cells) # append num_cells to his_pos 
     history=sc.array(his_pos) 


mean_his=history.mean(1) 
means.append(mean_his) 

更新2 ----------------------------- .....

if (i+1) % 10 == 0: 

      number=1000-sc.sum(grid) # count the number of "visited positions" in grid 
      num_cells.append(number) # append it to num_cells 

    his_pos.append(num_cells) # append num_cells to his_pos 
    history=sc.array(his_pos) 
mean_his=history.mean(0) 

plt.plot(mean_his,'bo') 
plt.show() 

謝謝！

Answer 1

1）是的，10000步指的是期望的準確性 - 您必須獲得10000次隨機行走時的t = 100, 200 ... 1000的平均訪問單元數。要獲取數據，您必須爲每次隨機遊走（10000次）累計訪問的單元格數量。要做到這一點，你必須初始化你的問題（即，初始化his_pos和means）以外的p循環。在p循環的內部，你應該初始化你的隨機遊走 - 獲取你的網格，初始位置和你要寫結果的列表。所以，問題的init看起來像

import scipy as sc 
import matplotlib.pyplot as plt 

plegma=1000 
his_pos=[] # list which holds the position of the particle in the grid 
means=[] #list which holds the means 

for p in range(10000): 
    #-------Initialize problem-------- 
    in_pos = int(sc.random.randint(0,len(grid),1)) #initial position of particle 
    grid=sc.ones(plegma) # grid full of available positions(ones)  
    his_pos.append([]) 

初始化後，我們需要進行隨機行走 - i循環。目前你只做10步隨機遊走（len(range(0,1000,100)) == 10），但步行應該包含1000步。因此，i迴路應

for i in range(1000): 

期間，您必須以某種方式紀念參觀細胞的步行路程。最簡單的方法是將grid[in_pos]更改爲0.然後，每隔100步計算訪問單元的數量。實現這一目標的方式是像

 if i % 100 == 0: 
      # count the number of 0s in grid and append it to his_pos[p] 

最後，在結束的1000隨機遊動的his_pos將列出的（10000×10）名單，其中有列明智的手段應該採取。

更新：

爲了不失去整個運行的信息，我們應該追加對地方第p運行結果被存儲，與所有結果的列表清單。後者是his_pos。要做到這一點，我們可以追加空單his_pos並在p個運行與結果填充它，或之前p個運行初始化空列表後p個運行追加到his_pos。然後該算法將如下所示：

#-------Initialize problem-------- 
plegma=1000 
his_pos=[] # list which holds the number of visited cells in the grid 
means=[] #list which holds the means 

for p in range(10000): 
    #-------Initialize run-------- 
    in_pos = int(sc.random.randint(0,len(grid),1)) #initial position of particle 
    grid=sc.ones(plegma) # grid full of available positions(ones)  
    num_cells = [] # list which holds number of visited cells during run 
    for i in range(1000): 
     # make i-th step, get particle position 
     # mark particle position on grid as "visited" 
     if (i+1) % 100 == 0: # on each 100th step (steps count from 0, thus i+1) 
      # count the number of "visited positions" in grid 
      # append it to num_cells 
    # append num_cells to his_pos 
# find column-wise means for his_pos 

Answer 2

我不確定是否理解問題的要求（時間t在您的方程式中考慮，point是指什麼？），但相對於您嘗試執行的操作，我瞭解您必須在最終的10000x10 means數組中包含每個迭代產生的每個或10個平均值陣列。

每個mean_his陣列從100個步驟的1D陣列準備。

>>> his_pos = [1,2,3,4,5,6,7,8,9,10] #the ten positions 
>>> history = np.array(his_pos) 
>>> history 
array([ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]) 
>>> ar2 = np.reshape(history, (-1,2)) # group in two in order to calculate mean 
>>> ar2 
array([[ 1, 2], 
     [ 3, 4], 
     [ 5, 6], 
     [ 7, 8], 
     [ 9, 10]]) 
>>> mean_his = ar2.mean(1) 
>>> mean_his 
array([ 1.5, 3.5, 5.5, 7.5, 9.5]) 
>>> 

則追加mean_his到means 10000倍，同樣計算平均值（注意，手段有：我會用十步一陣列已經被平均（而不是1000，每100）每兩個舉例說明這種在外部循環之外進行初始化，以便在每次重複時不被初始化）。