CODEWITHSUNDEEP
pythonalgorithm
Paul kingser
Paul kingser

Reputation: 61

Various list concatenation method and their performance

I was working on an algorithm and in that, we are trying to write every line in the code such that it adds up a good performance to the final code.

In one situation we have to add lists (more than two specifically). I know some of the ways to join more than two lists also I have looked upon StackOverflow but none of the answers are giving account on the performance of the method.

Can anyone show, what are the ways we can join more than two lists and their respective performance?

Edit : The size of the list is varying from 2 to 13 (to be specific). Edit Duplicate : I have been specifically asking for the ways we can add and their respected questions and in duplicate question its limited to only 4 methods

Upvotes: 5

Views: 351

Answers (3)

0xPrateek
0xPrateek

Reputation: 1188

There are multiples ways using which you can join more than two list.

Assuming that we have three list,

a = ['1']
b = ['2']
c = ['3']

Then, for joining two or more lists in python,

1) You can simply concatenate them,

 output = a + b + c

2) You can do it using list comprehension as well,

res_list = [y for x in [a,b,c] for y in x]

3) You can do it using extend() as well,

a.extend(b)
a.extend(c)
print(a)

4) You can do it by using * operator as well,

res = [*a,*b,*c]

For calculating performance, I have used timeit module present in python.

The performance of the following methods are;

4th method < 1st method < 3rd method < 2nd [method on the basis of time]

That means If you are going to use " * operator " for concatenation of more than two lists then you will get the best performance.

Hope you got what you were looking for.

Edit:: Image showing performance of all the methods (Calculated using timeit)

enter image description here

Upvotes: 5

Alain T.
Alain T.

Reputation: 42129

If you are going to concatenate a variable number of lists together, your input is going to be a list of lists (or some equivalent collection). The performance tests need to take this into account because you are not going to be able to do things like list1+list2+list3.

Here are some test results (1000 repetitions):

option1 += loop          0.00097 [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 0, 1, 2, 3, 4]
option2 itertools.chain  0.00138 [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 0, 1, 2, 3, 4]
option3 functools.reduce 0.00174 [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 0, 1, 2, 3, 4]
option4 comprehension    0.00188 [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 0, 1, 2, 3, 4]
option5 extend loop      0.00127 [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 0, 1, 2, 3, 4]
option6 deque            0.00180 [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 0, 1, 2, 3, 4]

This would indicate that a += loop through the list of lists is the fastest approach

And the source to produce them:

allLists = [ list(range(10)) for _ in range(5) ]

def option1():
    result = allLists[0].copy()
    for lst in allLists[1:]:
        result += lst
    return result

from itertools import chain
def option2(): return list(chain(*allLists))

from functools import reduce
def option3():
    return list(reduce(lambda a,b:a+b,allLists))

def option4(): return [ e for l in allLists for e in l ]

def option5():
    result = allLists[0].copy()
    for lst in allLists[1:]:
        result.extend(lst)
    return result

from collections import deque
def option6():
    result = deque()
    for lst in allLists:
        result.extend(lst)
    return list(result)


from timeit import timeit
count = 1000

t = timeit(lambda:option1(), number = count)
print(f"option1 += loop          {t:.5f}",option1()[:15])

t = timeit(lambda:option2(), number = count)
print(f"option2 itertools.chain  {t:.5f}",option2()[:15])

t = timeit(lambda:option3(), number = count)
print(f"option3 functools.reduce {t:.5f}",option3()[:15])

t = timeit(lambda:option4(), number = count)
print(f"option4 comprehension    {t:.5f}",option4()[:15])

t = timeit(lambda:option5(), number = count)
print(f"option5 extend loop      {t:.5f}",option5()[:15])

t = timeit(lambda:option6(), number = count)
print(f"option6 deque            {t:.5f}",option6()[:15])

Upvotes: 2

Andrej Kesely
Andrej Kesely

Reputation: 195613

I did some simple measurements, here are my results:

import timeit

from itertools import chain

a = [*range(1, 10)]
b = [*range(1, 10)]
c = [*range(1, 10)]

tests = ("""output = list(chain(a, b, c))""",
"""output = a + b + c""",
"""output = [*chain(a, b, c)]""",
"""output = a.copy();output.extend(b);output.extend(c);""",
"""output = [*a, *b, *c]""",
"""output = a.copy();output+=b;output+=c;""",
"""output = a.copy();output+=[*b, *c]""",
"""output = a.copy();output += b + c""")

results = sorted((timeit.timeit(stmt=test, number=1, globals=globals()), test) for test in tests)

for i, (t, stmt) in enumerate(results, 1):
    print(f'{i}.\t{t}\t{stmt}')

Prints on my machine (AMD 2400G, Python 3.6.7):

1.  6.010000106471125e-07   output = [*a, *b, *c]
2.  7.109999842214165e-07   output = a.copy();output += b + c
3.  7.720000212430023e-07   output = a.copy();output+=b;output+=c;
4.  7.820001428626711e-07   output = a + b + c
5.  1.0520000159885967e-06  output = a.copy();output+=[*b, *c]
6.  1.4030001693754457e-06  output = a.copy();output.extend(b);output.extend(c);
7.  1.4820000160398195e-06  output = [*chain(a, b, c)]
8.  2.525000127207022e-06   output = list(chain(a, b, c))

Upvotes: 3

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