Python array manipulation

Question

I have array [1,2,1,2,3,4,3,4,1,2]

I want to loop it x times, each times moving array every element 1 position forward:

So next loop will be:

2.[2,1,2,3,4,3,4,1,2,1]

3. [1,2,3,4,3,4,1,2,1,2]

etc....

How can I manipulate array like this way?

EDIT:

What I thought, but maybe some better tricks:

Just go over the array with while loop and create new array with for cycle.

for i in range(11)
 array[i] = array[i-1]

etc etc..its pseudo code

• Jeffrey Kevin Pry: sounds like HW? You have to try to learn! :)
• Jaanus: @JeffreyKevinPry no I just develop all stuff, look my other questions, I am just like to learn new stuff.
• Mark Reed: Your loop won't work, because by the time you try to give a[1] the value of a[0], the old value of a[0] has already been overwritten with the value of a[-1], so you wind up with the whole array being copies of the last element.
• Martijn Pieters: possible duplicate of Efficient way to shift a list in python

Answer 1

Using a List data structure isn't an efficient way to do this. A Queue would be more appropriate. In any case:

Using a Queue

As I suggested, using a Queue (collections.deque):

>>> q = collections.deque([1,2,3,4,5,6,7,8])
>>> for _ in xrange(5):
... q.rotate(-1)
... 
>>> q
deque([6, 7, 8, 1, 2, 3, 4, 5])

Keeping the List

>>> a = [1,2,3,4,5,6,7,8]
>>> for _ in xrange(5):
... a = a[1:] + a[:1]
... 
>>> a
[6, 7, 8, 1, 2, 3, 4, 5]

Alternatively (faster than the previous one):

>>> a = [1,2,3,4,5,6,7,8]
>>> for _ in xrange(5):
... a.append(a.pop(0))
... 
>>> a
[6, 7, 8, 1, 2, 3, 4, 5]

Here you can change xrange for whatever you want to iterate over.

Timeit analysis:

Pop-Append

>>> timeit.timeit('a.append(a.pop(0))', setup='a = [0,1,2,3,4,5,6,7,8,9]', number=1000000)
0.24548697471618652
>>> timeit.timeit('a.append(a.pop(0))', setup='a = [0,1,2,3,4,5,6,7,8,9]', number=100000000)
23.65538215637207

Slicing

>>> timeit.timeit('a=a[1:] + a[:1]', setup='a = [0,1,2,3,4,5,6,7,8,9]', number=1000000)
0.36037278175354004
>>> timeit.timeit('a=a[1:] + a[:1]', setup='a = [0,1,2,3,4,5,6,7,8,9]', number=100000000)
35.06173801422119

Queue

>>> timeit.timeit('q.rotate(-1)', setup='import collections; q = collections.deque([0,1,2,3,4,5,6,7,8])', number=1000000)
0.16829514503479004
>>> timeit.timeit('q.rotate(-1)', setup='import collections; q = collections.deque([0,1,2,3,4,5,6,7,8])', number=100000000)
16.012277841567993

With a little optimization, basically removing the __getattr__ call for append, pop and rotate:

Pop-Append

>>> timeit.timeit('aa(ap(0))', setup='a = [0,1,2,3,4,5,6,7,8,9]; aa=a.append; ap=a.pop', number=1000000)
0.15255093574523926
>>> timeit.timeit('aa(ap(0))', setup='a = [0,1,2,3,4,5,6,7,8,9]; aa=a.append; ap=a.pop', number=100000000)
14.50795292854309

Queue

>>> timeit.timeit('r(-1)', setup='import collections; q = collections.deque([0,1,2,3,4,5,6,7,8]); r=q.rotate', number=1000000)
0.13374090194702148
>>> timeit.timeit('r(-1)', setup='import collections; q = collections.deque([0,1,2,3,4,5,6,7,8]); r=q.rotate', number=100000000)
11.435136079788208

Answer 2

I would use a deque because it has the built in method rotate:

import collections
d = collections.deque([1,2,1,2,3,4,3,4,1,2])
for _ in xrange(number_of_shifts):
 d.rotate(-1)
 print list(d)

Answer 3

my_list = [1,2,1,2,3,4,3,4,1,2]
for i in range(10):
 my_list.insert(0,my_list.pop())
 print my_list

Answer 4

You question finds its answer in this question (which answers help in improving ...) Efficient way to shift a list in python

To make it short, make a function to shift your array:

def shift(l, n):
 return l[n:] + l[:n]

Then call this function in a loop:

myarray=[1,2,3,4,5,6,7]
for ii in range(10):
 myarray=shift(myarray, 1)
 print myarray

• Steven Rumbalski: This is also quite inefficient. Three new lists created for each shift!
• Bruno von Paris: Sure. But it answers the question (that was not about optimization). I though it would have some educational value for a beginner.