Python elegant assignment based on True/False values

Question

I have a variable I want to set depending on the values in three booleans. The most straight-forward way is an if statement followed by a series of elifs:

if a a         


        

Answer 1

To measure speeds:

from time import clock
a,b,c = True,False,False

A,B,C,D,E,F,G,H = [],[],[],[],[],[],[],[]


for j in xrange(30):


    te = clock()
    for i in xrange(10000):
        name = (a and b and c and 'first' or
                a and b and not c and 'second' or
                a and not b and c and 'third' or
                a and not b and not c and 'fourth' or
                not a and b and c and 'fifth' or
                not a and b and not c and 'sixth' or
                not a and not b and c and 'seventh' or
                not a and not b and not c and 'eighth')
    A.append(clock()-te)



    te = clock()
    for i in xrange(10000):
        if a and b and c:
            name = 'first'
        elif a and b and not c:
            name = 'second'
        elif a and not b and c:
            name = 'third'
        elif a and not b and not c:
            name = 'fourth'
        elif not a and b and c:
            name = 'fifth'
        elif not a and b and not c:
            name = 'sixth'
        elif not a and not b and c:
            name = 'seventh'
        elif not a and not b and not c:
            name = 'eighth'
    B.append(clock()-te)

    #=====================================================================================

    li = ['eighth', 'seventh', 'sixth', 'fifth', 'fourth', 'third', 'second', 'first']
    te = clock()
    for i in xrange(10000):
        name = li[a*4 + b*2 + c]
    C.append(clock()-te)

    #=====================================================================================

    nth = ['eighth', 'seventh', 'sixth', 'fifth', 'fourth', 'third', 'second', 'first']
    te = clock()
    for i in xrange(10000):
        name = nth[(a and 4 or 0) | (b and 2 or 0) | (c and 1 or 0)]
    D.append(clock()-te)


    nth = ['eighth', 'seventh', 'sixth', 'fifth', 'fourth', 'third', 'second', 'first']
    te = clock()
    for i in xrange(10000):
        name = nth[(a and 4 or 0) + (b and 2 or 0) + (c and 1 or 0)]
    E.append(clock()-te)

    #=====================================================================================

    values = ['eighth', 'seventh', 'sixth', 'fifth', 'fourth', 'third', 'second', 'first']
    te = clock()
    for i in xrange(10000):
        name = values[( 4 if a else 0 )| ( 2 if b else 0 ) | ( 1 if c else 0 )]
    F.append(clock()-te)


    values = ['eighth', 'seventh', 'sixth', 'fifth', 'fourth', 'third', 'second', 'first']
    te = clock()
    for i in xrange(10000):
        name = values[( 4 if a else 0 ) + ( 2 if b else 0 ) + ( 1 if c else 0 )]
    G.append(clock()-te)

    #=====================================================================================

    dic = {(True, True, True): "first",
           (True, True, False): "second",
           (True, False, True): "third",
           (True, False, False): "fourth",
           (False, True, True): "fifth",
           (False, True, False): "sixth",
           (False, False, True): "seventh",
           (False, False, False): "eighth"}
    te = clock()
    for i in xrange(10000):
        name = dic[a,b,c]
    H.append(clock()-te)




print min(A),'\n', min(B),'\n\n', min(C),'\n\n', min(D),'\n',min(E),'\n\n',min(F),'\n', min(G),'\n\n', min(H)

Result

0.0480533140385 
0.0450973517584 

0.0309056039245 

0.0295291720037 
0.0286550385594 

0.0280122194301 
0.0266760160858 

0.0249769174574