深入理解NumPy简明教程---数组3（组合）

student= dtype({'names':['name', 'age', 'weight'], 'formats':['S32', 'i','f']}, align = True) 

a= array([(“Zhang”, 32, 65.5), (“Wang”, 24, 55.2)], dtype =student) 

> a = arange(9).reshape(3,3) 
> a 
array([[0, 1, 2], 
   [3, 4, 5], 
   [6, 7, 8]]) 
> b = 2 * a 
> b 
array([[ 0, 2, 4], 
  [ 6, 8, 10], 
  [12, 14, 16]]) 

> hstack((a, b)) 
array([[ 0, 1, 2, 0, 2, 4], 
  [ 3, 4, 5, 6, 8, 10], 
  [ 6, 7, 8, 12, 14, 16]]) 

> concatenate((a, b), axis=1) 
array([[ 0, 1, 2, 0, 2, 4], 
  [ 3, 4, 5, 6, 8, 10], 
  [ 6, 7, 8, 12, 14, 16]]) 

> vstack((a, b)) 
array([[ 0, 1, 2], 
  [ 3, 4, 5], 
  [ 6, 7, 8], 
  [ 0, 2, 4], 
  [ 6, 8, 10], 
  [12, 14, 16]]) 

> concatenate((a, b), axis=0) 
array([[ 0, 1, 2], 
  [ 3, 4, 5], 
  [ 6, 7, 8], 
  [ 0, 2, 4], 
  [ 6, 8, 10], 
  [12, 14, 16]]) 

> dstack((a, b)) 
array([[[ 0, 0], 
  [ 1, 2], 
  [ 2, 4]], 
 
  [[ 3, 6], 
  [ 4, 8], 
  [ 5, 10]], 
 
  [[ 6, 12], 
  [ 7, 14], 
  [ 8, 16]]]) 

> one = arange(2) 
> one 
array([0, 1]) 
> two = one + 2 
> two 
array([2, 3]) 
> row_stack((one, two)) 
array([[0, 1], 
  [2, 3]]) 

> column_stack((oned, twiceoned)) 
array([[0, 2], 
  [1, 3]]) 

> a = arange(9).reshape(3,3) 
> a 
array([[0, 1, 2], 
  [3, 4, 5], 
  [6, 7, 8]]) 
> hsplit(a, 3) 
[array([[0], 
  [3], 
  [6]]), 
 array([[1], 
  [4], 
  [7]]), 
 array([[2], 
  [5], 
  [8]])] 

split(a, 3, axis=1) 

> vsplit(a, 3) 
> [array([[0, 1, 2]]), array([[3, 4, 5]]), array([[6, 7, 8]])] 

> split(a, 3, axis=0) 

> c = arange(27).reshape(3, 3, 3) 
> c 
array([[[ 0, 1, 2], 
  [ 3, 4, 5], 
  [ 6, 7, 8]], 
 
  [[ 9, 10, 11], 
  [12, 13, 14], 
  [15, 16, 17]], 
 
  [[18, 19, 20], 
  [21, 22, 23], 
  [24, 25, 26]]]) 
> dsplit(c, 3) 
[array([[[ 0], 
  [ 3], 
  [ 6]], 
 
  [[ 9], 
  [12], 
  [15]], 
 
  [[18], 
  [21], 
  [24]]]), 
 array([[[ 1], 
  [ 4], 
  [ 7]], 
 
  [[10], 
  [13], 
  [16]], 
 
  [[19], 
  [22], 
  [25]]]), 
 array([[[ 2], 
  [ 5], 
  [ 8]], 
 
  [[11], 
  [14], 
  [17]], 
 
  [[20], 
  [23], 
  [26]]])] 

> a = arange(12) 
> b = a  #不创建新对象 
> b is a   # a和b是同一个数组对象的两个名字 
True 
> b.shape = 3,4 #也改变了a的形状 
> a.shape 
(3, 4) 
 　　　Python 传递不定对象作为参考4，所以函数调用不拷贝数组。
 > def f(x): 
...  print id(x) 
... 
> id(a)  #id是一个对象的唯一标识 
148293216 
> f(a) 
148293216 

> c = a.view() 
> c is a 
False 
> c.base is a  #c是a持有数据的镜像 
True 
> c.flags.owndata 
False 
> 
> c.shape = 2,6 # a的形状没变 
> a.shape 
(3, 4) 
> c[0,4] = 1234  #a的数据改变了 
> a 
array([[ 0, 1, 2, 3], 
  [1234, 5, 6, 7], 
  [ 8, 9, 10, 11]]) 

> s = a[ : , 1:3]  # 获得每一行1，2处的元素 
> s[:] = 10   # s[:] 是s的镜像。注意区别s=10 and s[:]=10 
> a 
array([[ 0, 10, 10, 3], 
  [1234, 10, 10, 7], 
  [ 8, 10, 10, 11]]) 

 > d = a.copy()  #创建了一个含有新数据的新数组对象 
> d is a 
False 
> d.base is a  #d和a现在没有任何关系 
False 
> d[0,0] = 9999 
> a 
array([[ 0, 10, 10, 3], 
  [1234, 10, 10, 7], 
  [ 8, 10, 10, 11]])