How to toggle a value in Python

Question

What is the most efficient way to toggle between 0 and 1?

Answer 1

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

Answer 2

How about an imaginary toggle that stores not only the current toggle, but a couple other values associated with it?

toggle = complex.conjugate

Store any + or - value on the left, and any unsigned value on the right:

>>> x = 2 - 3j
>>> toggle(x)
(2+3j)

Zero works, too:

>>> y = -2 - 0j
>>> toggle(y)
(-2+0j)

Easily retrieve the current toggle value (True and False represent + and -), LHS (real) value, or RHS (imaginary) value:

>>> import math
>>> curr = lambda i: math.atan2(i.imag, -abs(i.imag)) > 0
>>> lhs = lambda i: i.real
>>> rhs = lambda i: abs(i.imag)
>>> x = toggle(x)
>>> curr(x)
True
>>> lhs(x)
2.0
>>> rhs(x)
3.0

Easily swap LHS and RHS (but note that the sign of the both values must not be important):

>>> swap = lambda i: i/-1j
>>> swap(2+0j)
2j
>>> swap(3+2j)
(2+3j)

Easily swap LHS and RHS and also toggle at the same time:

>>> swaggle = lambda i: i/1j
>>> swaggle(2+0j)
-2j
>>> swaggle(3+2j)
(2-3j)

Guards against errors:

>>> toggle(1)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: descriptor 'conjugate' requires a 'complex' object but received a 'int'

Perform changes to LHS and RHS:

>>> x += 1+2j
>>> x
(3+5j)

...but be careful manipulating the RHS:

>>> z = 1-1j
>>> z += 2j
>>> z
(1+1j) # whoops! toggled it!

Answer 3

I always use:

p^=True

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

Answer 4

If you are dealing with an integer variable, you can increment 1 and limit your set to 0 and 1 (mod)

X = 0  # or X = 1
X = (X + 1)%2

Answer 5

You can make use of the index of lists.

def toggleValues(values, currentValue):
    return values[(values.index(currentValue) + 1) % len(values)]

> toggleValues( [0,1] , 1 )
> 0
> toggleValues( ["one","two","three"] , "one" )
> "two"
> toggleValues( ["one","two","three"] , "three")
> "one"

Pros: No additional libraries, self.explanatory code and working with arbitrary data types.

Cons: not duplicate-save. toggleValues(["one","two","duped", "three", "duped", "four"], "duped") will always return "three"

Answer 6

Just between 1 and 0, do this

1-x 

x can take 1 or 0