numpy深入理解(1) indexing: view or copy
View ? Copy?
对View这个概念感到困惑吗，不妨搞搞清楚~
先看numpy 官方定义的view
An array that does not own its data, but refers to another array’s data instead. For example, we may create a view that only shows every second element of another array
官方定义就是清晰！
所以说，View也是一个array，但是这个array是建立在别人的数据上的，并没有独立的memory存放data。（通过共享数据，可以在一些场合节省无意义的memory copy开销。）请看下面例子，y就是建立在x的data上的一个view.
>>> x = np.arange(5)
>>> x
array([0, 1, 2, 3, 4])
>>> y = x[::2]
>>> y
array([0, 2, 4])
>>> x[0] = 3 # changing x changes y as well, since y is a view on x
>>> y
array([3, 2, 4])显然，上述代码中x和y共享了一段data。但是，我们能不能把所有的array[index]都看做是array的view呢？请看下面代码
>>> x = np.arange(5)
>>> y_copy = x[[0, 2, 4]] # index is a list, not a tuple!
>>> y_copy
array([0, 2, 4])
>>> x[0] = 3
>>> x
array([3, 1, 2, 3, 4])
>>> y_copy
array([0, 2, 4]) # x[0] changed, y[0] not affected!是不是有点儿奇怪呢，看来还需要透彻理解一下numpy的indexing功能~
Indexing
ndarray_object [index] ，如果作为“右值”，会触发
ndarray_object.__getitem__(index)
这个函数的逻辑根据index的类型不同，会有很大的区别：
- simple indexing
- single element indexing (view)
- slice indexing (view)
- Advanced indexing
- integer array indexing (copy)
- bool array indexing (copy)
ndarray_object [index] 的返回值，从array data memory layout的角度，有view和copy的区别。
让我们看几个实例，结合问题加深理解。
假定已经有
import numpy as npQ1 生成矩阵Mij，矩阵长相如下
[0, 1, 0, 1, 0]
[1, 0, 1, 0, 1]
[0, 1, 0, 1, 0]
[1, 0, 1, 0, 1]
[0, 1, 0, 1, 0]
下面这种方法（并非最佳），先生成5*5的zero矩阵，然后在合适的位置上填1。用index slice的方法，可以容易的找出填1的位置。
>> matrix = np.zeros((5, 5)，dtype=int)
array([[0, 0, 0, 0, 0],
[0, 0, 0, 0, 0],
[0, 0, 0, 0, 0],
[0, 0, 0, 0, 0],
[0, 0, 0, 0, 0]])
>> matrix[0::2, 1::2] = 1
>> matrix[1::2, 0::2] = 1
>> matrix
array([[0, 1, 0, 1, 0],
[1, 0, 1, 0, 1],
[0, 1, 0, 1, 0],
[1, 0, 1, 0, 1],
[0, 1, 0, 1, 0]])Q2 用一行定义矩阵的“侧转置”，“侧转置”定义如下，输入时
[0, 1, 2]
[3, 4, 5]
[6, 7, 8]
“侧转置”后，得到
[8, 7, 6]
[5, 4, 3]
[2, 1, 0]
(array indexing; hint: slicing using -1)
Q3 Let x be the following 2-D array with shape (3, 3)
[0, 1, 2]
[3, 4, 5]
[6, 7, 8]
what is x[0, 1]?
what is x[(0, 1)]?
what is x[[0, 1]]?
what is x[[0, 1], :]?
(array indexing: single element indexing; index arrays)
Q4 Let x be the following 2-D array with shape (3, 3)
[0, 1, 2]
[3, 4, 5]
[6, 7, 8]
# what is x after executing the following code
y = x[[2]] # Not x[(2)]!
y[:] = 0# what is x after executing the following code
y = x[2] # same as x[(2)]
y[:] = 0(array indexing VIEW & COPY)
Q5 Let x be the following 2-D array with shape (3, 3)
[0, 1, 2]
[3, 4, 5]
[6, 7, 8]
# define "indexing by slice" and "indexing by array"
idx_slice = slice(0, 3, 2)
idx_array = [0, 2]# what is x after executing the following code
x[idx_slice, :][:, idx_array] = 0# what is x after executing the following code
x[idx_array, :][:, idx_slice] = 0Q5 这一题是难度最大的，涉及到对getitem(), setitem(), numpy index by slice & by array的深入理解。这里有篇文章ViewsVsCopies，讲的很好。
Reference Solution
# Solution of Q1 By GongPing @2016-10-26
# more elegent solution
matrix = np.fromfunc(lambda i,j: (i+j)%2, (5, 5), dtype=int)
# yet another solution
matrix = np.arange(25).reshape(5, 5) % 2参考文档：numpy.fromfunction
# Solution of Q2 By GongPing @2016-10-26
side_transpose = lambda a: a[::-1, ::-1]Q3, Q4, Q5的结果大家可以很容易的试出来，建议动手try一下。
如果要透彻理解的话，建议通读一下官方quick start教程