Does Python have a ternary conditional operator?


You can index into a tuple:

(falseValue, trueValue)[test]

test needs to return True or False.
It might be safer to always implement it as:

(falseValue, trueValue)[test == True]

or you can use the built-in bool() to assure a Boolean value:

(falseValue, trueValue)[bool(<expression>)]

If Python does not have a ternary conditional operator, is it possible to simulate one using other language constructs?

expression1 if condition else expression2

>>> a = 1
>>> b = 2
>>> 1 if a > b else -1 
>>> 1 if a > b else -1 if a < b else 0

You might often find

cond and on_true or on_false

but this lead to problem when on_true == 0

>>> x = 0
>>> print x == 0 and 0 or 1 
>>> x = 1
>>> print x == 0 and 0 or 1 

where you would expect for a normal ternary operator this result

>>> x = 0
>>> print 0 if x == 0 else 1 
>>> x = 1
>>> print 0 if x == 0 else 1 


>>> b = (True if 5 > 4 else False)
>>> print b

For Python 2.5 and newer there is a specific syntax:

[on_true] if [cond] else [on_false]

In older Pythons a ternary operator is not implemented but it's possible to simulate it.

cond and on_true or on_false

Though, there is a potential problem, which if cond evaluates to True and on_true evaluates to False then on_false is returned instead of on_true. If you want this behavior the method is OK, otherwise use this:

{True: on_true, False: on_false}[cond is True] # is True, not == True

which can be wrapped by:

def q(cond, on_true, on_false)
    return {True: on_true, False: on_false}[cond is True]

and used this way:

q(cond, on_true, on_false)

It is compatible with all Python versions.

In [1]: a = 1 if False else 0

In [2]: a
Out[2]: 0

In [3]: b = 1 if True else 0

In [4]: b
Out[4]: 1

An operator for a conditional expression in Python was added in 2006 as part of Python Enhancement Proposal 308. Its form differ from common ?: operator and it's:

<expression1> if <condition> else <expression2>

which is equivalent to:

if <condition>: <expression1> else: <expression2>

Here is an example:

result = x if a > b else y

Another syntax which can be used (compatible with versions before 2.5):

result = (lambda:y, lambda:x)[a > b]()

where operands are lazily evaluated.

Another way is by indexing a tuple (which isn't consistent with the conditional operator of most other languages):

result = (y, x)[a > b]

or explicitly constructed dictionary:

result = {True: x, False: y}[a > b]

Another (less reliable), but simpler method is to use and and or operators:

result = (a > b) and x or y

however this won't work if x would be False.

A possible workaround is to make x and y lists or tuples as in the following:

result = ((a > b) and [x] or [y])[0]


result = ((a > b) and (x,) or (y,))[0]

If you're working with dictionaries, instead of using a ternary conditional, you can take advantage of get(key, default), for example:

shell = os.environ.get('SHELL', "/bin/sh")

Source: ?: in Python at Wikipedia

Does Python have a ternary conditional operator?

Yes. From the grammar file:

test: or_test ['if' or_test 'else' test] | lambdef

The part of interest is:

or_test ['if' or_test 'else' test]

So, a ternary conditional operation is of the form:

expression1 if expression2 else expression3

expression3 will be lazily evaluated (that is, evaluated only if expression2 is false in a boolean context). And because of the recursive definition, you can chain them indefinitely (though it may considered bad style.)

expression1 if expression2 else expression3 if expression4 else expression5 # and so on

A note on usage:

Note that every if must be followed with an else. People learning list comprehensions and generator expressions may find this to be a difficult lesson to learn - the following will not work, as Python expects a third expression for an else:

[expression1 if expression2 for element in iterable]
#                          ^-- need an else here

which raises a SyntaxError: invalid syntax. So the above is either an incomplete piece of logic (perhaps the user expects a no-op in the false condition) or what may be intended is to use expression2 as a filter - notes that the following is legal Python:

[expression1 for element in iterable if expression2]

expression2 works as a filter for the list comprehension, and is not a ternary conditional operator.

Alternative syntax for a more narrow case:

You may find it somewhat painful to write the following:

expression1 if expression1 else expression2

expression1 will have to be evaluated twice with the above usage. It can limit redundancy if it is simply a local variable. However, a common and performant Pythonic idiom for this use-case is to use or's shortcutting behavior:

expression1 or expression2

which is equivalent in semantics. Note that some style-guides may limit this usage on the grounds of clarity - it does pack a lot of meaning into very little syntax.

There is a ternary option as stated in other answers, but you can also simulate it using "or" if you are checking against a boolean or None value:

>>> a = False
>>> b = 5
>>> a or b

>>> a = None
>>> a or b

More a tip than an answer (don't need to repeat the obvious for the hundreth time), but I sometimes use it as a oneliner shortcut in such constructs:

if conditionX:

, becomes:

print('yes') if conditionX else print('nah')

Some (many :) may frown upon it as unpythonic (even, ruby-ish :), but I personally find it more natural - i.e. how you'd express it normally, plus a bit more visually appealing in large blocks of code.