How to toggle a value?

Solution 1:

Solution using NOT

If the values are boolean, the fastest approach is to use the not operator:

>>> x = True
>>> x = not x        # toggle
>>> x
False
>>> x = not x        # toggle
>>> x
True
>>> x = not x        # toggle
>>> x
False

Solution using subtraction

If the values are numerical, then subtraction from the total is a simple and fast way to toggle values:

>>> A = 5
>>> B = 3
>>> total = A + B
>>> x = A
>>> x = total - x    # toggle
>>> x
3
>>> x = total - x    # toggle
>>> x
5
>>> x = total - x    # toggle
>>> x
3

Solution using XOR

If the value toggles between 0 and 1, you can use a bitwise exclusive-or:

>>> x = 1
>>> x ^= 1
>>> x
0
>>> x ^= 1
>>> x
1

The technique generalizes to any pair of integers. The xor-by-one step is replaced with a xor-by-precomputed-constant:

>>> A = 205
>>> B = -117
>>> t = A ^ B        # precomputed toggle constant
>>> x = A
>>> x ^= t           # toggle
>>> x
-117
>>> x ^= t           # toggle
>>> x
205
>>> x ^= t           # toggle
>>> x
-117

(This idea was submitted by Nick Coghlan and later generalized by @zxxc.)

Solution using a dictionary

If the values are hashable, you can use a dictionary:

>>> A = 'xyz'
>>> B = 'pdq'
>>> d = {A:B, B:A}
>>> x = A
>>> x = d[x]         # toggle
>>> x
'pdq'
>>> x = d[x]         # toggle
>>> x
'xyz'
>>> x = d[x]         # toggle
>>> x
'pdq'

Solution using a conditional expression

The slowest way is to use a conditional expression:

>>> A = [1,2,3]
>>> B = [4,5,6]
>>> x = A
>>> x = B if x == A else A
>>> x
[4, 5, 6]
>>> x = B if x == A else A
>>> x
[1, 2, 3]
>>> x = B if x == A else A
>>> x
[4, 5, 6]

Solution using itertools

If you have more than two values, the itertools.cycle() function provides a generic fast way to toggle between successive values:

>>> import itertools
>>> toggle = itertools.cycle(['red', 'green', 'blue']).next
>>> toggle()
'red'
>>> toggle()
'green'
>>> toggle()
'blue'
>>> toggle()
'red'
>>> toggle()
'green'
>>> toggle()
'blue'

Note that in Python 3 the next() method was changed to __next__(), so the first line would be now written as toggle = itertools.cycle(['red', 'green', 'blue']).__next__

Solution 2:

I always use:

p^=True

If p is a boolean, this switches between true and false.

Solution 3:

Here is another non intuitive way. The beauty is you can cycle over multiple values and not just two [0,1]

For Two values (toggling)

>>> x=[1,0]
>>> toggle=x[toggle]

For Multiple Values (say 4)

>>> x=[1,2,3,0]
>>> toggle=x[toggle]

I didn't expect this solution to be almost the fastest too

>>> stmt1="""
toggle=0
for i in xrange(0,100):
    toggle = 1 if toggle == 0 else 0
"""
>>> stmt2="""
x=[1,0]
toggle=0
for i in xrange(0,100):
    toggle=x[toggle]
"""
>>> t1=timeit.Timer(stmt=stmt1)
>>> t2=timeit.Timer(stmt=stmt2)
>>> print "%.2f usec/pass" % (1000000 * t1.timeit(number=100000)/100000)
7.07 usec/pass
>>> print "%.2f usec/pass" % (1000000 * t2.timeit(number=100000)/100000)
6.19 usec/pass
stmt3="""
toggle = False
for i in xrange(0,100):
    toggle = (not toggle) & 1
"""
>>> t3=timeit.Timer(stmt=stmt3)
>>> print "%.2f usec/pass" % (1000000 * t3.timeit(number=100000)/100000)
9.84 usec/pass
>>> stmt4="""
x=0
for i in xrange(0,100):
    x=x-1
"""
>>> t4=timeit.Timer(stmt=stmt4)
>>> print "%.2f usec/pass" % (1000000 * t4.timeit(number=100000)/100000)
6.32 usec/pass