用numpy中的數組填充數組？

Question

一個相當簡單的問題，但是我似乎沒有找到任何解決方案：

我正在嘗試使用數組中兩個數組的微分，並將其作為新結果插入到新數組中...我希望代碼能夠澄清這種意圖，即使它不起作用，因為python告訴我它不能結合序列和數組計算...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

這是我得到的錯誤：

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

嘗試在容器數組中添加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

dtype參數指示數組所需的數據類型，例如numpy.int8。 默認值為numpy.float64。

Answer 2

為了消除錯誤，我認為您需要在數組r增加一個大小為2的維：

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

由於XYPos[x,:] - XYPos[i,:] for循環中的XYPos[x,:] - XYPos[i,:]的形狀為(2,)

這將返回一個形狀為(5, 5, 2) 5，5，2）的數組。