Python:Numpy切片深度分解 [英] Python: Numpy slicing indepth explnation

问题描述

我在conway的生活克隆游戏中找到了以下代码. 我不明白以下代码是如何运行的.有人可以对代码的执行方式进行深入的解释吗?

I found the following code in a conway's game of life clone. I don't understand how exactly the following code is run. Can someone give an in depth explanation in how the Code is executed?

def iterate(Z):
    # find number of neighbors that each square has
    N = np.zeros(Z.shape)
    N[1:, 1:] += Z[:-1, :-1]
    N[1:, :-1] += Z[:-1, 1:]
    N[:-1, 1:] += Z[1:, :-1]
    N[:-1, :-1] += Z[1:, 1:]
    N[:-1, :] += Z[1:, :]
    N[1:, :] += Z[:-1, :]
    N[:, :-1] += Z[:, 1:]
    N[:, 1:] += Z[:, :-1]
    # a live cell is killed if it has fewer than 2 or more than 3 neighbours.
    part1 = ((Z == 1) & (N < 4) & (N > 1))
    # a new cell forms if a square has exactly three members
    part2 = ((Z == 0) & (N == 3))
    return (part1 | part2).astype(int)

推荐答案

在所有N+=行之后添加print(N)行，并尝试各种Z数组.

Add a print(N) line after all the N+= lines, and try various Z arrays.

例如

定义一个小的z，中间用1 s的块表示:

Define a small z with a block of 1s in the middle:

In [29]: z = np.zeros((10,10),int)
In [31]: z[4:6,4:6]=1
In [34]: z[4:8,5]=1

In [35]: z
Out[35]: 
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, 0, 0, 0, 0, 0],
       [0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
       [0, 0, 0, 0, 1, 1, 0, 0, 0, 0],
       [0, 0, 0, 0, 1, 1, 0, 0, 0, 0],
       [0, 0, 0, 0, 0, 1, 0, 0, 0, 0],
       [0, 0, 0, 0, 0, 1, 0, 0, 0, 0],
       [0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
       [0, 0, 0, 0, 0, 0, 0, 0, 0, 0]])

将其传递给函数:

In [36]: iterate(z)
[[ 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.  0.  0.  0.  0.  0.]
 [ 0.  0.  0.  1.  2.  2.  1.  0.  0.  0.]
 [ 0.  0.  0.  2.  3.  3.  2.  0.  0.  0.]
 [ 0.  0.  0.  2.  4.  4.  3.  0.  0.  0.]
 [ 0.  0.  0.  1.  4.  3.  3.  0.  0.  0.]
 [ 0.  0.  0.  0.  2.  1.  2.  0.  0.  0.]
 [ 0.  0.  0.  0.  1.  1.  1.  0.  0.  0.]
 [ 0.  0.  0.  0.  0.  0.  0.  0.  0.  0.]]

N计算出的邻居数为1.自己检查一下计数.

N has counted the number of neighbors that are 1. Check the counts yourself.

Out[36]: 
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, 0, 0, 0, 0, 0],
       [0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
       [0, 0, 0, 0, 1, 1, 0, 0, 0, 0],
       [0, 0, 0, 0, 0, 0, 1, 0, 0, 0],
       [0, 0, 0, 0, 0, 1, 1, 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, 0, 0, 0, 0, 0, 0, 0, 0]])

尝试各种图案，重复进行迭代，然后观察图案变化.有些人死了，有些人以命令的方式移动，有些眨眼"等.

Try various patterns, repeat the iterate and watch the pattern change. Some die away, some move in an order fashion, some 'blink', etc.

在这样的行中:

N[1:, 1:] += Z[:-1, :-1]

RHS是左上部分(此处为9x9)； LHS是右下角，同样是9x9.有8个N+=表达式，计算8个邻居(在3x3块中减去中心).使用此偏移量切片，它可以立即对Z中的所有点进行计数.

the RHS is an upper left portion (here 9x9); LHS is a bottom right, again 9x9. There are 8 N+= expressions, calculating the 8 neighbors (in a 3x3 block, minus the center). With this offset slicing it can do the count for all points in Z at once.

一开始，1行数组可能更容易可视化

For a start, a 1 row array might be easier to visualize

In [47]: z = np.zeros((1,10),int)
In [49]: z[0,4:7]=1
In [50]: z
Out[50]: array([[0, 0, 0, 0, 1, 1, 1, 0, 0, 0]])
In [51]: iterate(z)
[[ 0.  0.  0.  1.  1.  2.  1.  1.  0.  0.]]
Out[51]: array([[0, 0, 0, 0, 0, 1, 0, 0, 0, 0]])

我认为，如果z的所有边值均为0，则效果最好.

I think this works best if all the edge values of z are 0.

此数组创建glider，该动画在 https://en上进行动画处理.wikipedia.org/wiki/Glider_(Conway％27s_Life)

This array creates the glider that is animated on https://en.wikipedia.org/wiki/Glider_(Conway%27s_Life)

In [64]: z = np.zeros((10,10),int)
In [65]: z[1,2]=1;z[2,3]=1;z[3,1:4]=1

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