用numpy數組填充數組？

Question

一個相當簡單的問題，但我似乎沒有找到任何解決方案：

我想在數組中使用兩個數組的差異，並將其作爲新的結果插入到一個新的數組中......我希望代碼澄清這個意圖，儘管它不工作，因爲python告訴我它不能將序列與數組計算結合起來......我該如何解決這個問題？提前致謝！

MeteorNum = 5 
#Coordinate Setup: 

XPos = np.random.randint(0, 600, (MeteorNum,1)) 
YPos = np.random.randint(0, 800, (MeteorNum,1)) 

XYPos = np.hstack((XPos,YPos)) 
XYPos = np.vstack((XYPos, XYPos[0])) 
print XYPos 

#Distances: 
r = np.zeros((MeteorNum,MeteorNum)) 

for i in range(0, MeteorNum): 
    for x in range(0, MeteorNum): 
     r[x,i] = (XYPos[x,:] - XYPos[i,:]) 
r = r*10**5 
print r 

，並在這裏的錯誤，我得到：

runfile('C:/Users/Marco DS/Dropbox/1_TUDelft/4Q/AE1205 Python/python codes  2015/bonus assigments/week_5 Gravity simu.py', wdir='C:/Users/Marco DS/Dropbox/1_TUDelft/4Q/AE1205 Python/python codes 2015/bonus assigments') 
[[249 660] 
[256 605] 
[119 423] 
[422 398] 
[480 504] 
[249 660]] 
Traceback (most recent call last): 

    File "<ipython-input-57-4cd9cc29ece0>", line 1, in <module> 
    runfile('C:/Users/Marco DS/Dropbox/1_TUDelft/4Q/AE1205 Python/python  codes 2015/bonus assigments/week_5 Gravity simu.py', wdir='C:/Users/Marco  DS/Dropbox/1_TUDelft/4Q/AE1205 Python/python codes 2015/bonus assigments') 

    File "C:\Users\Marco DS\Anaconda\lib\site-packages\spyderlib\widgets\externalshell\sitecustomize.py", line 682, in runfile 
execfile(filename, namespace) 

File "C:\Users\Marco DS\Anaconda\lib\site- packages\spyderlib\widgets\externalshell\sitecustomize.py", line 71, in execfile 
exec(compile(scripttext, filename, 'exec'), glob, loc) 

    File "C:/Users/Marco DS/Dropbox/1_TUDelft/4Q/AE1205 Python/python codes 2015/bonus assigments/week_5 Gravity simu.py", line 37, in <module> 
r[x,i] = (XYPos[x,:] - XYPos[i,:]) 

ValueError: setting an array element with a sequence. 

Answer 1

嘗試在容器陣列中添加dtype=：

MeteorNum = 5 
#Coordinate Setup: 

XPos = nprand.randint(0, 600, (MeteorNum,1)) 
YPos = nprand.randint(0, 800, (MeteorNum,1)) 

XYPos = np.hstack((XPos,YPos)) 
XYPos = np.vstack((XYPos, XYPos[0])) 
print XYPos 

#Distances: 
r = np.zeros((MeteorNum,MeteorNum),dtype=np.ndarray) 

for i in range(0, MeteorNum): 
    for x in range(0, MeteorNum): 
     r[x,i] = (XYPos[x,:] - XYPos[i,:]) 
r = r*10**5 
print r 

編輯

的D型參數秈稻獲取陣列所需的數據類型，例如， numpy.int8。默認是numpy.float64。

Answer 2

要刪除的錯誤，我想你需要一個額外的維度，大小爲2，你的陣列r在：

r = np.zeros((MeteorNum,MeteorNum,2)) 

由於XYPos[x,:] - XYPos[i,:]在for循環有(2,)

這是一個形狀返回形狀(5, 5, 2)的單個數組。