python list concatenation efficiency

Question

What is the most efficient way to concatenate two lists list_a and list_b when:

• list_b items have to be placed before

Answer 1

itertools.chain just makes a generator, so if you can get away with using a generator instead of a list, it's constant time to generate but you pay the cost when you access each element. Otherwise list_a[0:0] = list_b is about 6 times faster than list_a = list_b + list_a

I think that list_a = list_b + list_a is the most readable choice and it's already pretty fast.

The two methods that you mentioned that use append() in a for loop are unusably slow so I didn't bother including them.

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Ran with Python 3.7.5 [Clang 11.0.0 (clang-1100.0.33.8)] on darwin on a 1.6 GHz Dual-Core Intel Core i5 with 16 GB of 2133 MHz LPDDR3 RAM using the following code:

from timeit import timeit
import random
import matplotlib.pyplot as plt

num_data_points = 1000
step = 10
methods = [
    # ordered from slowest to fastest to make the key easier to read
    # """for item in list_a: list_b.append(item); list_a = list_b""",
    # """for item in list_b: list_a.insert(0, item)""",
    # "list_a = list(itertools.chain(list_b, list_a))",
    "list_a = list_b + list_a",
    "list_a[0:0] = list_b",
    "list_a = itertools.chain(list_b, list_a)",
]

x = list(range(0, num_data_points * step, step))
y = [[] for _ in methods]
for i in x:
    list_a = list(range(i))
    list_b = list(range(i))
    random.shuffle(list_a)
    random.shuffle(list_b)
    setup = f"list_a = {list_a}; list_b = {list_b}"
    for method_index, method in enumerate(methods):
        y[method_index].append(timeit(method, setup=setup, number=30))
    print(i, "out of", num_data_points * step)

ax = plt.axes()
for method_index, method in enumerate(methods):
    ax.plot(x, y[method_index], label=method)
ax.set(xlabel="number of elements in both lists", ylabel="time (s) (lower is better)")
ax.legend()
plt.show()