rotate - python shift()




Efficient way to shift a list in python (16)

A collections.deque is optimized for pulling and pushing on both ends. They even have a dedicated rotate() method.

from collections import deque
items = deque([1, 2])
items.append(3)        # deque == [1, 2, 3]
items.rotate(1)        # The deque is now: [3, 1, 2]
items.rotate(-1)       # Returns deque to original state: [1, 2, 3]
item = items.popleft() # deque == [2, 3]

What is the most efficient way to shift a list in python? Right now I have something like this:

>>> def shift(l, n):
...     return l[n:] + l[:n]
... 
>>> l = [1,2,3,4]
>>> shift(l,1)
[2, 3, 4, 1]
>>> shift(l,2)
[3, 4, 1, 2]
>>> shift(l,0)
[1, 2, 3, 4]
>>> shift(l,-1)
[4, 1, 2, 3]

Is there a better way?


Another alternative:

def move(arr, n):
    return [arr[(idx-n) % len(arr)] for idx,_ in enumerate(arr)]

For a list X = ['a', 'b', 'c', 'd', 'e', 'f'] and a desired shift value of shift less than list length, we can define the function list_shift() as below

def list_shift(my_list, shift):
    assert shift < len(my_list)
    return my_list[shift:] + my_list[:shift]

Examples,

list_shift(X,1) returns ['b', 'c', 'd', 'e', 'f', 'a'] list_shift(X,3) returns ['d', 'e', 'f', 'a', 'b', 'c']


For an immutable implementation, you could use something like this:

def shift(seq, n):
    shifted_seq = []
    for i in range(len(seq)):
        shifted_seq.append(seq[(i-n) % len(seq)])
    return shifted_seq

print shift([1, 2, 3, 4], 1)

I don't know if this is 'efficient', but it also works:

x = [1,2,3,4]
x.insert(0,x.pop())

EDIT: Hello again, I just found a big problem with this solution! Consider the following code:

class MyClass():
    def __init__(self):
        self.classlist = []

    def shift_classlist(self): # right-shift-operation
        self.classlist.insert(0, self.classlist.pop())

if __name__ == '__main__':
    otherlist = [1,2,3]
    x = MyClass()

    # this is where kind of a magic link is created...
    x.classlist = otherlist

    for ii in xrange(2): # just to do it 2 times
        print '\n\n\nbefore shift:'
        print '     x.classlist =', x.classlist
        print '     otherlist =', otherlist
        x.shift_classlist() 
        print 'after shift:'
        print '     x.classlist =', x.classlist
        print '     otherlist =', otherlist, '<-- SHOULD NOT HAVE BIN CHANGED!'

The shift_classlist() method executes the same code as my x.insert(0,x.pop())-solution, otherlist is a list indipendent from the class. After passing the content of otherlist to the MyClass.classlist list, calling the shift_classlist() also changes the otherlist list:

CONSOLE OUTPUT:

before shift:
     x.classlist = [1, 2, 3]
     otherlist = [1, 2, 3]
after shift:
     x.classlist = [3, 1, 2]
     otherlist = [3, 1, 2] <-- SHOULD NOT HAVE BIN CHANGED!



before shift:
     x.classlist = [3, 1, 2]
     otherlist = [3, 1, 2]
after shift:
     x.classlist = [2, 3, 1]
     otherlist = [2, 3, 1] <-- SHOULD NOT HAVE BIN CHANGED!

I use Python 2.7. I don't know if thats a bug, but I think it's more likely that I missunderstood something here.

Does anyone of you know why this happens?


I have similar thing. For example, to shift by two...

def Shift(*args):
    return args[len(args)-2:]+args[:len(args)-2]

I think you are looking for this:

a.insert(0, x)

I think you've got the most efficient way

def shift(l,n):
    n = n % len(l)  
    return l[-U:] + l[:-U]

If you just want to iterate over these sets of elements rather than construct a separate data structure, consider using iterators to construct a generator expression:

def shift(l,n):
    return itertools.islice(itertools.cycle(l),n,n+len(l))

>>> list(shift([1,2,3],1))
[2, 3, 1]

It depends on what you want to have happen when you do this:

>>> shift([1,2,3], 14)

You might want to change your:

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

to:

def shift(seq, n):
    n = n % len(seq)
    return seq[n:] + seq[:n]

Just some notes on timing:

If you're starting with a list, l.append(l.pop(0)) is the fastest method you can use. This can be shown with time complexity alone:

  • deque.rotate is O(k) (k=number of elements)
  • list to deque conversion is O(n)
  • list.append and list.pop are both O(1)

So if you are starting with deque objects, you can deque.rotate() at the cost of O(k). But, if the starting point is a list, the time complexity of using deque.rotate() is O(n). l.append(l.pop(0) is faster at O(1).

Just for the sake of illustration, here are some sample timings on 1M iterations:

Methods which require type conversion:

  • deque.rotate with deque object: 0.12380790710449219 seconds (fastest)
  • deque.rotate with type conversion: 6.853878974914551 seconds
  • np.roll with nparray: 6.0491721630096436 seconds
  • np.roll with type conversion: 27.558452129364014 seconds

List methods mentioned here:

  • l.append(l.pop(0)): 0.32483696937561035 seconds (fastest)
  • "shiftInPlace": 4.819645881652832 seconds
  • ...

Timing code used is below.


collections.deque

Showing that creating deques from lists is O(n):

from collections import deque
import big_o

def create_deque_from_list(l):
     return deque(l)

best, others = big_o.big_o(create_deque_from_list, lambda n: big_o.datagen.integers(n, -100, 100))
print best

# --> Linear: time = -2.6E-05 + 1.8E-08*n

If you need to create deque objects:

1M iterations @ 6.853878974914551 seconds

setup_deque_rotate_with_create_deque = """
from collections import deque
import random
l = [random.random() for i in range(1000)]
"""

test_deque_rotate_with_create_deque = """
dl = deque(l)
dl.rotate(-1)
"""
timeit.timeit(test_deque_rotate_with_create_deque, setup_deque_rotate_with_create_deque)

If you already have deque objects:

1M iterations @ 0.12380790710449219 seconds

setup_deque_rotate_alone = """
from collections import deque
import random
l = [random.random() for i in range(1000)]
dl = deque(l)
"""

test_deque_rotate_alone= """
dl.rotate(-1)
"""
timeit.timeit(test_deque_rotate_alone, setup_deque_rotate_alone)

np.roll

If you need to create nparrays

1M iterations @ 27.558452129364014 seconds

setup_np_roll_with_create_npa = """
import numpy as np
import random
l = [random.random() for i in range(1000)]
"""

test_np_roll_with_create_npa = """
np.roll(l,-1) # implicit conversion of l to np.nparray
"""

If you already have nparrays:

1M iterations @ 6.0491721630096436 seconds

setup_np_roll_alone = """
import numpy as np
import random
l = [random.random() for i in range(1000)]
npa = np.array(l)
"""

test_roll_alone = """
np.roll(npa,-1)
"""
timeit.timeit(test_roll_alone, setup_np_roll_alone)

"Shift in place"

Requires no type conversion

1M iterations @ 4.819645881652832 seconds

setup_shift_in_place="""
import random
l = [random.random() for i in range(1000)]
def shiftInPlace(l, n):
    n = n % len(l)
    head = l[:n]
    l[:n] = []
    l.extend(head)
    return l
"""

test_shift_in_place="""
shiftInPlace(l,-1)
"""

timeit.timeit(test_shift_in_place, setup_shift_in_place)

l.append(l.pop(0))

Requires no type conversion

1M iterations @ 0.32483696937561035

setup_append_pop="""
import random
l = [random.random() for i in range(1000)]
"""

test_append_pop="""
l.append(l.pop(0))
"""
timeit.timeit(test_append_pop, setup_append_pop)

Numpy can do this using the roll command:

>>> import numpy
>>> a=numpy.arange(1,10) #Generate some data
>>> numpy.roll(a,1)
array([9, 1, 2, 3, 4, 5, 6, 7, 8])
>>> numpy.roll(a,-1)
array([2, 3, 4, 5, 6, 7, 8, 9, 1])
>>> numpy.roll(a,5)
array([5, 6, 7, 8, 9, 1, 2, 3, 4])
>>> numpy.roll(a,9)
array([1, 2, 3, 4, 5, 6, 7, 8, 9])

Simplest way I can think of:

a.append(a.pop(0))

The following method is O(n) in place with constant auxiliary memory:

def rotate(arr, shift):
  pivot = shift % len(arr)
  dst = 0
  src = pivot
  while (dst != src):
    arr[dst], arr[src] = arr[src], arr[dst]
    dst += 1
    src += 1
    if src == len(arr):
      src = pivot
    elif dst == pivot:
      pivot = src

Note that in python, this approach is horribly inefficient compared to others as it can't take advantage of native implementations of any of the pieces.


What about just using pop(0)?

list.pop([i])

Remove the item at the given position in the list, and return it. If no index is specified, a.pop() removes and returns the last item in the list. (The square brackets around the i in the method signature denote that the parameter is optional, not that you should type square brackets at that position. You will see this notation frequently in the Python Library Reference.)


What is the use case? Often, we don't actually need a fully shifted array --we just need to access a handful of elements in the shifted array.

Getting Python slices is runtime O(k) where k is the slice, so a sliced rotation is runtime N. The deque rotation command is also O(k). Can we do better?

Consider an array that is extremely large (let's say, so large it would be computationally slow to slice it). An alternative solution would be to leave the original array alone and simply calculate the index of the item that would have existed in our desired index after a shift of some kind.

Accessing a shifted element thus becomes O(1).

def get_shifted_element(original_list, shift_to_left, index_in_shifted):
    # back calculate the original index by reversing the left shift
    idx_original = (index_in_shifted + shift_to_left) % len(original_list)
    return original_list[idx_original]

my_list = [1, 2, 3, 4, 5]

print get_shifted_element(my_list, 1, 2) ----> outputs 4

print get_shifted_element(my_list, -2, 3) -----> outputs 2 




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