What is the most efficient way to rotate a list in python? Right now I have something like this:
>>> def rotate(l, n):
... return l[n:] + l[:n]
...
>>> l = [1,2,3,4]
>>> rotate(l,1)
[2, 3, 4, 1]
>>> rotate(l,2)
[3, 4, 1, 2]
>>> rotate(l,0)
[1, 2, 3, 4]
>>> rotate(l,-1)
[4, 1, 2, 3]
Is there a better way?
The answer is
I was looking for in place solution to this problem. This solves the purpose in O(k).
def solution(self, list, k):
r=len(list)-1
i = 0
while i<k:
temp = list[0]
list[0:r] = list[1:r+1]
list[r] = temp
i+=1
return list
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]
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]
If efficiency is your goal, (cycles? memory?) you may be better off looking at the array module: http://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.
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)
Following function copies sent list to a templist, so that pop function does not affect the original list:
def shift(lst, n, toreverse=False):
templist = []
for i in lst: templist.append(i)
if toreverse:
for i in range(n): templist = [templist.pop()]+templist
else:
for i in range(n): templist = templist+[templist.pop(0)]
return templist
Testing:
lst = [1,2,3,4,5]
print("lst=", lst)
print("shift by 1:", shift(lst,1))
print("lst=", lst)
print("shift by 7:", shift(lst,7))
print("lst=", lst)
print("shift by 1 reverse:", shift(lst,1, True))
print("lst=", lst)
print("shift by 7 reverse:", shift(lst,7, True))
print("lst=", lst)
Output:
lst= [1, 2, 3, 4, 5]
shift by 1: [2, 3, 4, 5, 1]
lst= [1, 2, 3, 4, 5]
shift by 7: [3, 4, 5, 1, 2]
lst= [1, 2, 3, 4, 5]
shift by 1 reverse: [5, 1, 2, 3, 4]
lst= [1, 2, 3, 4, 5]
shift by 7 reverse: [4, 5, 1, 2, 3]
lst= [1, 2, 3, 4, 5]
I think you are looking for this:
a.insert(0, x)
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 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
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 theiin 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.)
I take this cost model as a reference:
http://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)
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 also got interested in this and compared some of the suggested solutions with perfplot (a small project of mine).
It turns out that
for _ in range(n):
data.append(data.pop(0))
is by far the fastest method for small shifts n.
For larger n,
data[n:] + data[:n]
isn't bad.
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 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):
for _ in range(shift):
data.append(data.pop(0))
return data
perfplot.save(
"shift100.png",
setup=lambda n: numpy.random.rand(n).tolist(),
kernels=[list_append, roll, shift_concatenate, collections_deque, pop_append],
n_range=[2 ** k for k in range(7, 20)],
xlabel="len(data)",
)
I think you've got the most efficient way
def shift(l,n):
n = n % len(l)
return l[-U:] + l[:-U]
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 http://gist.github.com/288272
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.rotatewith deque object: 0.12380790710449219 seconds (fastest)deque.rotatewith type conversion: 6.853878974914551 secondsnp.rollwith nparray: 6.0491721630096436 secondsnp.rollwith 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)
Simplest way I can think of:
a.append(a.pop(0))
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])
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'm "old school" I define efficiency in lowest latency, processor time and memory usage, our nemesis are the bloated libraries. So there is exactly one right way:
def rotatel(nums):
back = nums.pop(0)
nums.append(back)
return nums
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]
I have similar thing. For example, to shift by two...
def Shift(*args):
return args[len(args)-2:]+args[:len(args)-2]
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']
Another alternative:
def move(arr, n):
return [arr[(idx-n) % len(arr)] for idx,_ in enumerate(arr)]
Jon Bentley in Programming Pearls (Column 2) describes an elegant and efficient algorithm for rotating an n-element vector x left by i positions:
Let's view the problem as transforming the array
abinto the arrayba, but let's also assume that we have a function that reverses the elements in a specified portion of the array. Starting withab, we reverseato getarb, reversebto getarbr, and then reverse the whole thing to get(arbr)r, which is exactlyba. This results in the following code for rotation:reverse(0, i-1) reverse(i, n-1) reverse(0, n-1)
This can be translated to Python as follows:
def rotate(x, i):
i %= len(x)
x[:i] = reversed(x[:i])
x[i:] = reversed(x[i:])
x[:] = reversed(x)
return x
Demo:
>>> def rotate(x, i):
... i %= len(x)
... x[:i] = reversed(x[:i])
... x[i:] = reversed(x[i:])
... x[:] = reversed(x)
... return x
...
>>> rotate(list('abcdefgh'), 1)
['b', 'c', 'd', 'e', 'f', 'g', 'h', 'a']
>>> rotate(list('abcdefgh'), 3)
['d', 'e', 'f', 'g', 'h', 'a', 'b', 'c']
>>> rotate(list('abcdefgh'), 8)
['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h']
>>> rotate(list('abcdefgh'), 9)
['b', 'c', 'd', 'e', 'f', 'g', 'h', 'a']
Source: Stackoverflow.com