https://stackoverflow.com/questions/51597284/invert-edge-values-in-python-boolean-list

results.md

Execution time in seconds

original list with 11 items

jpp: 0.038900841027498245
Laurent H.: 0.09572448302060366
Kevin: 0.007683080970309675
davidak: 0.027118309983052313

list with 100 items

jpp: 0.085919318953529
Laurent H.: 0.8218901020009071
Kevin: 0.007853235001675785
davidak: 0.046300466055981815

list with 1000 items

jpp: 0.2682207649340853
Laurent H.: 5.869488948956132
Kevin: 0.008967073052190244
davidak: 0.26018757303245366

list with 10000 items

jpp: 2.4661101199453697
Laurent H.: 62.00245025800541
Kevin: 0.009241680963896215
davidak: 2.5198689920362085

list with 100000 items

jpp: 24.208682479104027
Laurent H.: 680.6267002499662
Kevin: 0.009258356061764061
davidak: 26.868416464072652

test.py

#!/usr/bin/env python3

import numpy as np
import matplotlib
matplotlib.use('AGG')
import matplotlib.pyplot as plt
import copy
import timeit


iterations = 10000


def invert_edge_values_jpp(l):
    idx1 = next(i for i, j in enumerate(l) if j)
    idx2 = next(i for i, j in enumerate(l[::-1]) if j)

    l = np.array(l)
    l[:idx1] = True
    l[len(l)-idx2:] = True

    return l


def invert_edge_values_laurent_h(l):
    l = copy.deepcopy(l)
    tempList = l[:]

    # Iterate from the beginning (and invert) until the element differs from the first one
    for i,v in enumerate(tempList):
        if v == tempList[0]: l[i] = not v
        else:                break

    # Iterate from the end (and invert) until the element differs from the last one
    for i,v in reversed(list(enumerate(tempList))):
        if v == tempList[-1]: l[i] = not v
        else:                 break

    return l


def invert_edge_values_kevin(l):
    for i in range(len(l)):
        if l[i]:
            break
        l[i] = True

    for i in range(len(l)-1, -1, -1):
        if l[i]:
            break
        l[i] = True

    return l


def invert_edge_values_davidak(l):
    l = np.array(l)
    ind = np.where(l)[0]
    first, last = ind[0], ind[-1]

    l[:first] = True
    l[last + 1:] = True

    return l


if __name__ == '__main__':
    # create long random list of booleans
    # with some False on the edges
    l = []
    list_items = 100000

    number_pre_false = np.random.randint(5)
    number_past_false = np.random.randint(5)
    number_random_bool = list_items - (number_pre_false + number_past_false)

    for i in range(number_pre_false):
        l.append(False)

    for i in range(number_random_bool):
        l.append(np.random.rand() > 0.5)

    for i in range(number_past_false):
        l.append(False)

    # l = [False, False, False, True, True, True, False, False, True, False, False]
    # list_items = len(l)

    jpp = timeit.timeit("invert_edge_values_jpp(l)", number=iterations, globals=globals())
    laurent_h = timeit.timeit("invert_edge_values_laurent_h(l)", number=iterations, globals=globals())
    kevin = timeit.timeit("invert_edge_values_kevin(l)", number=iterations, globals=globals())
    davidak = timeit.timeit("invert_edge_values_davidak(l)", number=iterations, globals=globals())

    print("jpp:", jpp)
    print("Laurent H.:", laurent_h)
    print("Kevin:", kevin)
    print("davidak:", davidak)

    dpi = 96
    fig = plt.figure(num=None, figsize=(1024/dpi, 768/dpi), dpi=dpi)
    ax = fig.add_subplot(111)

    plt.title("Comparison of functions to invert edge values in python boolean list (with {} items)".format(list_items))
    ax.set_ylabel('execution time of {} iterations in seconds'.format(iterations))

    x = ['jpp', 'Laurent H.', 'Kevin', 'davidak']
    y = [jpp, laurent_h, kevin, davidak]
    ax.bar(x, y)

    plt.savefig("test.svg")