numpy - 如何反转 numpy.roll？

Question

我有一个重复调用 roll 的 for 循环，我想反转创建数组的顺序。

我想我忽略了一些琐碎的方法来做到这一点，但到目前为止我只找到了 10000 3 5 种不这样做的方法。

In [1]: from numpy import roll  
In [2]: c = range(5)

## The code I want to invert
In [3]: for i in range(len(c)):
   ...:         c = roll(c, 1)
   ...:         print c
[4 0 1 2 3]
[3 4 0 1 2]
[2 3 4 0 1]
[1 2 3 4 0]
[0 1 2 3 4]    

## The result I want
[0 1 2 3 4]
[1 2 3 4 0]
[2 3 4 0 1]
[3 4 0 1 2]    
[4 0 1 2 3]    


## What I've tried:
In [4]: for i in range(len(c)):
   ...:         c = roll(c, -1)
   ...:         print c
[1 2 3 4 0]
[2 3 4 0 1] # <- false
[3 4 0 1 2]
[4 0 1 2 3]
[0 1 2 3 4]    
In [5]: for i in reversed(range(len(c))):
   ...:         c = roll(c, -i)
   ...:         print c
[4 0 1 2 3] # <- false
[2 3 4 0 1]
[4 0 1 2 3]
[0 1 2 3 4]
[0 1 2 3 4]    

In [6]: for i in reversed(range(len(c))):
        c = roll(c, i)
        print c
   ...:     
[1 2 3 4 0]
[3 4 0 1 2] # <- false
[1 2 3 4 0]
[0 1 2 3 4]
[0 1 2 3 4]

In [7]: for i in range(len(c)):
   ...:    c = roll(c, i)
   ...:    print c
   ...:     
[0 1 2 3 4]
[4 0 1 2 3] # <- false
[2 3 4 0 1]
[4 0 1 2 3]
[0 1 2 3 4]

In [8]: for i in range(len(c)):
   ...:         c = roll(c, -i)
   ...:         print c
   ...:     
[0 1 2 3 4]
[1 2 3 4 0]
[3 4 0 1 2] # <- false
[1 2 3 4 0]
[0 1 2 3 4]

Answer 1

怎么样

for i in range(len(c)):
    print c
    c = roll(c, len(c) - 1)

[0 1 2 3 4]
[1 2 3 4 0]
[2 3 4 0 1]
[3 4 0 1 2]
[4 0 1 2 3]

一直滚动所有内容(除了一个)并在第一卷之前打印(所以第一行的 c 为 range(5))。

或者甚至是你的第一个解决方案，如果你先打印 c

for i in range(len(c)):
    print c
    c = roll(c, -1)

[0 1 2 3 4]
[1 2 3 4 0]
[2 3 4 0 1]
[3 4 0 1 2]
[4 0 1 2 3]