Are these list-comprehensions written the fastest possible way?

Question

This is a simple repeatable question regarding the usage of Python3's comprehensions:

Could the Python3 syntax be used in another way to speed up the process, further so the gap between Python3 and numpy would be reduced? (see chart below).

The results & code: Above: Runtime test of Python3's list comprehension, dict comprehension, python for loop and numpy Update 1 Above: Runtime test without power-functions including Janne Karila suggestion to break down the power function to x * x * x Update2 Above: Runtime test without powerfunctions and with Justin Peels recommendation to use return list(a.values())

#/usr/bin/python3 -


import sys
from datetime import datetime
import numpy
import matplotlib.pyplot as pyplot

def doingnothing(n):
    for i in range(n):
        pass
    return []

def numpysum(n): #NPY
    a = numpy.arange(n) ** 2.
    b = numpy.arange(n) ** 3.
    c = a + b
    return c

def listexpression(n):  #LE
    #return [x**2+x**3 for x in range(n)]
    return [x*x+x*x*x for x in range(n)]

def dictcomprehension(n):  #DC
    # a = {x:x**2+x**3 for x in range(n)}
    a = {x:x*x+x*x*x for x in range(n)}
    # return [a[key] for key in a]
    return list(a.values())

def pythonsum(n): #PS
    a = list(range(n))
    b = list(range(n))
    c = []

    for i in range(len(a)):
            # a[i] = i ** 2.
            a[i] = i * i 
            # b[i] = i ** 3.
            b[i] = i * i * i
            c.append(a[i] + b[i])
    return c

def runtimetest(size,verbose=True):
    v = verbose
    d = {'DoingNothing':None,
         'NPY':None,
         'LE':None,
         'DC':None,
         'PS':None}

    start = datetime.now()
    c = doingnothing(size)
    d['DoingNothing'] = datetime.now() - start

    start = datetime.now()
    c = numpysum(size)
    d['NPY'] = datetime.now() - start

    start = datetime.now()
    c = listexpression(size)
    d['LE'] = datetime.now() - start

    start = datetime.now()
    c = dictcomprehension(size)
    d['DC'] = datetime.now() - start

    start = datetime.now()
    c = pythonsum(size)
    d['PS'] = datetime.now() - start

    if v:
        for key in sorted(d):
            print(key, "took", round(d[key].seconds + d[key].microseconds/10**6,6), " sec.")
    else:
        return d

def view(results):
    """
    result['header']=['DoingNothing','NPY','LE','DC','PS']
    result[3000]=[0.0, 0.0, 0.002, 0.003, 0.003001] 
    """
    if 'header' in results.keys():
        results.pop('header')

    steps,DN,NPY,LE,DC,PS=[],[],[],[],[],[] #I love multiple assignment!

    for step in sorted(results):
        steps.append(step)
        DN.append(results[step][0])
        NPY.append(results[step][4])
        LE.append(results[step][5])
        DC.append(results[step][6])
        PS.append(results[step][4])

    pyplot.plot(steps,DN)
    pyplot.plot(steps,NPY)
    pyplot.plot(steps,LE)
    pyplot.plot(steps,DC)
    pyplot.plot(steps,PS)
    pyplot.legend(['Empty Loop', 'Numpy.Arange', 'List Comprehension', 'Dict Comprehension', 'Python for Loop'], loc='upper left')
    scale = 'linear'
    pyplot.xscale(scale)
    pyplot.yscale(scale)
    pyplot.title('runtime test')
    pyplot.xlabel('length of list')
    pyplot.ylabel('runtime in seconds')

    pyplot.show()

def longruntimetest(length):
    if length<10**4:
        length=10**4

    result = {}
    result['header']=['DoingNothing','NPY','LE','DC','PS']
    for step in range(10**3, length+1,10**3):
        t=runtimetest(step,verbose=False)        
        result[step]=[t['DoingNothing'],t['NPY'],t['LE'],t['DC'],t['PS']]
        for i in range(len(result[step])):
            result[step][i]=round(result[step][i].seconds
                                  +result[step][i].microseconds/10**6,6)

    view(result)

def main():
    if sys.version_info <= (3, 2):
        sys.stdout.write("Sorry, requires Python 3.2.x, not Python 2.x\n")
        sys.exit(1)

    testsize=10**5
    runtimetest(testsize)   
    longruntimetest(testsize)

if __name__=='__main__':
    main()

Answer 1

def listexpression(n):  #LE
    return [x * x * (1 + x) for x in range(n)]

is over 3 times faster than your version with x**2+x**3 (from 8.47 ms to 2.31 ms with n = 10000).

Though, NumPy will also benefit from regrouping the expression this way.