# How to toggle a value?

## Question:

What is the most efficient way to toggle between `0`

and `1`

?

## Answers:

The `not`

operator negates your variable (converting it into a boolean if it isn’t already one). You can *probably* use `1`

and `0`

interchangeably with `True`

and `False`

, so just negate it:

```
toggle = not toggle
```

But if you are using two arbitrary values, use an inline `if`

:

```
toggle = 'a' if toggle == 'b' else 'b'
```

Just between 1 and 0, do this

```
1-x
```

x can take 1 or 0

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
```

## 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__`

one way to toggle is by using Multiple assignment

```
>>> a = 5
>>> b = 3
>>> t = a, b = b, a
>>> t[0]
3
>>> t = a, b = b, a
>>> t[0]
5
```

Using itertools:

```
In [12]: foo = itertools.cycle([1, 2, 3])
In [13]: next(foo)
Out[13]: 1
In [14]: next(foo)
Out[14]: 2
In [15]: next(foo)
Out[15]: 3
In [16]: next(foo)
Out[16]: 1
In [17]: next(foo)
Out[17]: 2
```

Surprisingly nobody mention good old division modulo 2:

```
In : x = (x + 1) % 2 ; x
Out: 1
In : x = (x + 1) % 2 ; x
Out: 0
In : x = (x + 1) % 2 ; x
Out: 1
In : x = (x + 1) % 2 ; x
Out: 0
```

Note that it is equivalent to `x = x - 1`

, but the advantage of modulo technique is that the size of the group or length of the interval can be bigger then just 2 elements, thus giving you a similar to round-robin interleaving scheme to loop over.

Now just for 2, toggling can be a bit shorter (using bit-wise operator):

```
x = x ^ 1
```

I always use:

```
p^=True
```

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

**Using exception handler**

```
>>> def toogle(x):
... try:
... return x/x-x/x
... except ZeroDivisionError:
... return 1
...
>>> x=0
>>> x=toogle(x)
>>> x
1
>>> x=toogle(x)
>>> x
0
>>> x=toogle(x)
>>> x
1
>>> x=toogle(x)
>>> x
0
```

Ok, I’m the worst:

```
import math
import sys
d={1:0,0:1}
l=[1,0]
def exception_approach(x):
try:
return x/x-x/x
except ZeroDivisionError:
return 1
def cosinus_approach(x):
return abs( int( math.cos( x * 0.5 * math.pi ) ) )
def module_approach(x):
return (x + 1) % 2
def subs_approach(x):
return x - 1
def if_approach(x):
return 0 if x == 1 else 1
def list_approach(x):
global l
return l[x]
def dict_approach(x):
global d
return d[x]
def xor_approach(x):
return x^1
def not_approach(x):
b=bool(x)
p=not b
return int(p)
funcs=[ exception_approach, cosinus_approach, dict_approach, module_approach, subs_approach, if_approach, list_approach, xor_approach, not_approach ]
f=funcs[int(sys.argv[1])]
print "nnn", f.func_name
x=0
for _ in range(0,100000000):
x=f(x)
```

The easiest way to toggle between 1 and 0 is to subtract from 1.

```
def toggle(value):
return 1 - value
```

I use abs function, very useful on loops

```
x = 1
for y in range(0, 3):
x = abs(x - 1)
```

x will be 0.

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!
```

Let’s do some frame hacking. Toggle a variable by name. Note: This may not work with every Python runtime.

Say you have a variable “x”

```
>>> import inspect
>>> def toggle(var_name):
>>> frame = inspect.currentframe().f_back
>>> vars = frame.f_locals
>>> vars[var_name] = 0 if vars[var_name] == 1 else 1
>>> x = 0
>>> toggle('x')
>>> x
1
>>> toggle('x')
>>> x
0
```

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
```

Variables a and b can be ANY two values, like 0 and 1, or 117 and 711, or “heads” and “tails”. No math is used, just a quick swap of the values each time a toggle is desired.

```
a = True
b = False
a,b = b,a # a is now False
a,b = b,a # a is now True
```

Switching between -1 and +1 can be obtained by inline multiplication; used for calculation of pi the ‘Leibniz’ way (or similar):

```
sign = 1
result = 0
for i in range(100000):
result += 1 / (2*i + 1) * sign
sign *= -1
print("pi (estimate): ", result*4)
```

You can make use of the `index`

of `list`

s.

```
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"`