How to move array elements in python

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]

codeforester

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answered Jan 27, 2010 at 20:46

4

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.]

tobias_k

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answered Dec 5, 2011 at 20:20

JamgoldJamgold

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7

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]]

answered Oct 13, 2012 at 10:57

RichardRichard

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3

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]

answered Jan 27, 2010 at 21:48

jcdyerjcdyer

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4

Simplest way I can think of:

a.append[a.pop[0]]

runDOSrun

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answered Jul 8, 2014 at 12:15

ThijsThijs

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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]

answered Jul 4, 2017 at 9:12

PurrellPurrell

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4

I also got interested in this and compared some of the suggested solutions with perfplot [a small project of mine].

It turns out that Kelly Bundy's suggestion

tmp = data[shift:]
tmp += data[:shift]

performs very well for all shifts.

Essentially, perfplot performs the shift for increasing large arrays and measures the time. Here are the results:

shift = 1:

shift = 100:

Code to reproduce the plot:

import numpy
import perfplot
import collections


shift = 100


def list_append[data]:
    return data[shift:] + data[:shift]


def list_append2[data]:
    tmp = data[shift:]
    tmp += data[:shift]
    return tmp


def shift_concatenate[data]:
    return numpy.concatenate[[data[shift:], data[:shift]]]


def roll[data]:
    return numpy.roll[data, -shift]


def collections_deque[data]:
    items = collections.deque[data]
    items.rotate[-shift]
    return items


def pop_append[data]:
    data = data.copy[]
    for _ in range[shift]:
        data.append[data.pop[0]]
    return data


b = perfplot.bench[
    setup=lambda n: numpy.random.rand[n].tolist[],
    kernels=[
        list_append,
        list_append2,
        roll,
        shift_concatenate,
        collections_deque,
        pop_append,
    ],
    n_range=[2 ** k for k in range[7, 20]],
    xlabel="len[data]",
]
b.show[]
b.save["shift100.png"]

answered Jul 20, 2018 at 15:04

Nico SchlömerNico Schlömer

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6

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]

answered Oct 29, 2012 at 15:34

Phil HPhil H

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This also depends on if you want to shift the list in place [mutating it], or if you want the function to return a new list. Because, according to my tests, something like this is at least twenty times faster than your implementation that adds two lists:

def shiftInPlace[l, n]:
    n = n % len[l]
    head = l[:n]
    l[:n] = []
    l.extend[head]
    return l

In fact, even adding a l = l[:] to the top of that to operate on a copy of the list passed in is still twice as fast.

Various implementations with some timing at //gist.github.com/288272

answered Jan 27, 2010 at 23:10

keturnketurn

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3

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]

dabuno

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answered Feb 25, 2012 at 21:49

BittercoderBittercoder

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Possibly a ringbuffer is more suitable. It is not a list, although it is likely that it can behave enough like a list for your purposes.

The problem is that the efficiency of a shift on a list is O[n], which becomes significant for large enough lists.

Shifting in a ringbuffer is simply updating the head location which is O[1]

answered Jan 27, 2010 at 21:59

John La RooyJohn La Rooy

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If efficiency is your goal, [cycles? memory?] you may be better off looking at the array module: //docs.python.org/library/array.html

Arrays do not have the overhead of lists.

As far as pure lists go though, what you have is about as good as you can hope to do.

answered Jan 27, 2010 at 20:47

recursiverecursive

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0

I think you are looking for this:

a.insert[0, x]

answered Mar 25, 2014 at 9:46

1

Another alternative:

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

answered Apr 9, 2016 at 17:05

damiodamio

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def solution[A, K]:
    if len[A] == 0:
        return A

    K = K % len[A]

    return A[-K:] + A[:-K]

# use case
A = [1, 2, 3, 4, 5, 6]
K = 3
print[solution[A, K]]

For example, given

A = [3, 8, 9, 7, 6]
K = 3

the function should return [9, 7, 6, 3, 8]. Three rotations were made:

[3, 8, 9, 7, 6] -> [6, 3, 8, 9, 7]
[6, 3, 8, 9, 7] -> [7, 6, 3, 8, 9]
[7, 6, 3, 8, 9] -> [9, 7, 6, 3, 8]

For another example, given

A = [0, 0, 0]
K = 1

the function should return [0, 0, 0]

Given

A = [1, 2, 3, 4]
K = 4

the function should return [1, 2, 3, 4]

RobC

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answered Nov 15, 2019 at 8:41

I take this cost model as a reference:

//scripts.mit.edu/~6.006/fall07/wiki/index.php?title=Python_Cost_Model

Your method of slicing the list and concatenating two sub-lists are linear-time operations. I would suggest using pop, which is a constant-time operation, e.g.:

def shift[list, n]:
    for i in range[n]
        temp = list.pop[]
        list.insert[0, temp]

shanethehat

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answered Feb 21, 2012 at 22:32

herrfzherrfz

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2

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, '